contestId int64 0 1.01k | index stringclasses 57 values | name stringlengths 2 58 | type stringclasses 2 values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522 values | time-limit stringclasses 8 values | memory-limit stringclasses 8 values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3 values | verdict stringclasses 14 values | testset stringclasses 12 values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
659 | A | Round House | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to *n*. Entrance *n* and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance *a* and he decided that during his walk he will move around the house *b* entrances in the direction of increasing numbers (in this order entrance *n* should be followed by entrance 1). The negative value of *b* corresponds to moving |*b*| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance *n*). If *b*<==<=0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk. | The single line of the input contains three space-separated integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*a*<=≤<=*n*,<=<=-<=100<=≤<=*b*<=≤<=100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively. | Print a single integer *k* (1<=≤<=*k*<=≤<=*n*) — the number of the entrance where Vasya will be at the end of his walk. | [
"6 2 -5\n",
"5 1 3\n",
"3 2 7\n"
] | [
"3\n",
"4\n",
"3\n"
] | The first example is illustrated by the picture in the statements. | 500 | [
{
"input": "6 2 -5",
"output": "3"
},
{
"input": "5 1 3",
"output": "4"
},
{
"input": "3 2 7",
"output": "3"
},
{
"input": "1 1 0",
"output": "1"
},
{
"input": "1 1 -1",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "100 ... | 1,503,859,206 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 62 | 0 | n,a,b=map(int,input().split(' '))
z=(((a+b)%n)+n)%n
if(z==0):
print(n)
else:
print(z) | Title: Round House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to *n*. Entrance *n* and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance *a* and he decided that during his walk he will move around the house *b* entrances in the direction of increasing numbers (in this order entrance *n* should be followed by entrance 1). The negative value of *b* corresponds to moving |*b*| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance *n*). If *b*<==<=0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
Input Specification:
The single line of the input contains three space-separated integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*a*<=≤<=*n*,<=<=-<=100<=≤<=*b*<=≤<=100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
Output Specification:
Print a single integer *k* (1<=≤<=*k*<=≤<=*n*) — the number of the entrance where Vasya will be at the end of his walk.
Demo Input:
['6 2 -5\n', '5 1 3\n', '3 2 7\n']
Demo Output:
['3\n', '4\n', '3\n']
Note:
The first example is illustrated by the picture in the statements. | ```python
n,a,b=map(int,input().split(' '))
z=(((a+b)%n)+n)%n
if(z==0):
print(n)
else:
print(z)
``` | 3 | |
811 | B | Vladik and Complicated Book | PROGRAMMING | 1,200 | [
"implementation",
"sortings"
] | null | null | Vladik had started reading a complicated book about algorithms containing *n* pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation *P*<==<=[*p*1,<=*p*2,<=...,<=*p**n*], where *p**i* denotes the number of page that should be read *i*-th in turn.
Sometimes Vladik’s mom sorted some subsegment of permutation *P* from position *l* to position *r* inclusive, because she loves the order. For every of such sorting Vladik knows number *x* — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has *p**x* changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other. | First line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=104) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.
Second line contains *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — permutation *P*. Note that elements in permutation are distinct.
Each of the next *m* lines contains three space-separated integers *l**i*, *r**i*, *x**i* (1<=≤<=*l**i*<=≤<=*x**i*<=≤<=*r**i*<=≤<=*n*) — left and right borders of sorted subsegment in *i*-th sorting and position that is interesting to Vladik. | For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise. | [
"5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3\n",
"6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3\n"
] | [
"Yes\nNo\nYes\nYes\nNo\n",
"Yes\nNo\nYes\nNo\nYes\n"
] | Explanation of first test case:
1. [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No". 1. [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes". 1. [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No". | 1,000 | [
{
"input": "5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3",
"output": "Yes\nNo\nYes\nYes\nNo"
},
{
"input": "6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3",
"output": "Yes\nNo\nYes\nNo\nYes"
},
{
"input": "10 10\n10 1 6 7 9 8 4 3 5 2\n1 1 1\n4 4 4\n7 7 7\n3 3 3\n1 6 5\n2 6 2\n6... | 1,496,294,133 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 204,800 | def swap(A, i, j):
temp = A[i]
A[i] = A[j]
A[j] = temp
def Partition(A, p, r, x):
i = p - 1
j = r + 1
while True:
j-=1
while A[j] > x:
j-=1
i+=1
while A[i] < x:
i+=1
if i < j:
swap(A, i, j)
else:
return j
def median_of_3(a,b,c):
if (a<=b and b<=c) or (c<=b and b<=a):
return b
elif (b<=a and a<=c) or (c<=a and a<=b):
return a
else:
return c
def insertion_sort(A, p, r):
for i in range(p, r + 1):
key = A[i]
j = i - 1
while j>=0 and key<A[j]:
A[j + 1] = A[j]
j-=1
A[j + 1] = key
def _median_of_3_quick_sort(A, f, b):
if b - f + 1 <= 10:
insertion_sort(A, f, b)
else:
p = Partition(A, f, b, median_of_3(A[f], \
A[f + (b-f)//2 + 1], A[b]))
_median_of_3_quick_sort(A, f, p)
_median_of_3_quick_sort(A, p+1, b)
def median_of_3_quick_sort(A):
p = 0
r = len(A) - 1
_median_of_3_quick_sort(A, p, r)
def main():
t=0
n=int(input())
m=int(input())
p=[]
for i in range(n):
a=int(input())
p.append(a)
while t<m:
l=int(input())
r=int(input())
x=int(input())
le=r-l+1
pos=p[x-1]
s=[]
for i in range(le):
s.append(p[l+i-1])
median_of_3_quick_sort(s)
pos2 = s[x-l]
if pos2==pos:
print("Yes")
else:
print("No")
t+=1
if __name__ == '__main__':
main() | Title: Vladik and Complicated Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vladik had started reading a complicated book about algorithms containing *n* pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation *P*<==<=[*p*1,<=*p*2,<=...,<=*p**n*], where *p**i* denotes the number of page that should be read *i*-th in turn.
Sometimes Vladik’s mom sorted some subsegment of permutation *P* from position *l* to position *r* inclusive, because she loves the order. For every of such sorting Vladik knows number *x* — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has *p**x* changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.
Input Specification:
First line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=104) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.
Second line contains *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — permutation *P*. Note that elements in permutation are distinct.
Each of the next *m* lines contains three space-separated integers *l**i*, *r**i*, *x**i* (1<=≤<=*l**i*<=≤<=*x**i*<=≤<=*r**i*<=≤<=*n*) — left and right borders of sorted subsegment in *i*-th sorting and position that is interesting to Vladik.
Output Specification:
For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise.
Demo Input:
['5 5\n5 4 3 2 1\n1 5 3\n1 3 1\n2 4 3\n4 4 4\n2 5 3\n', '6 5\n1 4 3 2 5 6\n2 4 3\n1 6 2\n4 5 4\n1 3 3\n2 6 3\n']
Demo Output:
['Yes\nNo\nYes\nYes\nNo\n', 'Yes\nNo\nYes\nNo\nYes\n']
Note:
Explanation of first test case:
1. [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No". 1. [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 1. [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes". 1. [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No". | ```python
def swap(A, i, j):
temp = A[i]
A[i] = A[j]
A[j] = temp
def Partition(A, p, r, x):
i = p - 1
j = r + 1
while True:
j-=1
while A[j] > x:
j-=1
i+=1
while A[i] < x:
i+=1
if i < j:
swap(A, i, j)
else:
return j
def median_of_3(a,b,c):
if (a<=b and b<=c) or (c<=b and b<=a):
return b
elif (b<=a and a<=c) or (c<=a and a<=b):
return a
else:
return c
def insertion_sort(A, p, r):
for i in range(p, r + 1):
key = A[i]
j = i - 1
while j>=0 and key<A[j]:
A[j + 1] = A[j]
j-=1
A[j + 1] = key
def _median_of_3_quick_sort(A, f, b):
if b - f + 1 <= 10:
insertion_sort(A, f, b)
else:
p = Partition(A, f, b, median_of_3(A[f], \
A[f + (b-f)//2 + 1], A[b]))
_median_of_3_quick_sort(A, f, p)
_median_of_3_quick_sort(A, p+1, b)
def median_of_3_quick_sort(A):
p = 0
r = len(A) - 1
_median_of_3_quick_sort(A, p, r)
def main():
t=0
n=int(input())
m=int(input())
p=[]
for i in range(n):
a=int(input())
p.append(a)
while t<m:
l=int(input())
r=int(input())
x=int(input())
le=r-l+1
pos=p[x-1]
s=[]
for i in range(le):
s.append(p[l+i-1])
median_of_3_quick_sort(s)
pos2 = s[x-l]
if pos2==pos:
print("Yes")
else:
print("No")
t+=1
if __name__ == '__main__':
main()
``` | -1 | |
732 | A | Buy a Shovel | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. | The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. | Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. | [
"117 3\n",
"237 7\n",
"15 2\n"
] | [
"9\n",
"1\n",
"2\n"
] | In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | 500 | [
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
... | 1,691,372,387 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 54 | 62 | 0 | k,r=map(int,input().split())
for i in range(1,11):
if (k*i)%10 in [0,r]:
print(i)
quit() | Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | ```python
k,r=map(int,input().split())
for i in range(1,11):
if (k*i)%10 in [0,r]:
print(i)
quit()
``` | 3 | |
732 | A | Buy a Shovel | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. | The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. | Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. | [
"117 3\n",
"237 7\n",
"15 2\n"
] | [
"9\n",
"1\n",
"2\n"
] | In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | 500 | [
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
... | 1,684,145,748 | 2,147,483,647 | PyPy 3 | OK | TESTS | 54 | 77 | 0 | price, coin = map(int, input().split())
for i in range(1, 11):
if (price * i) % 10 == 0 or coin == (price * i) % 10:
print(i)
break
| Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | ```python
price, coin = map(int, input().split())
for i in range(1, 11):
if (price * i) % 10 == 0 or coin == (price * i) % 10:
print(i)
break
``` | 3 | |
868 | A | Bark to Unlock | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not. | The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct. | Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"ya\n4\nah\noy\nto\nha\n",
"hp\n2\nht\ntp\n",
"ah\n1\nha\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring. | 250 | [
{
"input": "ya\n4\nah\noy\nto\nha",
"output": "YES"
},
{
"input": "hp\n2\nht\ntp",
"output": "NO"
},
{
"input": "ah\n1\nha",
"output": "YES"
},
{
"input": "bb\n4\nba\nab\naa\nbb",
"output": "YES"
},
{
"input": "bc\n4\nca\nba\nbb\ncc",
"output": "YES"
},
{
... | 1,677,129,120 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 14 | 46 | 0 | a=input();l=[]
for _ in range(int(input())):
b=input()
if b[0] == a[-1]:
l.append(b)
elif b[-1] == a[0]:
l.append(b)
print("YES" if l else "NO") | Title: Bark to Unlock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
Input Specification:
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
Output Specification:
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['ya\n4\nah\noy\nto\nha\n', 'hp\n2\nht\ntp\n', 'ah\n1\nha\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring. | ```python
a=input();l=[]
for _ in range(int(input())):
b=input()
if b[0] == a[-1]:
l.append(b)
elif b[-1] == a[0]:
l.append(b)
print("YES" if l else "NO")
``` | 0 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,696,065,925 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | s=input()
u=0
l=0
o=""
for i in range(0,len(s)):
a=s[i]
if(a.isupper()):
u=u+1
else:
l=l+1
if u>l:
o=s.upper()
elif u<l:
o=s.lower()
else:
o=s.lower()
print(o) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
s=input()
u=0
l=0
o=""
for i in range(0,len(s)):
a=s[i]
if(a.isupper()):
u=u+1
else:
l=l+1
if u>l:
o=s.upper()
elif u<l:
o=s.lower()
else:
o=s.lower()
print(o)
``` | 3.977 |
411 | A | Password Check | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q. | The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_". | If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes). | [
"abacaba\n",
"X12345\n",
"CONTEST_is_STARTED!!11\n"
] | [
"Too weak\n",
"Too weak\n",
"Correct\n"
] | none | 0 | [
{
"input": "abacaba",
"output": "Too weak"
},
{
"input": "X12345",
"output": "Too weak"
},
{
"input": "CONTEST_is_STARTED!!11",
"output": "Correct"
},
{
"input": "1zA__",
"output": "Correct"
},
{
"input": "1zA_",
"output": "Too weak"
},
{
"input": "zA_... | 1,670,746,054 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | # -*- coding: utf-8 -*-
"""
Created on Sun Dec 11 09:35:06 2022
@author: ahmed abdelaziz foaud
#ID: 120210044
"""
print("this program takes password and check if it is strong enough")
pass_word=input("enter password: ")
letter=0
cap=0
small=0
nums=0
for i in pass_word:
if ord(i) in range(65,91):
cap+=1
letter+=1
elif ord(i) in range(97,123):
small+=1
letter+=1
elif ord(i) in range(48,58):
nums+=1
if letter>=5 and nums>=1 and cap>=1 and small>=1:
print("correct")
else:
print("too weak")
| Title: Password Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input Specification:
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output Specification:
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Demo Input:
['abacaba\n', 'X12345\n', 'CONTEST_is_STARTED!!11\n']
Demo Output:
['Too weak\n', 'Too weak\n', 'Correct\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""
Created on Sun Dec 11 09:35:06 2022
@author: ahmed abdelaziz foaud
#ID: 120210044
"""
print("this program takes password and check if it is strong enough")
pass_word=input("enter password: ")
letter=0
cap=0
small=0
nums=0
for i in pass_word:
if ord(i) in range(65,91):
cap+=1
letter+=1
elif ord(i) in range(97,123):
small+=1
letter+=1
elif ord(i) in range(48,58):
nums+=1
if letter>=5 and nums>=1 and cap>=1 and small>=1:
print("correct")
else:
print("too weak")
``` | 0 | |
166 | E | Tetrahedron | PROGRAMMING | 1,500 | [
"dp",
"math",
"matrices"
] | null | null | You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7). | The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path. | Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7). | [
"2\n",
"4\n"
] | [
"3\n",
"21\n"
] | The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D* | 1,000 | [
{
"input": "2",
"output": "3"
},
{
"input": "4",
"output": "21"
},
{
"input": "1",
"output": "0"
},
{
"input": "3",
"output": "6"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "183"
},
{
"input": "7",
"output": "546"
... | 1,622,419,623 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 2,000 | 4,096,000 | mod = 1000000007
n = int(input())
dp = [1, 0, 0, 0]
tot = 1
for i in range(n):
tmp = []
for k in range(4):
tmp.append((sum(dp) + mod - dp[k]) % mod)
dp = tmp
print(dp[0]) | Title: Tetrahedron
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path.
Output Specification:
Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7).
Demo Input:
['2\n', '4\n']
Demo Output:
['3\n', '21\n']
Note:
The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D* | ```python
mod = 1000000007
n = int(input())
dp = [1, 0, 0, 0]
tot = 1
for i in range(n):
tmp = []
for k in range(4):
tmp.append((sum(dp) + mod - dp[k]) % mod)
dp = tmp
print(dp[0])
``` | 0 | |
735 | D | Taxes | PROGRAMMING | 1,600 | [
"math",
"number theory"
] | null | null | Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to *n* (*n*<=≥<=2) burles and the amount of tax he has to pay is calculated as the maximum divisor of *n* (not equal to *n*, of course). For example, if *n*<==<=6 then Funt has to pay 3 burles, while for *n*<==<=25 he needs to pay 5 and if *n*<==<=2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial *n* in several parts *n*1<=+<=*n*2<=+<=...<=+<=*n**k*<==<=*n* (here *k* is arbitrary, even *k*<==<=1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition *n**i*<=≥<=2 should hold for all *i* from 1 to *k*.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split *n* in parts. | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·109) — the total year income of mr. Funt. | Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax. | [
"4\n",
"27\n"
] | [
"2\n",
"3\n"
] | none | 1,750 | [
{
"input": "4",
"output": "2"
},
{
"input": "27",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "2000000000",
"output": "2"
},
{
"input": "26",
"output": ... | 1,561,148,769 | 2,147,483,647 | PyPy 3 | OK | TESTS | 85 | 155 | 1,228,800 | from math import sqrt
def prime(n):
for d in range(3, int(sqrt(n)) + 1, 2):
if n % d == 0:
return 0
return 1
n = int(input())
if n == 2:
print(1)
elif n % 2 == 0:
print(2)
elif prime(n):
print(1)
elif prime(n - 2):
print(2)
else:
print(3)
| Title: Taxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to *n* (*n*<=≥<=2) burles and the amount of tax he has to pay is calculated as the maximum divisor of *n* (not equal to *n*, of course). For example, if *n*<==<=6 then Funt has to pay 3 burles, while for *n*<==<=25 he needs to pay 5 and if *n*<==<=2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial *n* in several parts *n*1<=+<=*n*2<=+<=...<=+<=*n**k*<==<=*n* (here *k* is arbitrary, even *k*<==<=1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition *n**i*<=≥<=2 should hold for all *i* from 1 to *k*.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split *n* in parts.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·109) — the total year income of mr. Funt.
Output Specification:
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Demo Input:
['4\n', '27\n']
Demo Output:
['2\n', '3\n']
Note:
none | ```python
from math import sqrt
def prime(n):
for d in range(3, int(sqrt(n)) + 1, 2):
if n % d == 0:
return 0
return 1
n = int(input())
if n == 2:
print(1)
elif n % 2 == 0:
print(2)
elif prime(n):
print(1)
elif prime(n - 2):
print(2)
else:
print(3)
``` | 3 | |
70 | C | Lucky Tickets | PROGRAMMING | 2,200 | [
"binary search",
"data structures",
"sortings",
"two pointers"
] | C. Lucky Tickets | 1 | 256 | In Walrusland public transport tickets are characterized by two integers: by the number of the series and by the number of the ticket in the series. Let the series number be represented by *a* and the ticket number — by *b*, then a ticket is described by the ordered pair of numbers (*a*,<=*b*).
The walruses believe that a ticket is lucky if *a*<=*<=*b*<==<=*rev*(*a*)<=*<=*rev*(*b*). The function *rev*(*x*) reverses a number written in the decimal system, at that the leading zeroes disappear. For example, *rev*(12343)<==<=34321, *rev*(1200)<==<=21.
The Public Transport Management Committee wants to release *x* series, each containing *y* tickets, so that at least *w* lucky tickets were released and the total number of released tickets (*x*<=*<=*y*) were minimum. The series are numbered from 1 to *x* inclusive. The tickets in each series are numbered from 1 to *y* inclusive. The Transport Committee cannot release more than *max**x* series and more than *max**y* tickets in one series. | The first line contains three integers *max**x*, *max**y*, *w* (1<=≤<=*max**x*,<=*max**y*<=≤<=105, 1<=≤<=*w*<=≤<=107). | Print on a single line two space-separated numbers, the *x* and the *y*. If there are several possible variants, print any of them. If such *x* and *y* do not exist, print a single number <=-<=1. | [
"2 2 1\n",
"132 10 35\n",
"5 18 1000\n",
"48 132 235\n"
] | [
"1 1",
"7 5",
"-1\n",
"22 111"
] | none | 1,500 | [
{
"input": "2 2 1",
"output": "1 1"
},
{
"input": "132 10 35",
"output": "7 5"
},
{
"input": "5 18 1000",
"output": "-1"
},
{
"input": "48 132 235",
"output": "22 111"
},
{
"input": "119 69 169",
"output": "101 9"
},
{
"input": "49 24 83",
"output"... | 1,690,580,160 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | print("_RANDOM_GUESS_1690580160.0545058")# 1690580160.0545278 | Title: Lucky Tickets
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
In Walrusland public transport tickets are characterized by two integers: by the number of the series and by the number of the ticket in the series. Let the series number be represented by *a* and the ticket number — by *b*, then a ticket is described by the ordered pair of numbers (*a*,<=*b*).
The walruses believe that a ticket is lucky if *a*<=*<=*b*<==<=*rev*(*a*)<=*<=*rev*(*b*). The function *rev*(*x*) reverses a number written in the decimal system, at that the leading zeroes disappear. For example, *rev*(12343)<==<=34321, *rev*(1200)<==<=21.
The Public Transport Management Committee wants to release *x* series, each containing *y* tickets, so that at least *w* lucky tickets were released and the total number of released tickets (*x*<=*<=*y*) were minimum. The series are numbered from 1 to *x* inclusive. The tickets in each series are numbered from 1 to *y* inclusive. The Transport Committee cannot release more than *max**x* series and more than *max**y* tickets in one series.
Input Specification:
The first line contains three integers *max**x*, *max**y*, *w* (1<=≤<=*max**x*,<=*max**y*<=≤<=105, 1<=≤<=*w*<=≤<=107).
Output Specification:
Print on a single line two space-separated numbers, the *x* and the *y*. If there are several possible variants, print any of them. If such *x* and *y* do not exist, print a single number <=-<=1.
Demo Input:
['2 2 1\n', '132 10 35\n', '5 18 1000\n', '48 132 235\n']
Demo Output:
['1 1', '7 5', '-1\n', '22 111']
Note:
none | ```python
print("_RANDOM_GUESS_1690580160.0545058")# 1690580160.0545278
``` | 0 |
233 | A | Perfect Permutation | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*. | A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size. | If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces. | [
"1\n",
"2\n",
"4\n"
] | [
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] | none | 500 | [
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
... | 1,649,937,084 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | # -*- coding: utf-8 -*-
"""
Created on Thu Apr 14 12:45:15 2022
@author: dehon
"""
n = int(input())
if(n%2 == 1):
print(-1)
else:
for i in range(1,n+1):
if(i%2 == 1):
print(i+1, end=' ')
else:
print(i-1, end=' ') | Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""
Created on Thu Apr 14 12:45:15 2022
@author: dehon
"""
n = int(input())
if(n%2 == 1):
print(-1)
else:
for i in range(1,n+1):
if(i%2 == 1):
print(i+1, end=' ')
else:
print(i-1, end=' ')
``` | 3 | |
665 | C | Simple Strings | PROGRAMMING | 1,300 | [
"dp",
"greedy",
"strings"
] | null | null | zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task! | The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters. | Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters. | [
"aab\n",
"caaab\n",
"zscoder\n"
] | [
"bab\n",
"cabab\n",
"zscoder\n"
] | none | 0 | [
{
"input": "aab",
"output": "bab"
},
{
"input": "caaab",
"output": "cabab"
},
{
"input": "zscoder",
"output": "zscoder"
},
{
"input": "u",
"output": "u"
},
{
"input": "h",
"output": "h"
},
{
"input": "dtottttotd",
"output": "dtotataotd"
},
{
... | 1,664,936,705 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 27 | 109 | 15,769,600 | def add(a, b):
for j in range(26):
c = chr(j + ord('a'))
if(c != a and c != b):
return c
s = list(input())
n = len(s)
for i in range(1, n - 1):
if(s[i] == s[i - 1]):
s[i] = add(s[i - 1], s[i + 1])
if(n > 1):
if(s[-1] == s[-2]):
for j in range(26):
c = chr(j + ord('a'))
if(c != s[-2]):
s[-1] = c
break
print(''.join(s))
| Title: Simple Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task!
Input Specification:
The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters.
Output Specification:
Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters.
Demo Input:
['aab\n', 'caaab\n', 'zscoder\n']
Demo Output:
['bab\n', 'cabab\n', 'zscoder\n']
Note:
none | ```python
def add(a, b):
for j in range(26):
c = chr(j + ord('a'))
if(c != a and c != b):
return c
s = list(input())
n = len(s)
for i in range(1, n - 1):
if(s[i] == s[i - 1]):
s[i] = add(s[i - 1], s[i + 1])
if(n > 1):
if(s[-1] == s[-2]):
for j in range(26):
c = chr(j + ord('a'))
if(c != s[-2]):
s[-1] = c
break
print(''.join(s))
``` | 3 | |
863 | A | Quasi-palindrome | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number. | The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes. | Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes). | [
"131\n",
"320\n",
"2010200\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 0 | [
{
"input": "131",
"output": "YES"
},
{
"input": "320",
"output": "NO"
},
{
"input": "2010200",
"output": "YES"
},
{
"input": "1",
"output": "YES"
},
{
"input": "1000000000",
"output": "YES"
},
{
"input": "999999999",
"output": "YES"
},
{
"i... | 1,674,300,241 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 46 | 0 | inputStr = input()
tempRev = inputStr[::-1]
ok = False
if(inputStr == tempRev):
ok = True
else:
zerosAdded = 0
temp = list(inputStr)
for i in reversed(temp):
if(i != '0'):
break
else:
zerosAdded += 1
inputStr =inputStr.rjust(len(inputStr) + zerosAdded, "0")
tempRev = inputStr[::-1]
if(inputStr == tempRev):
ok = True
print("YES" if ok else "NO") | Title: Quasi-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
Input Specification:
The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes.
Output Specification:
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
Demo Input:
['131\n', '320\n', '2010200\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
inputStr = input()
tempRev = inputStr[::-1]
ok = False
if(inputStr == tempRev):
ok = True
else:
zerosAdded = 0
temp = list(inputStr)
for i in reversed(temp):
if(i != '0'):
break
else:
zerosAdded += 1
inputStr =inputStr.rjust(len(inputStr) + zerosAdded, "0")
tempRev = inputStr[::-1]
if(inputStr == tempRev):
ok = True
print("YES" if ok else "NO")
``` | 3 | |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output"... | 1,481,643,073 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 122 | 307,200 | n=int(input())
nlist=[int(i) for i in input().split()]
a0=nlist[0]
#n==3??#
if n==3:
if nlist[0]>nlist[1] and nlist[2]<nlist[1]:
print("2")
elif nlist[0]==nlist[1] or (nlist[0]>nlist[1] and nlist[2]>nlist[1]):
print("3")
elif nlist[0]<nlist[1]:
print("1")
#find evenness#
else:
dif1=(nlist[-1]-a0)/(n-1)
dif2=(nlist[-1]-nlist[1])/(n-2)
dif3=(nlist[-1]-nlist[2])/(n-3)
dif4=nlist[2]-nlist[1]
dif5=(nlist[2]-a0)/2
dif6=nlist[1]-a0
#allright#
if dif1==dif2==dif3==dif4==dif5==dif6:
dif=dif1
for i in range(3,n):
if (nlist[i]-nlist[i-1])!=dif:
result=i+1
print(result)
#a0wrong#
elif dif2==dif3==dif4:
print("1")
#a1wrong#
elif dif1==dif3==dif5:
print("2")
elif dif6==dif4==dif5:
print(len(nlist))
elif dif1==dif2==dif6:
print("3")
| Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n=int(input())
nlist=[int(i) for i in input().split()]
a0=nlist[0]
#n==3??#
if n==3:
if nlist[0]>nlist[1] and nlist[2]<nlist[1]:
print("2")
elif nlist[0]==nlist[1] or (nlist[0]>nlist[1] and nlist[2]>nlist[1]):
print("3")
elif nlist[0]<nlist[1]:
print("1")
#find evenness#
else:
dif1=(nlist[-1]-a0)/(n-1)
dif2=(nlist[-1]-nlist[1])/(n-2)
dif3=(nlist[-1]-nlist[2])/(n-3)
dif4=nlist[2]-nlist[1]
dif5=(nlist[2]-a0)/2
dif6=nlist[1]-a0
#allright#
if dif1==dif2==dif3==dif4==dif5==dif6:
dif=dif1
for i in range(3,n):
if (nlist[i]-nlist[i-1])!=dif:
result=i+1
print(result)
#a0wrong#
elif dif2==dif3==dif4:
print("1")
#a1wrong#
elif dif1==dif3==dif5:
print("2")
elif dif6==dif4==dif5:
print(len(nlist))
elif dif1==dif2==dif6:
print("3")
``` | 0 |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output"... | 1,532,606,696 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 218 | 0 | n = int(input())
m = list(map(int, input().split()))
k=0
z=0
for i in range(n):
if m[i] % 2 !=0:
k+=1
b=i+1
elif m[i] % 2 == 0:
z+=1
j=i+1
if k > z:
print(j)
else:
print(b)
| Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n = int(input())
m = list(map(int, input().split()))
k=0
z=0
for i in range(n):
if m[i] % 2 !=0:
k+=1
b=i+1
elif m[i] % 2 == 0:
z+=1
j=i+1
if k > z:
print(j)
else:
print(b)
``` | 3.9455 |
1,006 | C | Three Parts of the Array | PROGRAMMING | 1,200 | [
"binary search",
"data structures",
"two pointers"
] | null | null | You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible. | The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$. | Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$). | [
"5\n1 3 1 1 4\n",
"5\n1 3 2 1 4\n",
"3\n4 1 2\n"
] | [
"5\n",
"4\n",
"0\n"
] | In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$. | 0 | [
{
"input": "5\n1 3 1 1 4",
"output": "5"
},
{
"input": "5\n1 3 2 1 4",
"output": "4"
},
{
"input": "3\n4 1 2",
"output": "0"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "5\n1 3 5 4 5",
"output": ... | 1,691,890,894 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 27 | 109 | 16,998,400 | n = int(input())
arr = [int(x) for x in input().split(" ")]
soma1 = 0
soma2 = 0
l = 0
r = n-1
ans = 0
while l<=r:
if soma1 < soma2:
soma1+=arr[l]
l += 1
else:
soma2 += arr[r]
r -= 1
if soma1 == soma2:
ans = soma1
print(ans)
| Title: Three Parts of the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$.
Output Specification:
Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
Demo Input:
['5\n1 3 1 1 4\n', '5\n1 3 2 1 4\n', '3\n4 1 2\n']
Demo Output:
['5\n', '4\n', '0\n']
Note:
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$. | ```python
n = int(input())
arr = [int(x) for x in input().split(" ")]
soma1 = 0
soma2 = 0
l = 0
r = n-1
ans = 0
while l<=r:
if soma1 < soma2:
soma1+=arr[l]
l += 1
else:
soma2 += arr[r]
r -= 1
if soma1 == soma2:
ans = soma1
print(ans)
``` | 3 | |
1,005 | D | Polycarp and Div 3 | PROGRAMMING | 1,500 | [
"dp",
"greedy",
"number theory"
] | null | null | Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain? | The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0. | Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$. | [
"3121\n",
"6\n",
"1000000000000000000000000000000000\n",
"201920181\n"
] | [
"2\n",
"1\n",
"33\n",
"4\n"
] | In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | 0 | [
{
"input": "3121",
"output": "2"
},
{
"input": "6",
"output": "1"
},
{
"input": "1000000000000000000000000000000000",
"output": "33"
},
{
"input": "201920181",
"output": "4"
},
{
"input": "4",
"output": "0"
},
{
"input": "10",
"output": "1"
},
... | 1,685,470,795 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | def max_divisible_numbers(s):
n = len(s)
# Create a DP table to store the maximum number of divisible-by-3 numbers for each prefix
dp = [[0] * 3 for _ in range(n + 1)]
for i in range(1, n + 1):
# Calculate the remainder of the current digit when divided by 3
remainder = int(s[i - 1]) % 3
# Update the DP table based on the remainder and the previous state
dp[i][0] = dp[i - 1][0] + (remainder == 0)
dp[i][1] = max(dp[i - 1][1] + (remainder == 1), dp[i - 1][0] + (remainder == 2))
dp[i][2] = max(dp[i - 1][2] + (remainder == 2), dp[i - 1][0] + (remainder == 1))
# Return the maximum number of divisible-by-3 numbers for the entire string
return dp[n][0]
# Read the input number and call the function
s = input().strip()
result = max_divisible_numbers(s)
print(result) | Title: Polycarp and Div 3
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp likes numbers that are divisible by 3.
He has a huge number $s$. Polycarp wants to cut from it the maximum number of numbers that are divisible by $3$. To do this, he makes an arbitrary number of vertical cuts between pairs of adjacent digits. As a result, after $m$ such cuts, there will be $m+1$ parts in total. Polycarp analyzes each of the obtained numbers and finds the number of those that are divisible by $3$.
For example, if the original number is $s=3121$, then Polycarp can cut it into three parts with two cuts: $3|1|21$. As a result, he will get two numbers that are divisible by $3$.
Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.
What is the maximum number of numbers divisible by $3$ that Polycarp can obtain?
Input Specification:
The first line of the input contains a positive integer $s$. The number of digits of the number $s$ is between $1$ and $2\cdot10^5$, inclusive. The first (leftmost) digit is not equal to 0.
Output Specification:
Print the maximum number of numbers divisible by $3$ that Polycarp can get by making vertical cuts in the given number $s$.
Demo Input:
['3121\n', '6\n', '1000000000000000000000000000000000\n', '201920181\n']
Demo Output:
['2\n', '1\n', '33\n', '4\n']
Note:
In the first example, an example set of optimal cuts on the number is 3|1|21.
In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $3$.
In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $33$ digits 0. Each of the $33$ digits 0 forms a number that is divisible by $3$.
In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $0$, $9$, $201$ and $81$ are divisible by $3$. | ```python
def max_divisible_numbers(s):
n = len(s)
# Create a DP table to store the maximum number of divisible-by-3 numbers for each prefix
dp = [[0] * 3 for _ in range(n + 1)]
for i in range(1, n + 1):
# Calculate the remainder of the current digit when divided by 3
remainder = int(s[i - 1]) % 3
# Update the DP table based on the remainder and the previous state
dp[i][0] = dp[i - 1][0] + (remainder == 0)
dp[i][1] = max(dp[i - 1][1] + (remainder == 1), dp[i - 1][0] + (remainder == 2))
dp[i][2] = max(dp[i - 1][2] + (remainder == 2), dp[i - 1][0] + (remainder == 1))
# Return the maximum number of divisible-by-3 numbers for the entire string
return dp[n][0]
# Read the input number and call the function
s = input().strip()
result = max_divisible_numbers(s)
print(result)
``` | 0 | |
104 | A | Blackjack | PROGRAMMING | 800 | [
"implementation"
] | A. Blackjack | 2 | 256 | One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. | The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. | Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. | [
"12\n",
"20\n",
"10\n"
] | [
"4",
"15",
"0"
] | In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | 500 | [
{
"input": "12",
"output": "4"
},
{
"input": "20",
"output": "15"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "4"
},
{
"input": "15",
"output": "4"
},
{
"input": "18",
"output": "4"
},
{
"input": "25",
"output": "0"
... | 1,614,180,571 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 122 | 0 | import sys
from os import path
if(path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
import math
n = int(input())
n = n - 10
lst = [2,3,4,5,6,7,8,9]
if n == 0:
print(0)
else:
if n in lst:
print(4)
elif n == 10:
print(15)
else:
print(1) | Title: Blackjack
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.
Input Specification:
The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.
Output Specification:
Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.
Demo Input:
['12\n', '20\n', '10\n']
Demo Output:
['4', '15', '0']
Note:
In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | ```python
import sys
from os import path
if(path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
import math
n = int(input())
n = n - 10
lst = [2,3,4,5,6,7,8,9]
if n == 0:
print(0)
else:
if n in lst:
print(4)
elif n == 10:
print(15)
else:
print(1)
``` | 0 |
38 | A | Army | PROGRAMMING | 800 | [
"implementation"
] | A. Army | 2 | 256 | The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream. | The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated. | Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*. | [
"3\n5 6\n1 2\n",
"3\n5 6\n1 3\n"
] | [
"5\n",
"11\n"
] | none | 0 | [
{
"input": "3\n5 6\n1 2",
"output": "5"
},
{
"input": "3\n5 6\n1 3",
"output": "11"
},
{
"input": "2\n55\n1 2",
"output": "55"
},
{
"input": "3\n85 78\n1 3",
"output": "163"
},
{
"input": "4\n63 4 49\n2 3",
"output": "4"
},
{
"input": "5\n93 83 42 56\n... | 1,585,222,557 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 278 | 0 | ### A. Army
n=int(input())
arr=[int(x) for x in input().split()]
a,b=map(int,input().split())
total=0
for i in range(a,b):
total+=arr[i-1]
print(total) | Title: Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
Input Specification:
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Output Specification:
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
Demo Input:
['3\n5 6\n1 2\n', '3\n5 6\n1 3\n']
Demo Output:
['5\n', '11\n']
Note:
none | ```python
### A. Army
n=int(input())
arr=[int(x) for x in input().split()]
a,b=map(int,input().split())
total=0
for i in range(a,b):
total+=arr[i-1]
print(total)
``` | 3.9305 |
52 | C | Circular RMQ | PROGRAMMING | 2,200 | [
"data structures"
] | C. Circular RMQ | 1 | 256 | You are given circular array *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. There are two types of operations with it:
- *inc*(*lf*,<=*rg*,<=*v*) — this operation increases each element on the segment [*lf*,<=*rg*] (inclusively) by *v*; - *rmq*(*lf*,<=*rg*) — this operation returns minimal value on the segment [*lf*,<=*rg*] (inclusively).
Assume segments to be circular, so if *n*<==<=5 and *lf*<==<=3,<=*rg*<==<=1, it means the index sequence: 3,<=4,<=0,<=1.
Write program to process given sequence of operations. | The first line contains integer *n* (1<=≤<=*n*<=≤<=200000). The next line contains initial state of the array: *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (<=-<=106<=≤<=*a**i*<=≤<=106), *a**i* are integer. The third line contains integer *m* (0<=≤<=*m*<=≤<=200000), *m* — the number of operartons. Next *m* lines contain one operation each. If line contains two integer *lf*,<=*rg* (0<=≤<=*lf*,<=*rg*<=≤<=*n*<=-<=1) it means *rmq* operation, it contains three integers *lf*,<=*rg*,<=*v* (0<=≤<=*lf*,<=*rg*<=≤<=*n*<=-<=1;<=-<=106<=≤<=*v*<=≤<=106) — *inc* operation. | For each *rmq* operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d). | [
"4\n1 2 3 4\n4\n3 0\n3 0 -1\n0 1\n2 1\n"
] | [
"1\n0\n0\n"
] | none | 1,500 | [
{
"input": "4\n1 2 3 4\n4\n3 0\n3 0 -1\n0 1\n2 1",
"output": "1\n0\n0"
},
{
"input": "1\n-1\n10\n0 0 -1\n0 0\n0 0 1\n0 0\n0 0 1\n0 0\n0 0 0\n0 0\n0 0 -1\n0 0 1",
"output": "-2\n-1\n0\n0"
},
{
"input": "2\n-1 -1\n10\n0 0\n0 0\n0 0 1\n0 0\n1 1\n0 0 -1\n0 0 0\n0 0 1\n1 1 0\n0 0 -1",
"ou... | 1,673,061,035 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 109 | 2,764,800 |
from typing import List
class Node ():
def __init__ (self):
self.min=float('inf')
self.increase=0;
self.left=self.right=None
self.lChild=self.rChild=None
def recalc(root:Node):
if root.left==root.right :return
root.min=min(root.lChild.min+root.lChild.increase,root.rChild.min+root.rChild.increase)
def createSgTree(nums:List[int],l:int,r:int):
root=Node()
root.right=r
root.left=l
if l==r :
root.min=nums[l]
else :
mid=(l+r)//2
root.lChild=createSgTree(nums,l,mid)
root.rChild=createSgTree(nums,mid+1,r)
recalc(root)
return root
def propagation(root:Node) :
root.min+=root.increase
if root.left!=root.right :
root.lChild.increase=root.rChild.increase=root.increase
root.increase=0
def updateRange(root:Node,ql:int,qr:int,diff):
if ql<=root.left and root.right<=qr :
root.increase+=diff
return
if root.left>qr or root.right<ql : return
if root.increase!=0 :propagation(root)
updateRange(root.lChild,ql,qr,diff)
updateRange(root.rChild,ql,qr,diff)
recalc(root)
def rangeMin(root:Node,ql,qr) :
if ql<=root.left and root.right<=qr :return root.min +root.increase
if root.left>qr or root.right<ql : return float('inf')
if root.increase!=0 :
propagation(root)
return min(rangeMin(root.lChild,ql,qr),rangeMin(root.rChild,ql,qr))
class Solution():
def Circular_RMQ(self,n:int,nums:List[int]):
root=createSgTree(nums,0,n-1)
nq=int(input())
for i in range(nq):
request=list(map(int, input().split()))
if len(request)==2 :
print(min(rangeMin(root,request[0],n-1),rangeMin(root,0,request[1])))
else :
updateRange(root,request[0],n-1,request[2])
updateRange(root,0,request[1],request[2])
n=int(input())
nums = list(map(int, input().split()))
x=Solution()
x.Circular_RMQ(n,nums)
| Title: Circular RMQ
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given circular array *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. There are two types of operations with it:
- *inc*(*lf*,<=*rg*,<=*v*) — this operation increases each element on the segment [*lf*,<=*rg*] (inclusively) by *v*; - *rmq*(*lf*,<=*rg*) — this operation returns minimal value on the segment [*lf*,<=*rg*] (inclusively).
Assume segments to be circular, so if *n*<==<=5 and *lf*<==<=3,<=*rg*<==<=1, it means the index sequence: 3,<=4,<=0,<=1.
Write program to process given sequence of operations.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200000). The next line contains initial state of the array: *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (<=-<=106<=≤<=*a**i*<=≤<=106), *a**i* are integer. The third line contains integer *m* (0<=≤<=*m*<=≤<=200000), *m* — the number of operartons. Next *m* lines contain one operation each. If line contains two integer *lf*,<=*rg* (0<=≤<=*lf*,<=*rg*<=≤<=*n*<=-<=1) it means *rmq* operation, it contains three integers *lf*,<=*rg*,<=*v* (0<=≤<=*lf*,<=*rg*<=≤<=*n*<=-<=1;<=-<=106<=≤<=*v*<=≤<=106) — *inc* operation.
Output Specification:
For each *rmq* operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Demo Input:
['4\n1 2 3 4\n4\n3 0\n3 0 -1\n0 1\n2 1\n']
Demo Output:
['1\n0\n0\n']
Note:
none | ```python
from typing import List
class Node ():
def __init__ (self):
self.min=float('inf')
self.increase=0;
self.left=self.right=None
self.lChild=self.rChild=None
def recalc(root:Node):
if root.left==root.right :return
root.min=min(root.lChild.min+root.lChild.increase,root.rChild.min+root.rChild.increase)
def createSgTree(nums:List[int],l:int,r:int):
root=Node()
root.right=r
root.left=l
if l==r :
root.min=nums[l]
else :
mid=(l+r)//2
root.lChild=createSgTree(nums,l,mid)
root.rChild=createSgTree(nums,mid+1,r)
recalc(root)
return root
def propagation(root:Node) :
root.min+=root.increase
if root.left!=root.right :
root.lChild.increase=root.rChild.increase=root.increase
root.increase=0
def updateRange(root:Node,ql:int,qr:int,diff):
if ql<=root.left and root.right<=qr :
root.increase+=diff
return
if root.left>qr or root.right<ql : return
if root.increase!=0 :propagation(root)
updateRange(root.lChild,ql,qr,diff)
updateRange(root.rChild,ql,qr,diff)
recalc(root)
def rangeMin(root:Node,ql,qr) :
if ql<=root.left and root.right<=qr :return root.min +root.increase
if root.left>qr or root.right<ql : return float('inf')
if root.increase!=0 :
propagation(root)
return min(rangeMin(root.lChild,ql,qr),rangeMin(root.rChild,ql,qr))
class Solution():
def Circular_RMQ(self,n:int,nums:List[int]):
root=createSgTree(nums,0,n-1)
nq=int(input())
for i in range(nq):
request=list(map(int, input().split()))
if len(request)==2 :
print(min(rangeMin(root,request[0],n-1),rangeMin(root,0,request[1])))
else :
updateRange(root,request[0],n-1,request[2])
updateRange(root,0,request[1],request[2])
n=int(input())
nums = list(map(int, input().split()))
x=Solution()
x.Circular_RMQ(n,nums)
``` | 0 |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input":... | 1,673,327,831 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n = int(input())
while n > = 0 :
n -=1
k = input()
if len(k) > 10 :
print(k[0]+str(len(k)-2)+k[-1])
else:
print(k) | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
n = int(input())
while n > = 0 :
n -=1
k = input()
if len(k) > 10 :
print(k[0]+str(len(k)-2)+k[-1])
else:
print(k)
``` | -1 |
714 | A | Meeting of Old Friends | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | null | null | Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute *l*1 to minute *r*1 inclusive. Also, during the minute *k* she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute *l*2 to minute *r*2 inclusive.
Calculate the number of minutes they will be able to spend together. | The only line of the input contains integers *l*1, *r*1, *l*2, *r*2 and *k* (1<=≤<=*l*1,<=*r*1,<=*l*2,<=*r*2,<=*k*<=≤<=1018, *l*1<=≤<=*r*1, *l*2<=≤<=*r*2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks. | Print one integer — the number of minutes Sonya and Filya will be able to spend together. | [
"1 10 9 20 1\n",
"1 100 50 200 75\n"
] | [
"2\n",
"50\n"
] | In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | 500 | [
{
"input": "1 10 9 20 1",
"output": "2"
},
{
"input": "1 100 50 200 75",
"output": "50"
},
{
"input": "6 6 5 8 9",
"output": "1"
},
{
"input": "1 1000000000 1 1000000000 1",
"output": "999999999"
},
{
"input": "5 100 8 8 8",
"output": "0"
},
{
"input":... | 1,592,417,055 | 2,147,483,647 | PyPy 3 | OK | TESTS | 96 | 389 | 10,444,800 | import operator as op
import re
import sys
from bisect import bisect, bisect_left, insort, insort_left
from collections import Counter, defaultdict, deque
from copy import deepcopy
from decimal import Decimal
from functools import reduce
from itertools import (
accumulate, combinations, combinations_with_replacement, groupby,
permutations, product)
from math import (acos, asin, atan, ceil, cos, degrees, factorial, gcd, hypot,
log2, pi, radians, sin, sqrt, tan)
from operator import itemgetter, mul
from string import ascii_lowercase, ascii_uppercase, digits
def inp():
return(int(input()))
def inlist():
return(list(map(int, input().split())))
def instr():
s = input()
return(list(s[:len(s)]))
def invr():
return(map(int, input().split()))
def def_value():
return 0
# For getting input from input.txt file
#sys.stdin = open('input.txt', 'r')
# Printing the Output to output.txt file
#sys.stdout = open('output.txt', 'w')
l1, r1, l2, r2, k = invr()
if r2 < l1 or l2 > r1:
print(0)
sys.exit()
L = max(l1, l2)
R = min(r1, r2)
res = R - L + 1
if k >= L and k <= R:
res -= 1
if res < 0:
print(0)
else:
print(res)
| Title: Meeting of Old Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute *l*1 to minute *r*1 inclusive. Also, during the minute *k* she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute *l*2 to minute *r*2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input Specification:
The only line of the input contains integers *l*1, *r*1, *l*2, *r*2 and *k* (1<=≤<=*l*1,<=*r*1,<=*l*2,<=*r*2,<=*k*<=≤<=1018, *l*1<=≤<=*r*1, *l*2<=≤<=*r*2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output Specification:
Print one integer — the number of minutes Sonya and Filya will be able to spend together.
Demo Input:
['1 10 9 20 1\n', '1 100 50 200 75\n']
Demo Output:
['2\n', '50\n']
Note:
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100. | ```python
import operator as op
import re
import sys
from bisect import bisect, bisect_left, insort, insort_left
from collections import Counter, defaultdict, deque
from copy import deepcopy
from decimal import Decimal
from functools import reduce
from itertools import (
accumulate, combinations, combinations_with_replacement, groupby,
permutations, product)
from math import (acos, asin, atan, ceil, cos, degrees, factorial, gcd, hypot,
log2, pi, radians, sin, sqrt, tan)
from operator import itemgetter, mul
from string import ascii_lowercase, ascii_uppercase, digits
def inp():
return(int(input()))
def inlist():
return(list(map(int, input().split())))
def instr():
s = input()
return(list(s[:len(s)]))
def invr():
return(map(int, input().split()))
def def_value():
return 0
# For getting input from input.txt file
#sys.stdin = open('input.txt', 'r')
# Printing the Output to output.txt file
#sys.stdout = open('output.txt', 'w')
l1, r1, l2, r2, k = invr()
if r2 < l1 or l2 > r1:
print(0)
sys.exit()
L = max(l1, l2)
R = min(r1, r2)
res = R - L + 1
if k >= L and k <= R:
res -= 1
if res < 0:
print(0)
else:
print(res)
``` | 3 | |
200 | B | Drinks | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space. | Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4. | [
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] | [
"66.666666666667\n",
"37.500000000000\n"
] | Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | 500 | [
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39... | 1,686,755,111 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | drinks = int(input())
porcentagem = []
for i in range(drinks):
porcentagem.append(int(input()))
volumeTotal = sum(porcentagem)
media = volumeTotal / drinks
print(media)
| Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | ```python
drinks = int(input())
porcentagem = []
for i in range(drinks):
porcentagem.append(int(input()))
volumeTotal = sum(porcentagem)
media = volumeTotal / drinks
print(media)
``` | -1 | |
638 | B | Making Genome in Berland | PROGRAMMING | 1,500 | [
"*special",
"dfs and similar",
"strings"
] | null | null | Berland scientists face a very important task - given the parts of short DNA fragments, restore the dinosaur DNA! The genome of a berland dinosaur has noting in common with the genome that we've used to: it can have 26 distinct nucleotide types, a nucleotide of each type can occur at most once. If we assign distinct English letters to all nucleotides, then the genome of a Berland dinosaur will represent a non-empty string consisting of small English letters, such that each letter occurs in it at most once.
Scientists have *n* genome fragments that are represented as substrings (non-empty sequences of consecutive nucleotides) of the sought genome.
You face the following problem: help scientists restore the dinosaur genome. It is guaranteed that the input is not contradictory and at least one suitable line always exists. When the scientists found out that you are a strong programmer, they asked you in addition to choose the one with the minimum length. If there are multiple such strings, choose any string. | The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of genome fragments.
Each of the next lines contains one descriptions of a fragment. Each fragment is a non-empty string consisting of distinct small letters of the English alphabet. It is not guaranteed that the given fragments are distinct. Fragments could arbitrarily overlap and one fragment could be a substring of another one.
It is guaranteed that there is such string of distinct letters that contains all the given fragments as substrings. | In the single line of the output print the genome of the minimum length that contains all the given parts. All the nucleotides in the genome must be distinct. If there are multiple suitable strings, print the string of the minimum length. If there also are multiple suitable strings, you can print any of them. | [
"3\nbcd\nab\ncdef\n",
"4\nx\ny\nz\nw\n"
] | [
"abcdef\n",
"xyzw\n"
] | none | 1,000 | [
{
"input": "3\nbcd\nab\ncdef",
"output": "abcdef"
},
{
"input": "4\nx\ny\nz\nw",
"output": "xyzw"
},
{
"input": "25\nef\nfg\ngh\nhi\nij\njk\nkl\nlm\nmn\nno\nab\nbc\ncd\nde\nop\npq\nqr\nrs\nst\ntu\nuv\nvw\nwx\nxy\nyz",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "1\nf... | 1,458,544,191 | 68,991 | PyPy 3 | WRONG_ANSWER | PRETESTS | 2 | 108 | 23,244,800 | n = int(input())
a = []
b = []
for i in range(n) :
b.append(input())
b.sort(key = len,reverse = True)
for i in range(n):
s = b[i]
ok = False
#перевірка чи є рядок пудрядком введеного
for i in range(len(a)) :
if a[i].replace(s,"*")!=a[i] :
ok = True
break
if ok : continue
#перевірка,чи є кониць сього рядка початком другого
for l in range(len(s)) :
if ok : break
s1 = s[l:]
for i in range(len(a)) :
s2 = a[i][:len(s1)]
if s1==s2 :
a[i] = s[:l]+a[i]
ok = True
break
if ok : continue
#перевірка,чи є початок сього рядка кінцьом другого
l = len(s)
while l :
if ok : break
s1 = s[:l]
#print(s1)
for i in range(len(a)) :
s2 = a[i][len(a[i])-l:]
#print('=='+s2)
if s1==s2 :
a[i] = a[i]+s[l:]
ok = True
break
l-=1
if not ok :
a.append(s)
print("".join(a))
#обидві перевірки роблять жадно,починаючи з найбільшої довжини рядка
| Title: Making Genome in Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland scientists face a very important task - given the parts of short DNA fragments, restore the dinosaur DNA! The genome of a berland dinosaur has noting in common with the genome that we've used to: it can have 26 distinct nucleotide types, a nucleotide of each type can occur at most once. If we assign distinct English letters to all nucleotides, then the genome of a Berland dinosaur will represent a non-empty string consisting of small English letters, such that each letter occurs in it at most once.
Scientists have *n* genome fragments that are represented as substrings (non-empty sequences of consecutive nucleotides) of the sought genome.
You face the following problem: help scientists restore the dinosaur genome. It is guaranteed that the input is not contradictory and at least one suitable line always exists. When the scientists found out that you are a strong programmer, they asked you in addition to choose the one with the minimum length. If there are multiple such strings, choose any string.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of genome fragments.
Each of the next lines contains one descriptions of a fragment. Each fragment is a non-empty string consisting of distinct small letters of the English alphabet. It is not guaranteed that the given fragments are distinct. Fragments could arbitrarily overlap and one fragment could be a substring of another one.
It is guaranteed that there is such string of distinct letters that contains all the given fragments as substrings.
Output Specification:
In the single line of the output print the genome of the minimum length that contains all the given parts. All the nucleotides in the genome must be distinct. If there are multiple suitable strings, print the string of the minimum length. If there also are multiple suitable strings, you can print any of them.
Demo Input:
['3\nbcd\nab\ncdef\n', '4\nx\ny\nz\nw\n']
Demo Output:
['abcdef\n', 'xyzw\n']
Note:
none | ```python
n = int(input())
a = []
b = []
for i in range(n) :
b.append(input())
b.sort(key = len,reverse = True)
for i in range(n):
s = b[i]
ok = False
#перевірка чи є рядок пудрядком введеного
for i in range(len(a)) :
if a[i].replace(s,"*")!=a[i] :
ok = True
break
if ok : continue
#перевірка,чи є кониць сього рядка початком другого
for l in range(len(s)) :
if ok : break
s1 = s[l:]
for i in range(len(a)) :
s2 = a[i][:len(s1)]
if s1==s2 :
a[i] = s[:l]+a[i]
ok = True
break
if ok : continue
#перевірка,чи є початок сього рядка кінцьом другого
l = len(s)
while l :
if ok : break
s1 = s[:l]
#print(s1)
for i in range(len(a)) :
s2 = a[i][len(a[i])-l:]
#print('=='+s2)
if s1==s2 :
a[i] = a[i]+s[l:]
ok = True
break
l-=1
if not ok :
a.append(s)
print("".join(a))
#обидві перевірки роблять жадно,починаючи з найбільшої довжини рядка
``` | 0 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,694,771,897 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 35 | 122 | 0 | val = [int(x) for x in input().split()]
print((val[0] * val[1])//2 if val[0]*val[1] % 2 == 0 else ((val[0]*val[1]) - 1)//2) | Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
val = [int(x) for x in input().split()]
print((val[0] * val[1])//2 if val[0]*val[1] % 2 == 0 else ((val[0]*val[1]) - 1)//2)
``` | 3.9695 |
43 | A | Football | PROGRAMMING | 1,000 | [
"strings"
] | A. Football | 2 | 256 | One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. | Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. | [
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] | [
"ABC\n",
"A\n"
] | none | 500 | [
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
... | 1,596,554,131 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 280 | 20,172,800 | n = int(input())
cnt1, cnt2 = 1, 0
first, second = str(input()), ''
for i in range(0,n-1):
ip = str(input())
if ip==first:
cnt1+=1
else:
second = ip
cnt2+=1
out = first if cnt1>cnt2 else second
print(out) | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none | ```python
n = int(input())
cnt1, cnt2 = 1, 0
first, second = str(input()), ''
for i in range(0,n-1):
ip = str(input())
if ip==first:
cnt1+=1
else:
second = ip
cnt2+=1
out = first if cnt1>cnt2 else second
print(out)
``` | 3.892425 |
271 | A | Beautiful Year | PROGRAMMING | 800 | [
"brute force"
] | null | null | It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits. | The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number. | Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists. | [
"1987\n",
"2013\n"
] | [
"2013\n",
"2014\n"
] | none | 500 | [
{
"input": "1987",
"output": "2013"
},
{
"input": "2013",
"output": "2014"
},
{
"input": "1000",
"output": "1023"
},
{
"input": "1001",
"output": "1023"
},
{
"input": "1234",
"output": "1235"
},
{
"input": "5555",
"output": "5601"
},
{
"inp... | 1,695,644,624 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 92 | 0 | y = input()
y = str(int(y) + 1)
while y[0] == y[1] or y[0] == y[2] or y[0] == y[3] or y[1] == y[2] or y[1] == y[3] or y[2] == y[3]:
y = str(int(y)+1)
else:
print(y) | Title: Beautiful Year
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
Input Specification:
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
Output Specification:
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
Demo Input:
['1987\n', '2013\n']
Demo Output:
['2013\n', '2014\n']
Note:
none | ```python
y = input()
y = str(int(y) + 1)
while y[0] == y[1] or y[0] == y[2] or y[0] == y[3] or y[1] == y[2] or y[1] == y[3] or y[2] == y[3]:
y = str(int(y)+1)
else:
print(y)
``` | 3 | |
723 | A | The New Year: Meeting Friends | PROGRAMMING | 800 | [
"implementation",
"math",
"sortings"
] | null | null | There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer. | The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively. | Print one integer — the minimum total distance the friends need to travel in order to meet together. | [
"7 1 4\n",
"30 20 10\n"
] | [
"6\n",
"20\n"
] | In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4. | 500 | [
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"... | 1,690,548,503 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 46 | 0 | x1,x2,x3=map(int,input().split())
y=max(x1,x2,x3)
z=min(x1,x2,x3)
print(y-z) | Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4. | ```python
x1,x2,x3=map(int,input().split())
y=max(x1,x2,x3)
z=min(x1,x2,x3)
print(y-z)
``` | 3 | |
109 | A | Lucky Sum of Digits | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | A. Lucky Sum of Digits | 2 | 256 | Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. | The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. | Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. | [
"11\n",
"10\n"
] | [
"47\n",
"-1\n"
] | none | 500 | [
{
"input": "11",
"output": "47"
},
{
"input": "10",
"output": "-1"
},
{
"input": "64",
"output": "4477777777"
},
{
"input": "1",
"output": "-1"
},
{
"input": "4",
"output": "4"
},
{
"input": "7",
"output": "7"
},
{
"input": "12",
"outpu... | 1,655,803,929 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 1 | 124 | 3,379,200 | a=int(input())
ans=[]
temp=0
while(a>0):
if(a%7==0):
for i in range(a//7):
ans.append(str(7))
a=a-7
elif(a%4==0):
for i in range(a//4):
ans.append(str(4))
a=a-4
else:
if(a>8):
ans.append(str(7))
a=a-7
else:
print(-1)
temp=1
break
if(temp==0):
z="".join(sorted(ans))
z=z.replace("4444444","7777")
ans1=[]
for i in z:
if(i=="4"):
ans1.append("4")
for j in z:
if(j=="7"):
ans1.append("7")
print("".join(ans1))
| Title: Lucky Sum of Digits
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
Output Specification:
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
Demo Input:
['11\n', '10\n']
Demo Output:
['47\n', '-1\n']
Note:
none | ```python
a=int(input())
ans=[]
temp=0
while(a>0):
if(a%7==0):
for i in range(a//7):
ans.append(str(7))
a=a-7
elif(a%4==0):
for i in range(a//4):
ans.append(str(4))
a=a-4
else:
if(a>8):
ans.append(str(7))
a=a-7
else:
print(-1)
temp=1
break
if(temp==0):
z="".join(sorted(ans))
z=z.replace("4444444","7777")
ans1=[]
for i in z:
if(i=="4"):
ans1.append("4")
for j in z:
if(j=="7"):
ans1.append("7")
print("".join(ans1))
``` | -1 |
416 | A | Guess a number! | PROGRAMMING | 1,400 | [
"greedy",
"implementation",
"two pointers"
] | null | null | A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.
The rules are simple. The host thinks of an integer *y* and the participants guess it by asking questions to the host. There are four types of acceptable questions:
- Is it true that *y* is strictly larger than number *x*? - Is it true that *y* is strictly smaller than number *x*? - Is it true that *y* is larger than or equal to number *x*? - Is it true that *y* is smaller than or equal to number *x*?
On each question the host answers truthfully, "yes" or "no".
Given the sequence of questions and answers, find any integer value of *y* that meets the criteria of all answers. If there isn't such value, print "Impossible". | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10000) — the number of questions (and answers). Next *n* lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:
- ">" (for the first type queries), - "<" (for the second type queries), - ">=" (for the third type queries), - "<=" (for the fourth type queries).
All values of *x* are integer and meet the inequation <=-<=109<=≤<=*x*<=≤<=109. The answer is an English letter "Y" (for "yes") or "N" (for "no").
Consequtive elements in lines are separated by a single space. | Print any of such integers *y*, that the answers to all the queries are correct. The printed number *y* must meet the inequation <=-<=2·109<=≤<=*y*<=≤<=2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). | [
"4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N\n",
"2\n> 100 Y\n< -100 Y\n"
] | [
"17\n",
"Impossible\n"
] | none | 500 | [
{
"input": "4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N",
"output": "17"
},
{
"input": "2\n> 100 Y\n< -100 Y",
"output": "Impossible"
},
{
"input": "4\n< 1 N\n> 1 N\n> 1 N\n> 1 N",
"output": "1"
},
{
"input": "4\n<= 1 Y\n>= 1 Y\n>= 1 Y\n<= 1 Y",
"output": "1"
},
{
"input": ... | 1,491,326,086 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 5,427,200 | import random
n = input()
lowerbound = -9999999
upperbound = 9999999
rule = []
for i in range(int(n)):
rule = [str(x) for x in input().split()]
if rule[0] == '>=':
if rule[2] == 'Y':
if lowerbound < int(rule[1]):
lowerbound = int(rule[1])
else:
if upperbound > int(rule[1]):
upperbound = int(rule[1])
elif rule[0] == '>':
if rule[2] == 'Y':
if lowerbound < int(rule[1]):
lowerbound = int(rule[1]) - 1
else:
if upperbound > int(rule[1]):
upperbound = int(rule[1]) + 1
elif rule[0] == '<=':
if rule[2] == 'Y':
if upperbound < int(rule[1]):
upperbound = int(rule[1])
else:
if lowerbound > int(rule[1]):
lowerbound = int(rule[1])
elif rule[0] == '<':
if upperbound < int(rule[1]):
upperbound = int(rule[1]) + 1
else:
if lowerbound > int(rule[1]):
lowerbound = int(rule[1]) - 1
if lowerbound == upperbound:
print(lowerbound)
elif lowerbound > upperbound:
print(random.randint(int(lowerbound), int(upperbound)))
else:
print('Impossible')
| Title: Guess a number!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.
The rules are simple. The host thinks of an integer *y* and the participants guess it by asking questions to the host. There are four types of acceptable questions:
- Is it true that *y* is strictly larger than number *x*? - Is it true that *y* is strictly smaller than number *x*? - Is it true that *y* is larger than or equal to number *x*? - Is it true that *y* is smaller than or equal to number *x*?
On each question the host answers truthfully, "yes" or "no".
Given the sequence of questions and answers, find any integer value of *y* that meets the criteria of all answers. If there isn't such value, print "Impossible".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10000) — the number of questions (and answers). Next *n* lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:
- ">" (for the first type queries), - "<" (for the second type queries), - ">=" (for the third type queries), - "<=" (for the fourth type queries).
All values of *x* are integer and meet the inequation <=-<=109<=≤<=*x*<=≤<=109. The answer is an English letter "Y" (for "yes") or "N" (for "no").
Consequtive elements in lines are separated by a single space.
Output Specification:
Print any of such integers *y*, that the answers to all the queries are correct. The printed number *y* must meet the inequation <=-<=2·109<=≤<=*y*<=≤<=2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes).
Demo Input:
['4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N\n', '2\n> 100 Y\n< -100 Y\n']
Demo Output:
['17\n', 'Impossible\n']
Note:
none | ```python
import random
n = input()
lowerbound = -9999999
upperbound = 9999999
rule = []
for i in range(int(n)):
rule = [str(x) for x in input().split()]
if rule[0] == '>=':
if rule[2] == 'Y':
if lowerbound < int(rule[1]):
lowerbound = int(rule[1])
else:
if upperbound > int(rule[1]):
upperbound = int(rule[1])
elif rule[0] == '>':
if rule[2] == 'Y':
if lowerbound < int(rule[1]):
lowerbound = int(rule[1]) - 1
else:
if upperbound > int(rule[1]):
upperbound = int(rule[1]) + 1
elif rule[0] == '<=':
if rule[2] == 'Y':
if upperbound < int(rule[1]):
upperbound = int(rule[1])
else:
if lowerbound > int(rule[1]):
lowerbound = int(rule[1])
elif rule[0] == '<':
if upperbound < int(rule[1]):
upperbound = int(rule[1]) + 1
else:
if lowerbound > int(rule[1]):
lowerbound = int(rule[1]) - 1
if lowerbound == upperbound:
print(lowerbound)
elif lowerbound > upperbound:
print(random.randint(int(lowerbound), int(upperbound)))
else:
print('Impossible')
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | You have *n* devices that you want to use simultaneously.
The *i*-th device uses *a**i* units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·*a**i* units of power. The *i*-th device currently has *b**i* units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add *p* units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·*p* units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power. | The first line contains two integers, *n* and *p* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*p*<=≤<=109) — the number of devices and the power of the charger.
This is followed by *n* lines which contain two integers each. Line *i* contains the integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100<=000) — the power of the device and the amount of power stored in the device in the beginning. | If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=4.
Namely, let's assume that your answer is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if . | [
"2 1\n2 2\n2 1000\n",
"1 100\n1 1\n",
"3 5\n4 3\n5 2\n6 1\n"
] | [
"2.0000000000",
"-1\n",
"0.5000000000"
] | In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second. | 0 | [
{
"input": "2 1\n2 2\n2 1000",
"output": "2.0000000000"
},
{
"input": "1 100\n1 1",
"output": "-1"
},
{
"input": "3 5\n4 3\n5 2\n6 1",
"output": "0.5000000000"
},
{
"input": "1 1\n1 87",
"output": "-1"
},
{
"input": "1 1\n100 77",
"output": "0.7777777778"
},... | 1,608,759,563 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 77 | 0 | n, k = [int(x) for x in input().split()]
list = [int(x) for x in input().split()]
c = 0
x = 0
for i in list:
if i <= k:
c += 1
else:
x += 1
break
list.reverse()
if x > 0:
for i in list:
if i <= k:
c += 1
else:
break
print(c) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *n* devices that you want to use simultaneously.
The *i*-th device uses *a**i* units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·*a**i* units of power. The *i*-th device currently has *b**i* units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add *p* units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·*p* units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Input Specification:
The first line contains two integers, *n* and *p* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*p*<=≤<=109) — the number of devices and the power of the charger.
This is followed by *n* lines which contain two integers each. Line *i* contains the integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100<=000) — the power of the device and the amount of power stored in the device in the beginning.
Output Specification:
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=4.
Namely, let's assume that your answer is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['2 1\n2 2\n2 1000\n', '1 100\n1 1\n', '3 5\n4 3\n5 2\n6 1\n']
Demo Output:
['2.0000000000', '-1\n', '0.5000000000']
Note:
In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second. | ```python
n, k = [int(x) for x in input().split()]
list = [int(x) for x in input().split()]
c = 0
x = 0
for i in list:
if i <= k:
c += 1
else:
x += 1
break
list.reverse()
if x > 0:
for i in list:
if i <= k:
c += 1
else:
break
print(c)
``` | 0 | |
950 | B | Intercepted Message | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of *k* files of sizes *l*1,<=*l*2,<=...,<=*l**k* bytes, then the *i*-th file is split to one or more blocks *b**i*,<=1,<=*b**i*,<=2,<=...,<=*b**i*,<=*m**i* (here the total length of the blocks *b**i*,<=1<=+<=*b**i*,<=2<=+<=...<=+<=*b**i*,<=*m**i* is equal to the length of the file *l**i*), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct. | The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of blocks in the first and in the second messages.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=106) — the length of the blocks that form the first message.
The third line contains *m* integers *y*1,<=*y*2,<=...,<=*y**m* (1<=≤<=*y**i*<=≤<=106) — the length of the blocks that form the second message.
It is guaranteed that *x*1<=+<=...<=+<=*x**n*<==<=*y*1<=+<=...<=+<=*y**m*. Also, it is guaranteed that *x*1<=+<=...<=+<=*x**n*<=≤<=106. | Print the maximum number of files the intercepted array could consist of. | [
"7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8\n",
"3 3\n1 10 100\n1 100 10\n",
"1 4\n4\n1 1 1 1\n"
] | [
"3\n",
"2\n",
"1\n"
] | In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4. | 1,000 | [
{
"input": "7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8",
"output": "3"
},
{
"input": "3 3\n1 10 100\n1 100 10",
"output": "2"
},
{
"input": "1 4\n4\n1 1 1 1",
"output": "1"
},
{
"input": "1 1\n1000000\n1000000",
"output": "1"
},
{
"input": "3 5\n2 2 9\n2 1 4 2 4",
"outp... | 1,521,043,455 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 46 | 5,632,000 |
k=input()
k=k.split(" ")
counter=0
sumx=0
sumy=0
n=k[0]
m=k[1]
i=1
j=1
greater=0
x=input()
x=x.split(" ")
x=list(map(int,x))
y=input()
y=y.split(" ")
y=list(map(int,y))
sumx = x[0]
sumy = y[0]
n= int (n)
m = int (m)
if int(m)>int(n):
while i<m and j<=n:
if sumx==sumy:
counter+=1
if j<=n:
sumx=x[j]
j+=1
if i<=m:
sumy=y[i]
i+=1
elif sumx>sumy:
sumy=sumy +y[i]
i+=1
elif sumy>sumx:
sumx=sumx + x[j]
j+=1
else:
while i<n and j<=m:
if sumx==sumy:
counter+=1
if i<=n:
sumx=x[i]
i+=1
if j<=m:
sumy=y[j]
j+=1
elif sumx>sumy:
sumy = sumy+y[j]
j+=1
elif sumy>sumx:
sumx= sumx+x[i]
i+=1
if sumx==sumy:
counter= counter +1
print(counter)
| Title: Intercepted Message
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.
Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of *k* files of sizes *l*1,<=*l*2,<=...,<=*l**k* bytes, then the *i*-th file is split to one or more blocks *b**i*,<=1,<=*b**i*,<=2,<=...,<=*b**i*,<=*m**i* (here the total length of the blocks *b**i*,<=1<=+<=*b**i*,<=2<=+<=...<=+<=*b**i*,<=*m**i* is equal to the length of the file *l**i*), and after that all blocks are transferred through the network, maintaining the order of files in the archive.
Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.
You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of blocks in the first and in the second messages.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=106) — the length of the blocks that form the first message.
The third line contains *m* integers *y*1,<=*y*2,<=...,<=*y**m* (1<=≤<=*y**i*<=≤<=106) — the length of the blocks that form the second message.
It is guaranteed that *x*1<=+<=...<=+<=*x**n*<==<=*y*1<=+<=...<=+<=*y**m*. Also, it is guaranteed that *x*1<=+<=...<=+<=*x**n*<=≤<=106.
Output Specification:
Print the maximum number of files the intercepted array could consist of.
Demo Input:
['7 6\n2 5 3 1 11 4 4\n7 8 2 4 1 8\n', '3 3\n1 10 100\n1 100 10\n', '1 4\n4\n1 1 1 1\n']
Demo Output:
['3\n', '2\n', '1\n']
Note:
In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.
In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.
In the third example the only possibility is that the archive contains a single file of size 4. | ```python
k=input()
k=k.split(" ")
counter=0
sumx=0
sumy=0
n=k[0]
m=k[1]
i=1
j=1
greater=0
x=input()
x=x.split(" ")
x=list(map(int,x))
y=input()
y=y.split(" ")
y=list(map(int,y))
sumx = x[0]
sumy = y[0]
n= int (n)
m = int (m)
if int(m)>int(n):
while i<m and j<=n:
if sumx==sumy:
counter+=1
if j<=n:
sumx=x[j]
j+=1
if i<=m:
sumy=y[i]
i+=1
elif sumx>sumy:
sumy=sumy +y[i]
i+=1
elif sumy>sumx:
sumx=sumx + x[j]
j+=1
else:
while i<n and j<=m:
if sumx==sumy:
counter+=1
if i<=n:
sumx=x[i]
i+=1
if j<=m:
sumy=y[j]
j+=1
elif sumx>sumy:
sumy = sumy+y[j]
j+=1
elif sumy>sumx:
sumx= sumx+x[i]
i+=1
if sumx==sumy:
counter= counter +1
print(counter)
``` | 0 | |
229 | D | Towers | PROGRAMMING | 2,100 | [
"dp",
"greedy",
"two pointers"
] | null | null | The city of D consists of *n* towers, built consecutively on a straight line. The height of the tower that goes *i*-th (from left to right) in the sequence equals *h**i*. The city mayor decided to rebuild the city to make it beautiful. In a beautiful city all towers are are arranged in non-descending order of their height from left to right.
The rebuilding consists of performing several (perhaps zero) operations. An operation constitutes using a crane to take any tower and put it altogether on the top of some other neighboring tower. In other words, we can take the tower that stands *i*-th and put it on the top of either the (*i*<=-<=1)-th tower (if it exists), or the (*i*<=+<=1)-th tower (of it exists). The height of the resulting tower equals the sum of heights of the two towers that were put together. After that the two towers can't be split by any means, but more similar operations can be performed on the resulting tower. Note that after each operation the total number of towers on the straight line decreases by 1.
Help the mayor determine the minimum number of operations required to make the city beautiful. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — the number of towers in the city. The next line contains *n* space-separated integers: the *i*-th number *h**i* (1<=≤<=*h**i*<=≤<=105) determines the height of the tower that is *i*-th (from left to right) in the initial tower sequence. | Print a single integer — the minimum number of operations needed to make the city beautiful. | [
"5\n8 2 7 3 1\n",
"3\n5 2 1\n"
] | [
"3\n",
"2\n"
] | none | 1,000 | [
{
"input": "5\n8 2 7 3 1",
"output": "3"
},
{
"input": "3\n5 2 1",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 3 2",
"output": "1"
},
{
"i... | 1,595,442,576 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 140 | 20,172,800 |
def solve(arr):
if len(arr) <= 1:
return 0
i, j = 0, 1
last_val = arr[i]
ops = 0
while j < len(arr):
val = 0
while j < len(arr) and val < last_val:
val += arr[j]
j += 1
if val >= last_val:
ops += j - 2 - i
i = j - 1
last_val = val
else:
ops += j - i - 1
return ops
n = input()
arr = list(map(int, input().split()))
print(solve(arr))
| Title: Towers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The city of D consists of *n* towers, built consecutively on a straight line. The height of the tower that goes *i*-th (from left to right) in the sequence equals *h**i*. The city mayor decided to rebuild the city to make it beautiful. In a beautiful city all towers are are arranged in non-descending order of their height from left to right.
The rebuilding consists of performing several (perhaps zero) operations. An operation constitutes using a crane to take any tower and put it altogether on the top of some other neighboring tower. In other words, we can take the tower that stands *i*-th and put it on the top of either the (*i*<=-<=1)-th tower (if it exists), or the (*i*<=+<=1)-th tower (of it exists). The height of the resulting tower equals the sum of heights of the two towers that were put together. After that the two towers can't be split by any means, but more similar operations can be performed on the resulting tower. Note that after each operation the total number of towers on the straight line decreases by 1.
Help the mayor determine the minimum number of operations required to make the city beautiful.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — the number of towers in the city. The next line contains *n* space-separated integers: the *i*-th number *h**i* (1<=≤<=*h**i*<=≤<=105) determines the height of the tower that is *i*-th (from left to right) in the initial tower sequence.
Output Specification:
Print a single integer — the minimum number of operations needed to make the city beautiful.
Demo Input:
['5\n8 2 7 3 1\n', '3\n5 2 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
def solve(arr):
if len(arr) <= 1:
return 0
i, j = 0, 1
last_val = arr[i]
ops = 0
while j < len(arr):
val = 0
while j < len(arr) and val < last_val:
val += arr[j]
j += 1
if val >= last_val:
ops += j - 2 - i
i = j - 1
last_val = val
else:
ops += j - i - 1
return ops
n = input()
arr = list(map(int, input().split()))
print(solve(arr))
``` | 0 | |
652 | B | z-sort | PROGRAMMING | 1,000 | [
"sortings"
] | null | null | A student of *z*-school found a kind of sorting called *z*-sort. The array *a* with *n* elements are *z*-sorted if two conditions hold:
1. *a**i*<=≥<=*a**i*<=-<=1 for all even *i*, 1. *a**i*<=≤<=*a**i*<=-<=1 for all odd *i*<=><=1.
For example the arrays [1,2,1,2] and [1,1,1,1] are *z*-sorted while the array [1,2,3,4] isn’t *z*-sorted.
Can you make the array *z*-sorted? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array *a*.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. | If it's possible to make the array *a* *z*-sorted print *n* space separated integers *a**i* — the elements after *z*-sort. Otherwise print the only word "Impossible". | [
"4\n1 2 2 1\n",
"5\n1 3 2 2 5\n"
] | [
"1 2 1 2\n",
"1 5 2 3 2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 1",
"output": "1 2 1 2"
},
{
"input": "5\n1 3 2 2 5",
"output": "1 5 2 3 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1"
},
{
"input": "10\n1 9 7 6 2 4 7 8 1 3",
"output": "1 ... | 1,619,300,629 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 62 | 0 | n = int(input())
a = [int(x) for x in input().split()]
a = sorted(a)
saida = []
for i in range(n):
if i%2 == 0:
saida.append(a[i//2])
else:
saida.append(a[n-1-i//2])
print(*saida)
| Title: z-sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A student of *z*-school found a kind of sorting called *z*-sort. The array *a* with *n* elements are *z*-sorted if two conditions hold:
1. *a**i*<=≥<=*a**i*<=-<=1 for all even *i*, 1. *a**i*<=≤<=*a**i*<=-<=1 for all odd *i*<=><=1.
For example the arrays [1,2,1,2] and [1,1,1,1] are *z*-sorted while the array [1,2,3,4] isn’t *z*-sorted.
Can you make the array *z*-sorted?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array *a*.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Output Specification:
If it's possible to make the array *a* *z*-sorted print *n* space separated integers *a**i* — the elements after *z*-sort. Otherwise print the only word "Impossible".
Demo Input:
['4\n1 2 2 1\n', '5\n1 3 2 2 5\n']
Demo Output:
['1 2 1 2\n', '1 5 2 3 2\n']
Note:
none | ```python
n = int(input())
a = [int(x) for x in input().split()]
a = sorted(a)
saida = []
for i in range(n):
if i%2 == 0:
saida.append(a[i//2])
else:
saida.append(a[n-1-i//2])
print(*saida)
``` | 3 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4... | 1,656,688,656 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 8,704,000 | from sys import stdin
def get_data(n,array,m,queries):
vasya = 0
petya = 0
for q in queries:
pos = array.index(q) + 1
vasya += pos
petya += n - pos + 1
return [vasya,petya]
def main():
n = int(stdin.readline())
array = [int(x) for x in stdin.readline().split()]
m = int(stdin.readline())
queries = [int(x) for x in stdin.readline().split()]
results = get_data(n,array,m,queries)
print("{} {}".format(results[0],results[1]))
main()
| Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
from sys import stdin
def get_data(n,array,m,queries):
vasya = 0
petya = 0
for q in queries:
pos = array.index(q) + 1
vasya += pos
petya += n - pos + 1
return [vasya,petya]
def main():
n = int(stdin.readline())
array = [int(x) for x in stdin.readline().split()]
m = int(stdin.readline())
queries = [int(x) for x in stdin.readline().split()]
results = get_data(n,array,m,queries)
print("{} {}".format(results[0],results[1]))
main()
``` | 0 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4... | 1,644,447,959 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 9,318,400 | dizi_uzunlugu=int(input())
dizi=[int(x) for x in input().split()]
soru_uzunlugu=int(input())
sorular=[int(x) for x in input().split()]
anlik_uzunluk=0
bastan=0
sondan=0
while(len(sorular)>0):
bastan+=(sorular.count(sorular[0])*(dizi.index(sorular[0])+1))
sondan+=(dizi_uzunlugu-dizi.index(sorular[0]))*sorular.count(sorular[0])
w=sorular[0]
q=sorular.count(sorular[0])
for i in range(q):
sorular.remove(w)
print(bastan,sondan)
| Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
dizi_uzunlugu=int(input())
dizi=[int(x) for x in input().split()]
soru_uzunlugu=int(input())
sorular=[int(x) for x in input().split()]
anlik_uzunluk=0
bastan=0
sondan=0
while(len(sorular)>0):
bastan+=(sorular.count(sorular[0])*(dizi.index(sorular[0])+1))
sondan+=(dizi_uzunlugu-dizi.index(sorular[0]))*sorular.count(sorular[0])
w=sorular[0]
q=sorular.count(sorular[0])
for i in range(q):
sorular.remove(w)
print(bastan,sondan)
``` | 0 | |
757 | A | Gotta Catch Em' All! | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different. | Input contains a single line containing a string *s* (1<=<=≤<=<=|*s*|<=<=≤<=<=105) — the text on the front page of the newspaper without spaces and punctuation marks. |*s*| is the length of the string *s*.
The string *s* contains lowercase and uppercase English letters, i.e. . | Output a single integer, the answer to the problem. | [
"Bulbbasaur\n",
"F\n",
"aBddulbasaurrgndgbualdBdsagaurrgndbb\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | 500 | [
{
"input": "Bulbbasaur",
"output": "1"
},
{
"input": "F",
"output": "0"
},
{
"input": "aBddulbasaurrgndgbualdBdsagaurrgndbb",
"output": "2"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbuuuuuuuuuullllllllllssssssssssaaaaaaaaaarrrrrrrrrr",
"output": "5"
},
{
"input": "BBBBBBB... | 1,566,253,197 | 2,147,483,647 | PyPy 3 | OK | TESTS | 107 | 171 | 4,403,200 | #391_A
st = list(input())
#B = 1, u = 2, l = 1, b = 1, a = 2, s = 1, r = 1
dct = {"B": 0, "u": 0, "l": 0, "b": 0, "a": 0, "s": 0, "r": 0}
for i in range(0, len(st)):
if st[i] in dct:
dct[st[i]] += 1
n = min(dct["B"], dct["u"] // 2, dct["l"], dct["b"], dct["a"] // 2, dct["s"], dct["r"])
print(n)
| Title: Gotta Catch Em' All!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input Specification:
Input contains a single line containing a string *s* (1<=<=≤<=<=|*s*|<=<=≤<=<=105) — the text on the front page of the newspaper without spaces and punctuation marks. |*s*| is the length of the string *s*.
The string *s* contains lowercase and uppercase English letters, i.e. .
Output Specification:
Output a single integer, the answer to the problem.
Demo Input:
['Bulbbasaur\n', 'F\n', 'aBddulbasaurrgndgbualdBdsagaurrgndbb\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur". | ```python
#391_A
st = list(input())
#B = 1, u = 2, l = 1, b = 1, a = 2, s = 1, r = 1
dct = {"B": 0, "u": 0, "l": 0, "b": 0, "a": 0, "s": 0, "r": 0}
for i in range(0, len(st)):
if st[i] in dct:
dct[st[i]] += 1
n = min(dct["B"], dct["u"] // 2, dct["l"], dct["b"], dct["a"] // 2, dct["s"], dct["r"])
print(n)
``` | 3 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,645,470,037 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 124 | 0 | m,n=map(int,input().split())
if m%2==0 or n%2==0:
print(m*n//2)
else:
print(max(m*(n-1)//2+n//2,n*(m-1)//2+m//2))
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m,n=map(int,input().split())
if m%2==0 or n%2==0:
print(m*n//2)
else:
print(max(m*(n-1)//2+n//2,n*(m-1)//2+m//2))
``` | 0 |
584 | B | Kolya and Tanya | PROGRAMMING | 1,500 | [
"combinatorics"
] | null | null | Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3*n* gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3*n*<=-<=1, let the gnome sitting on the *i*-th place have *a**i* coins. If there is an integer *i* (0<=≤<=*i*<=<<=*n*) such that *a**i*<=+<=*a**i*<=+<=*n*<=+<=*a**i*<=+<=2*n*<=≠<=6, then Tanya is satisfied.
Count the number of ways to choose *a**i* so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109<=+<=7. Two ways, *a* and *b*, are considered distinct if there is index *i* (0<=≤<=*i*<=<<=3*n*), such that *a**i*<=≠<=*b**i* (that is, some gnome got different number of coins in these two ways). | A single line contains number *n* (1<=≤<=*n*<=≤<=105) — the number of the gnomes divided by three. | Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109<=+<=7. | [
"1\n",
"2\n"
] | [
"20",
"680"
] | 20 ways for *n* = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <img class="tex-graphics" src="https://espresso.codeforces.com/64df38b85ccb482cf88d02dc52e348e33313f9da.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "1",
"output": "20"
},
{
"input": "2",
"output": "680"
},
{
"input": "3",
"output": "19340"
},
{
"input": "4",
"output": "529040"
},
{
"input": "5",
"output": "14332100"
},
{
"input": "7",
"output": "459529590"
},
{
"input": "10"... | 1,664,464,355 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 42 | 77 | 2,457,600 | n = int(input())
print((3**(3*n)-7**n) % (int(1e9)+7)) | Title: Kolya and Tanya
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3*n* gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3*n*<=-<=1, let the gnome sitting on the *i*-th place have *a**i* coins. If there is an integer *i* (0<=≤<=*i*<=<<=*n*) such that *a**i*<=+<=*a**i*<=+<=*n*<=+<=*a**i*<=+<=2*n*<=≠<=6, then Tanya is satisfied.
Count the number of ways to choose *a**i* so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109<=+<=7. Two ways, *a* and *b*, are considered distinct if there is index *i* (0<=≤<=*i*<=<<=3*n*), such that *a**i*<=≠<=*b**i* (that is, some gnome got different number of coins in these two ways).
Input Specification:
A single line contains number *n* (1<=≤<=*n*<=≤<=105) — the number of the gnomes divided by three.
Output Specification:
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109<=+<=7.
Demo Input:
['1\n', '2\n']
Demo Output:
['20', '680']
Note:
20 ways for *n* = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <img class="tex-graphics" src="https://espresso.codeforces.com/64df38b85ccb482cf88d02dc52e348e33313f9da.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
n = int(input())
print((3**(3*n)-7**n) % (int(1e9)+7))
``` | 3 | |
424 | A | Squats | PROGRAMMING | 900 | [
"implementation"
] | null | null | Pasha has many hamsters and he makes them work out. Today, *n* hamsters (*n* is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well? | The first line contains integer *n* (2<=≤<=*n*<=≤<=200; *n* is even). The next line contains *n* characters without spaces. These characters describe the hamsters' position: the *i*-th character equals 'X', if the *i*-th hamster in the row is standing, and 'x', if he is sitting. | In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them. | [
"4\nxxXx\n",
"2\nXX\n",
"6\nxXXxXx\n"
] | [
"1\nXxXx\n",
"1\nxX\n",
"0\nxXXxXx\n"
] | none | 500 | [
{
"input": "4\nxxXx",
"output": "1\nXxXx"
},
{
"input": "2\nXX",
"output": "1\nxX"
},
{
"input": "6\nxXXxXx",
"output": "0\nxXXxXx"
},
{
"input": "4\nxXXX",
"output": "1\nxxXX"
},
{
"input": "2\nXx",
"output": "0\nXx"
},
{
"input": "22\nXXxXXxxXxXxXXXX... | 1,586,084,583 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 93 | 307,200 | # -*- coding: utf-8 -*-
"""
Created on Sun Apr 5 05:45:34 2020
@author: alexi
"""
#https://codeforces.com/problemset/problem/424/A --- Alexis Galvan
def hamster_squat():
total = int(input())
hamsters = input()
dic = {'x':0,'X':0}
for i in range(len(hamsters)):
dic[hamsters[i]] += 1
if dic['x'] == dic['X']:
print(0)
print(hamsters)
return
sat = dic['x']
stood = dic['X']
add_lower = False
if sat < stood:
add_lower = True
maximum = max(sat, stood)
add = maximum - (int((sat+stood)/2))
print(add)
if add_lower:
output = ''
hamsters = [i for i in hamsters]
for i in range(len(hamsters)):
if add > 0:
if hamsters[i] == 'X':
hamsters[i] = 'x'
add -= 1
output = output + hamsters[i]
else:
output = ''
hamsters = [i for i in hamsters]
for i in range(len(hamsters)):
if add > 0:
if hamsters[i] == 'x':
hamsters[i] = 'X'
add -= 1
output = output + hamsters[i]
print(output)
hamster_squat()
| Title: Squats
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has many hamsters and he makes them work out. Today, *n* hamsters (*n* is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=200; *n* is even). The next line contains *n* characters without spaces. These characters describe the hamsters' position: the *i*-th character equals 'X', if the *i*-th hamster in the row is standing, and 'x', if he is sitting.
Output Specification:
In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.
Demo Input:
['4\nxxXx\n', '2\nXX\n', '6\nxXXxXx\n']
Demo Output:
['1\nXxXx\n', '1\nxX\n', '0\nxXXxXx\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 5 05:45:34 2020
@author: alexi
"""
#https://codeforces.com/problemset/problem/424/A --- Alexis Galvan
def hamster_squat():
total = int(input())
hamsters = input()
dic = {'x':0,'X':0}
for i in range(len(hamsters)):
dic[hamsters[i]] += 1
if dic['x'] == dic['X']:
print(0)
print(hamsters)
return
sat = dic['x']
stood = dic['X']
add_lower = False
if sat < stood:
add_lower = True
maximum = max(sat, stood)
add = maximum - (int((sat+stood)/2))
print(add)
if add_lower:
output = ''
hamsters = [i for i in hamsters]
for i in range(len(hamsters)):
if add > 0:
if hamsters[i] == 'X':
hamsters[i] = 'x'
add -= 1
output = output + hamsters[i]
else:
output = ''
hamsters = [i for i in hamsters]
for i in range(len(hamsters)):
if add > 0:
if hamsters[i] == 'x':
hamsters[i] = 'X'
add -= 1
output = output + hamsters[i]
print(output)
hamster_squat()
``` | 3 | |
224 | A | Parallelepiped | PROGRAMMING | 1,100 | [
"brute force",
"geometry",
"math"
] | null | null | You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped. | The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement. | Print a single number — the sum of all edges of the parallelepiped. | [
"1 1 1\n",
"4 6 6\n"
] | [
"12\n",
"28\n"
] | In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3. | 500 | [
{
"input": "1 1 1",
"output": "12"
},
{
"input": "4 6 6",
"output": "28"
},
{
"input": "20 10 50",
"output": "68"
},
{
"input": "9 4 36",
"output": "56"
},
{
"input": "324 9 36",
"output": "184"
},
{
"input": "1333 93 129",
"output": "308"
},
{... | 1,692,183,711 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 92 | 0 | import math
z = []
z[0:] = map(int, input().split())
z.sort()
lst1 = []
lst2 = []
lst3 = []
lst = [1]
for i in range(1,z[1]):
if i not in lst and z[0] % i == 0 and z[2] % i == 0: #and i not in lst:# and z[2] % i == 0:
lst.append(i)
#break
if i not in lst and z[0] % i == 0 and z[1] % i == 0: #and i not in lst:# and z[2] % i == 0:
lst.append(i)
#break
#print(lst)
count = sum(z)
if len(lst1) == 0:
lst1.append(1)
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
else:
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
for i in range(len(lst)):
if (lst1[0] * lst1[1] + lst1[0] * lst1[2] + lst1[1] * lst1[2]) != count:
lst1.clear()
lst1.append(lst[len(lst1) - 1 -i])
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
else:
break
print(sum(lst1) * 4) | Title: Parallelepiped
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
Input Specification:
The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
Output Specification:
Print a single number — the sum of all edges of the parallelepiped.
Demo Input:
['1 1 1\n', '4 6 6\n']
Demo Output:
['12\n', '28\n']
Note:
In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3. | ```python
import math
z = []
z[0:] = map(int, input().split())
z.sort()
lst1 = []
lst2 = []
lst3 = []
lst = [1]
for i in range(1,z[1]):
if i not in lst and z[0] % i == 0 and z[2] % i == 0: #and i not in lst:# and z[2] % i == 0:
lst.append(i)
#break
if i not in lst and z[0] % i == 0 and z[1] % i == 0: #and i not in lst:# and z[2] % i == 0:
lst.append(i)
#break
#print(lst)
count = sum(z)
if len(lst1) == 0:
lst1.append(1)
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
else:
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
for i in range(len(lst)):
if (lst1[0] * lst1[1] + lst1[0] * lst1[2] + lst1[1] * lst1[2]) != count:
lst1.clear()
lst1.append(lst[len(lst1) - 1 -i])
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
else:
break
print(sum(lst1) * 4)
``` | 3 | |
337 | B | Routine Problem | PROGRAMMING | 1,400 | [
"greedy",
"math",
"number theory"
] | null | null | Manao has a monitor. The screen of the monitor has horizontal to vertical length ratio *a*:*b*. Now he is going to watch a movie. The movie's frame has horizontal to vertical length ratio *c*:*d*. Manao adjusts the view in such a way that the movie preserves the original frame ratio, but also occupies as much space on the screen as possible and fits within it completely. Thus, he may have to zoom the movie in or out, but Manao will always change the frame proportionally in both dimensions.
Calculate the ratio of empty screen (the part of the screen not occupied by the movie) to the total screen size. Print the answer as an irreducible fraction *p*<=/<=*q*. | A single line contains four space-separated integers *a*, *b*, *c*, *d* (1<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=1000). | Print the answer to the problem as "p/q", where *p* is a non-negative integer, *q* is a positive integer and numbers *p* and *q* don't have a common divisor larger than 1. | [
"1 1 3 2\n",
"4 3 2 2\n"
] | [
"1/3\n",
"1/4\n"
] | Sample 1. Manao's monitor has a square screen. The movie has 3:2 horizontal to vertical length ratio. Obviously, the movie occupies most of the screen if the width of the picture coincides with the width of the screen. In this case, only 2/3 of the monitor will project the movie in the horizontal dimension: <img class="tex-graphics" src="https://espresso.codeforces.com/ce823413ad27813e27496a0d8bd4231e94b47662.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Sample 2. This time the monitor's width is 4/3 times larger than its height and the movie's frame is square. In this case, the picture must take up the whole monitor in the vertical dimension and only 3/4 in the horizontal dimension: <img class="tex-graphics" src="https://espresso.codeforces.com/c2bcb3b1f64810812eee368ff180e3e148d24c67.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "1 1 3 2",
"output": "1/3"
},
{
"input": "4 3 2 2",
"output": "1/4"
},
{
"input": "3 4 2 3",
"output": "1/9"
},
{
"input": "4 4 5 5",
"output": "0/1"
},
{
"input": "1 1 1 1",
"output": "0/1"
},
{
"input": "1000 1000 1000 1000",
"output":... | 1,680,027,086 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 122 | 409,600 | # from fractions import Fraction
# from decimal import Decimal
import fractions
a,b,c,d=map(int,input().split())
if(a/b<c/d):
p=a*d
q=b*c
x=(q-p)/q
# print(x)
b = fractions.Fraction(q-p,q)
print(b)
quit()
if(a/b>c/d):
p=b*c
q=a*d
x=(p/q)
b = fractions.Fraction(q-p,q)
print(b)
quit()
if(a/b==c/d):
print(0)
| Title: Routine Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao has a monitor. The screen of the monitor has horizontal to vertical length ratio *a*:*b*. Now he is going to watch a movie. The movie's frame has horizontal to vertical length ratio *c*:*d*. Manao adjusts the view in such a way that the movie preserves the original frame ratio, but also occupies as much space on the screen as possible and fits within it completely. Thus, he may have to zoom the movie in or out, but Manao will always change the frame proportionally in both dimensions.
Calculate the ratio of empty screen (the part of the screen not occupied by the movie) to the total screen size. Print the answer as an irreducible fraction *p*<=/<=*q*.
Input Specification:
A single line contains four space-separated integers *a*, *b*, *c*, *d* (1<=≤<=*a*,<=*b*,<=*c*,<=*d*<=≤<=1000).
Output Specification:
Print the answer to the problem as "p/q", where *p* is a non-negative integer, *q* is a positive integer and numbers *p* and *q* don't have a common divisor larger than 1.
Demo Input:
['1 1 3 2\n', '4 3 2 2\n']
Demo Output:
['1/3\n', '1/4\n']
Note:
Sample 1. Manao's monitor has a square screen. The movie has 3:2 horizontal to vertical length ratio. Obviously, the movie occupies most of the screen if the width of the picture coincides with the width of the screen. In this case, only 2/3 of the monitor will project the movie in the horizontal dimension: <img class="tex-graphics" src="https://espresso.codeforces.com/ce823413ad27813e27496a0d8bd4231e94b47662.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Sample 2. This time the monitor's width is 4/3 times larger than its height and the movie's frame is square. In this case, the picture must take up the whole monitor in the vertical dimension and only 3/4 in the horizontal dimension: <img class="tex-graphics" src="https://espresso.codeforces.com/c2bcb3b1f64810812eee368ff180e3e148d24c67.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
# from fractions import Fraction
# from decimal import Decimal
import fractions
a,b,c,d=map(int,input().split())
if(a/b<c/d):
p=a*d
q=b*c
x=(q-p)/q
# print(x)
b = fractions.Fraction(q-p,q)
print(b)
quit()
if(a/b>c/d):
p=b*c
q=a*d
x=(p/q)
b = fractions.Fraction(q-p,q)
print(b)
quit()
if(a/b==c/d):
print(0)
``` | 0 | |
200 | B | Drinks | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space. | Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4. | [
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] | [
"66.666666666667\n",
"37.500000000000\n"
] | Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | 500 | [
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39... | 1,692,529,668 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 31 | 124 | 0 | n = int(input())
def sum(l,n):
sum1 = 0
for i in range(n):
sum1 += l[i]
return sum1
l = [int(x) for x in input().split()]
sum(l, n)
print(sum(l,n)/n) | Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | ```python
n = int(input())
def sum(l,n):
sum1 = 0
for i in range(n):
sum1 += l[i]
return sum1
l = [int(x) for x in input().split()]
sum(l, n)
print(sum(l,n)/n)
``` | 3 | |
242 | C | King's Path | PROGRAMMING | 1,800 | [
"dfs and similar",
"graphs",
"hashing",
"shortest paths"
] | null | null | The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the *i*-th row and *j*-th column as (*i*,<=*j*).
You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as *n* segments. Each segment is described by three integers *r**i*,<=*a**i*,<=*b**i* (*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed.
Your task is to find the minimum number of moves the king needs to get from square (*x*0,<=*y*0) to square (*x*1,<=*y*1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way.
Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point. | The first line contains four space-separated integers *x*0,<=*y*0,<=*x*1,<=*y*1 (1<=≤<=*x*0,<=*y*0,<=*x*1,<=*y*1<=≤<=109), denoting the initial and the final positions of the king.
The second line contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of segments of allowed cells. Next *n* lines contain the descriptions of these segments. The *i*-th line contains three space-separated integers *r**i*,<=*a**i*,<=*b**i* (1<=≤<=*r**i*,<=*a**i*,<=*b**i*<=≤<=109,<=*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily.
It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105. | If there is no path between the initial and final position along allowed cells, print -1.
Otherwise print a single integer — the minimum number of moves the king needs to get from the initial position to the final one. | [
"5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5\n",
"3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10\n",
"1 1 2 10\n2\n1 1 3\n2 6 10\n"
] | [
"4\n",
"6\n",
"-1\n"
] | none | 1,500 | [
{
"input": "5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5",
"output": "4"
},
{
"input": "3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10",
"output": "6"
},
{
"input": "1 1 2 10\n2\n1 1 3\n2 6 10",
"output": "-1"
},
{
"input": "9 8 7 8\n9\n10 6 6\n10 6 6\n7 7 8\n9 5 6\n8 9 9\n9 5 5\n9 8 8\n8 5 6\n9 10... | 1,638,729,291 | 2,147,483,647 | PyPy 3 | OK | TESTS | 43 | 686 | 11,980,800 | import sys
import math
from sys import stdin, stdout
# TAKE INPUT
def get_ints_in_variables():
return map(int, sys.stdin.readline().strip().split())
def get_int(): return int(input())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
def main():
# Write Your Code Here
x0,y0,x1,y1 = get_ints_in_variables()
n = int(input())
g = {}
for i in range(0, n):
r,a,b = get_ints_in_variables()
for i in range(a, b+1):
g[(r, i)] = -1
g[(x0, y0)] = 0
# {[i-1][0], [i-1][j-1], [i-1][j+1], [i][j-1], [i][j+1], [i+1][j], [i+1][j-1], [i+1][j+1]}
dx = [-1, -1, -1, 0, 0, 1, 1, 1]
dy = [0, -1, 1, -1, 1, 0, -1, 1]
queue = [(x0, y0)]
while len(queue):
node = queue.pop(0)
for j in range(8):
v = (node[0]+dx[j], node[1]+dy[j])
# print(v, "v")
if not v in g:
continue
if v in g and g[v] != -1:
continue
g[v] = g[node]+1
queue.append(v)
print(g[(x1, y1)])
# calling main Function
if __name__ == "__main__":
main() | Title: King's Path
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the *i*-th row and *j*-th column as (*i*,<=*j*).
You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as *n* segments. Each segment is described by three integers *r**i*,<=*a**i*,<=*b**i* (*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed.
Your task is to find the minimum number of moves the king needs to get from square (*x*0,<=*y*0) to square (*x*1,<=*y*1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way.
Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point.
Input Specification:
The first line contains four space-separated integers *x*0,<=*y*0,<=*x*1,<=*y*1 (1<=≤<=*x*0,<=*y*0,<=*x*1,<=*y*1<=≤<=109), denoting the initial and the final positions of the king.
The second line contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of segments of allowed cells. Next *n* lines contain the descriptions of these segments. The *i*-th line contains three space-separated integers *r**i*,<=*a**i*,<=*b**i* (1<=≤<=*r**i*,<=*a**i*,<=*b**i*<=≤<=109,<=*a**i*<=≤<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily.
It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105.
Output Specification:
If there is no path between the initial and final position along allowed cells, print -1.
Otherwise print a single integer — the minimum number of moves the king needs to get from the initial position to the final one.
Demo Input:
['5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5\n', '3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10\n', '1 1 2 10\n2\n1 1 3\n2 6 10\n']
Demo Output:
['4\n', '6\n', '-1\n']
Note:
none | ```python
import sys
import math
from sys import stdin, stdout
# TAKE INPUT
def get_ints_in_variables():
return map(int, sys.stdin.readline().strip().split())
def get_int(): return int(input())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
def main():
# Write Your Code Here
x0,y0,x1,y1 = get_ints_in_variables()
n = int(input())
g = {}
for i in range(0, n):
r,a,b = get_ints_in_variables()
for i in range(a, b+1):
g[(r, i)] = -1
g[(x0, y0)] = 0
# {[i-1][0], [i-1][j-1], [i-1][j+1], [i][j-1], [i][j+1], [i+1][j], [i+1][j-1], [i+1][j+1]}
dx = [-1, -1, -1, 0, 0, 1, 1, 1]
dy = [0, -1, 1, -1, 1, 0, -1, 1]
queue = [(x0, y0)]
while len(queue):
node = queue.pop(0)
for j in range(8):
v = (node[0]+dx[j], node[1]+dy[j])
# print(v, "v")
if not v in g:
continue
if v in g and g[v] != -1:
continue
g[v] = g[node]+1
queue.append(v)
print(g[(x1, y1)])
# calling main Function
if __name__ == "__main__":
main()
``` | 3 | |
910 | B | Door Frames | PROGRAMMING | 1,600 | [
"greedy",
"implementation"
] | null | null | Petya has equal wooden bars of length *n*. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length *a* and one top side of length *b*. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar). | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the length of each wooden bar.
The second line contains a single integer *a* (1<=≤<=*a*<=≤<=*n*) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer *b* (1<=≤<=*b*<=≤<=*n*) — the length of the upper side of a door frame. | Print the minimal number of wooden bars with length *n* which are needed to make the frames for two doors. | [
"8\n1\n2\n",
"5\n3\n4\n",
"6\n4\n2\n",
"20\n5\n6\n"
] | [
"1\n",
"6\n",
"4\n",
"2\n"
] | In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed. | 1,000 | [
{
"input": "8\n1\n2",
"output": "1"
},
{
"input": "5\n3\n4",
"output": "6"
},
{
"input": "6\n4\n2",
"output": "4"
},
{
"input": "20\n5\n6",
"output": "2"
},
{
"input": "1\n1\n1",
"output": "6"
},
{
"input": "3\n1\n2",
"output": "3"
},
{
"in... | 1,535,312,179 | 2,147,483,647 | Python 3 | OK | TESTS | 120 | 124 | 0 | import math
[n,a,b],r,i,j=[int(input())for x in range(3)],6,4,5
while i>=0:
l,c,o=[b if x in[i,j]else a for x in range(6)],0,n
for k in l:
if o<k:
o,c=n-k,c+1
else:o-=k
r=min(r,c if o==n else c+1)
j-=1
if i==j:i,j=i-1,5
print(r)
# Made By Mostafa_Khaled | Title: Door Frames
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has equal wooden bars of length *n*. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length *a* and one top side of length *b*. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the length of each wooden bar.
The second line contains a single integer *a* (1<=≤<=*a*<=≤<=*n*) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer *b* (1<=≤<=*b*<=≤<=*n*) — the length of the upper side of a door frame.
Output Specification:
Print the minimal number of wooden bars with length *n* which are needed to make the frames for two doors.
Demo Input:
['8\n1\n2\n', '5\n3\n4\n', '6\n4\n2\n', '20\n5\n6\n']
Demo Output:
['1\n', '6\n', '4\n', '2\n']
Note:
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed. | ```python
import math
[n,a,b],r,i,j=[int(input())for x in range(3)],6,4,5
while i>=0:
l,c,o=[b if x in[i,j]else a for x in range(6)],0,n
for k in l:
if o<k:
o,c=n-k,c+1
else:o-=k
r=min(r,c if o==n else c+1)
j-=1
if i==j:i,j=i-1,5
print(r)
# Made By Mostafa_Khaled
``` | 3 | |
214 | A | System of Equations | PROGRAMMING | 800 | [
"brute force"
] | null | null | Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system. | A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space. | On a single line print the answer to the problem. | [
"9 3\n",
"14 28\n",
"4 20\n"
] | [
"1\n",
"1\n",
"0\n"
] | In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | 500 | [
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10"... | 1,656,565,182 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 54 | 218 | 2,048,000 | n,m=map(int,input().split())
c=0
for a in range(300):
for b in range(300):
if a**2+b==n and a+b**2==m:
c+=1
print(c)
| Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | ```python
n,m=map(int,input().split())
c=0
for a in range(300):
for b in range(300):
if a**2+b==n and a+b**2==m:
c+=1
print(c)
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,696,070,026 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | # -*- coding: utf-8 -*-
"""
Created on Sat Sep 30 18:22:42 2023
@author: huozi
"""
word = input()
target = "hello"
pointer = 0
count = 0
for char in :
if char == target[pointer]:
pointer += 1
count += 1
if count == len(target):
break
if count == len(target):
print("YES")
else:
print("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""
Created on Sat Sep 30 18:22:42 2023
@author: huozi
"""
word = input()
target = "hello"
pointer = 0
count = 0
for char in :
if char == target[pointer]:
pointer += 1
count += 1
if count == len(target):
break
if count == len(target):
print("YES")
else:
print("NO")
``` | -1 |
8 | A | Train and Peter | PROGRAMMING | 1,200 | [
"strings"
] | A. Train and Peter | 1 | 64 | Peter likes to travel by train. He likes it so much that on the train he falls asleep.
Once in summer Peter was going by train from city A to city B, and as usual, was sleeping. Then he woke up, started to look through the window and noticed that every railway station has a flag of a particular colour.
The boy started to memorize the order of the flags' colours that he had seen. But soon he fell asleep again. Unfortunately, he didn't sleep long, he woke up and went on memorizing the colours. Then he fell asleep again, and that time he slept till the end of the journey.
At the station he told his parents about what he was doing, and wrote two sequences of the colours that he had seen before and after his sleep, respectively.
Peter's parents know that their son likes to fantasize. They give you the list of the flags' colours at the stations that the train passes sequentially on the way from A to B, and ask you to find out if Peter could see those sequences on the way from A to B, or from B to A. Remember, please, that Peter had two periods of wakefulness.
Peter's parents put lowercase Latin letters for colours. The same letter stands for the same colour, different letters — for different colours. | The input data contains three lines. The first line contains a non-empty string, whose length does not exceed 105, the string consists of lowercase Latin letters — the flags' colours at the stations on the way from A to B. On the way from B to A the train passes the same stations, but in reverse order.
The second line contains the sequence, written by Peter during the first period of wakefulness. The third line contains the sequence, written during the second period of wakefulness. Both sequences are non-empty, consist of lowercase Latin letters, and the length of each does not exceed 100 letters. Each of the sequences is written in chronological order. | Output one of the four words without inverted commas:
- «forward» — if Peter could see such sequences only on the way from A to B; - «backward» — if Peter could see such sequences on the way from B to A; - «both» — if Peter could see such sequences both on the way from A to B, and on the way from B to A; - «fantasy» — if Peter could not see such sequences. | [
"atob\na\nb\n",
"aaacaaa\naca\naa\n"
] | [
"forward\n",
"both\n"
] | It is assumed that the train moves all the time, so one flag cannot be seen twice. There are no flags at stations A and B. | 0 | [
{
"input": "atob\na\nb",
"output": "forward"
},
{
"input": "aaacaaa\naca\naa",
"output": "both"
},
{
"input": "aaa\naa\naa",
"output": "fantasy"
},
{
"input": "astalavista\nastla\nlavista",
"output": "fantasy"
},
{
"input": "abacabadabacaba\nabacaba\nabacaba",
... | 1,489,831,638 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 122 | 4,608,000 | a=input()
b=input()
c=input()
a=a.split(b,-1)
l=len(a)
f=0
b=0
for i in range(l):
t=a[i].find(c)
if t!=-1:
if i!=0:f=1
if i!=l-1:b=1
if f:
if b:print('both')
else:print('forward')
else:
if b:print('backward')
else:print('fantasy')
| Title: Train and Peter
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Peter likes to travel by train. He likes it so much that on the train he falls asleep.
Once in summer Peter was going by train from city A to city B, and as usual, was sleeping. Then he woke up, started to look through the window and noticed that every railway station has a flag of a particular colour.
The boy started to memorize the order of the flags' colours that he had seen. But soon he fell asleep again. Unfortunately, he didn't sleep long, he woke up and went on memorizing the colours. Then he fell asleep again, and that time he slept till the end of the journey.
At the station he told his parents about what he was doing, and wrote two sequences of the colours that he had seen before and after his sleep, respectively.
Peter's parents know that their son likes to fantasize. They give you the list of the flags' colours at the stations that the train passes sequentially on the way from A to B, and ask you to find out if Peter could see those sequences on the way from A to B, or from B to A. Remember, please, that Peter had two periods of wakefulness.
Peter's parents put lowercase Latin letters for colours. The same letter stands for the same colour, different letters — for different colours.
Input Specification:
The input data contains three lines. The first line contains a non-empty string, whose length does not exceed 105, the string consists of lowercase Latin letters — the flags' colours at the stations on the way from A to B. On the way from B to A the train passes the same stations, but in reverse order.
The second line contains the sequence, written by Peter during the first period of wakefulness. The third line contains the sequence, written during the second period of wakefulness. Both sequences are non-empty, consist of lowercase Latin letters, and the length of each does not exceed 100 letters. Each of the sequences is written in chronological order.
Output Specification:
Output one of the four words without inverted commas:
- «forward» — if Peter could see such sequences only on the way from A to B; - «backward» — if Peter could see such sequences on the way from B to A; - «both» — if Peter could see such sequences both on the way from A to B, and on the way from B to A; - «fantasy» — if Peter could not see such sequences.
Demo Input:
['atob\na\nb\n', 'aaacaaa\naca\naa\n']
Demo Output:
['forward\n', 'both\n']
Note:
It is assumed that the train moves all the time, so one flag cannot be seen twice. There are no flags at stations A and B. | ```python
a=input()
b=input()
c=input()
a=a.split(b,-1)
l=len(a)
f=0
b=0
for i in range(l):
t=a[i].find(c)
if t!=-1:
if i!=0:f=1
if i!=l-1:b=1
if f:
if b:print('both')
else:print('forward')
else:
if b:print('backward')
else:print('fantasy')
``` | 0 |
884 | B | Japanese Crosswords Strike Back | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it! | The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding. | Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO. | [
"2 4\n1 3\n",
"3 10\n3 3 2\n",
"2 10\n1 3\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "2 4\n1 3",
"output": "NO"
},
{
"input": "3 10\n3 3 2",
"output": "YES"
},
{
"input": "2 10\n1 3",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "YES"
},
{
"input": "1 10\n10",
"output": "YES"
},
{
"input": "1 10000\n10000",
"output":... | 1,584,995,645 | 2,147,483,647 | Python 3 | OK | TESTS | 66 | 139 | 6,963,200 | n,x=map(int,input().split(' '))
val=[int(i) for i in input().split(' ') ]
s=len(val)-1
for i in val:
s+=i
if s==x:
print('YES')
else:
print('NO')
| Title: Japanese Crosswords Strike Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input Specification:
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
Output Specification:
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Demo Input:
['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
n,x=map(int,input().split(' '))
val=[int(i) for i in input().split(' ') ]
s=len(val)-1
for i in val:
s+=i
if s==x:
print('YES')
else:
print('NO')
``` | 3 | |
996 | A | Hit the Lottery | PROGRAMMING | 800 | [
"dp",
"greedy"
] | null | null | Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance? | The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$). | Output the minimum number of bills that Allen could receive. | [
"125\n",
"43\n",
"1000000000\n"
] | [
"3\n",
"5\n",
"10000000\n"
] | In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills. | 500 | [
{
"input": "125",
"output": "3"
},
{
"input": "43",
"output": "5"
},
{
"input": "1000000000",
"output": "10000000"
},
{
"input": "4",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "1",
"output": "1"
},
{
"input": "74",
"o... | 1,698,825,161 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 46 | 0 | x = int(input(""))
bills = 0
bills = bills + x//100
x = x%100
bills = bills + x//20
x = x%20
bills = bills + x//10
x = x%10
bills = bills + x//5
x = x%5
bills = bills + x//1
x = x%1
print(bills)
| Title: Hit the Lottery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input Specification:
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
Output Specification:
Output the minimum number of bills that Allen could receive.
Demo Input:
['125\n', '43\n', '1000000000\n']
Demo Output:
['3\n', '5\n', '10000000\n']
Note:
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills. | ```python
x = int(input(""))
bills = 0
bills = bills + x//100
x = x%100
bills = bills + x//20
x = x%20
bills = bills + x//10
x = x%10
bills = bills + x//5
x = x%5
bills = bills + x//1
x = x%1
print(bills)
``` | 3 | |
61 | B | Hard Work | PROGRAMMING | 1,300 | [
"strings"
] | B. Hard Work | 2 | 256 | After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his *n* students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
- As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin. - As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs. - The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600 - My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes. - Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN - You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer. - I should remind you that none of the strings (initial strings or answers) are empty. - Finally, do these as soon as possible. You have less than 2 hours to complete this. | The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer *n* (0<=≤<=*n*<=≤<=1000), the number of students.
Next *n* lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively. | For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK. | [
"Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful\n",
"Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapu... | [
"ACC\nACC\nACC\nWA\nACC\nACC\nWA\n",
"WA\nACC\nACC\n"
] | none | 1,000 | [
{
"input": "Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful",
"output": "ACC\nACC\nACC\nWA\nACC\nACC\nWA"
},
... | 1,646,915,967 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 43 | 93 | 5,734,400 | import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
strings = []
for i in range(3):
s = S()
letter = ""
for j in range(len(s)):
if 97<=ord(s[j])<=122:
letter += chr(ord(s[j])-32)
elif 90<ord(s[j])<97 or ord(s[j])>122 or ord(s[j])<65:
continue
else:
letter+= s[j]
strings.append(letter)
possible = []
for i in range(3):
for j in range(3):
for k in range(3):
if i!=j and i!=k and j!=k:
m = strings[i] + strings[j] + strings[k]
possible.append(m)
# print(strings)
n = I()
for i in range(n):
t = S()
given = ""
for j in range(len(t)):
if 97<=ord(t[j])<=122:
given += chr(ord(t[j])-32)
elif 90<ord(t[j])<97 or ord(t[j])>122 or ord(t[j])<65:
continue
else:
given+= t[j]
if given in possible:
print("ACC")
else:
print("WA") | Title: Hard Work
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his *n* students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
- As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin. - As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs. - The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600 - My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes. - Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN - You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer. - I should remind you that none of the strings (initial strings or answers) are empty. - Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input Specification:
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer *n* (0<=≤<=*n*<=≤<=1000), the number of students.
Next *n* lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output Specification:
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Demo Input:
['Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful\n', 'Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;\n']
Demo Output:
['ACC\nACC\nACC\nWA\nACC\nACC\nWA\n', 'WA\nACC\nACC\n']
Note:
none | ```python
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
strings = []
for i in range(3):
s = S()
letter = ""
for j in range(len(s)):
if 97<=ord(s[j])<=122:
letter += chr(ord(s[j])-32)
elif 90<ord(s[j])<97 or ord(s[j])>122 or ord(s[j])<65:
continue
else:
letter+= s[j]
strings.append(letter)
possible = []
for i in range(3):
for j in range(3):
for k in range(3):
if i!=j and i!=k and j!=k:
m = strings[i] + strings[j] + strings[k]
possible.append(m)
# print(strings)
n = I()
for i in range(n):
t = S()
given = ""
for j in range(len(t)):
if 97<=ord(t[j])<=122:
given += chr(ord(t[j])-32)
elif 90<ord(t[j])<97 or ord(t[j])>122 or ord(t[j])<65:
continue
else:
given+= t[j]
if given in possible:
print("ACC")
else:
print("WA")
``` | 3.966069 |
427 | A | Police Recruits | PROGRAMMING | 800 | [
"implementation"
] | null | null | The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated. | The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time. | Print a single integer, the number of crimes which will go untreated. | [
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] | [
"2\n",
"1\n",
"8\n"
] | Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated. | 500 | [
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 ... | 1,681,128,954 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 93 | 5,324,800 | _ = input()
events = input().split(" ")
crime_counter = police_counter = 0
for event in events:
event = int(event)
if event > 0 or police_counter > 0:
police_counter += event
else:
crime_counter += 1
print(crime_counter) | Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated. | ```python
_ = input()
events = input().split(" ")
crime_counter = police_counter = 0
for event in events:
event = int(event)
if event > 0 or police_counter > 0:
police_counter += event
else:
crime_counter += 1
print(crime_counter)
``` | 3 | |
995 | B | Suit and Tie | PROGRAMMING | 1,400 | [
"greedy",
"implementation",
"math"
] | null | null | Allen is hosting a formal dinner party. $2n$ people come to the event in $n$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $2n$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this makes the picture more aesthetic.
Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line. | The first line contains a single integer $n$ ($1 \le n \le 100$), the number of pairs of people.
The second line contains $2n$ integers $a_1, a_2, \dots, a_{2n}$. For each $i$ with $1 \le i \le n$, $i$ appears exactly twice. If $a_j = a_k = i$, that means that the $j$-th and $k$-th people in the line form a couple. | Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions. | [
"4\n1 1 2 3 3 2 4 4\n",
"3\n1 1 2 2 3 3\n",
"3\n3 1 2 3 1 2\n"
] | [
"2\n",
"0\n",
"3\n"
] | In the first sample case, we can transform $1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$ in two steps. Note that the sequence $1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$ also works in the same number of steps.
The second sample case already satisfies the constraints; therefore we need $0$ swaps. | 750 | [
{
"input": "4\n1 1 2 3 3 2 4 4",
"output": "2"
},
{
"input": "3\n1 1 2 2 3 3",
"output": "0"
},
{
"input": "3\n3 1 2 3 1 2",
"output": "3"
},
{
"input": "8\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4",
"output": "27"
},
{
"input": "2\n1 2 1 2",
"output": "1"
},
{
... | 1,622,719,459 | 2,147,483,647 | Python 3 | OK | TESTS | 22 | 108 | 0 | n=int(input())
lst = list(map(int, input().strip().split(' ')))
c=0
while(len(lst)!=0):
p=lst[0]
del lst[0]
i=lst.index(p)
c+=i
del lst[i]
print(c)
| Title: Suit and Tie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen is hosting a formal dinner party. $2n$ people come to the event in $n$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $2n$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this makes the picture more aesthetic.
Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 100$), the number of pairs of people.
The second line contains $2n$ integers $a_1, a_2, \dots, a_{2n}$. For each $i$ with $1 \le i \le n$, $i$ appears exactly twice. If $a_j = a_k = i$, that means that the $j$-th and $k$-th people in the line form a couple.
Output Specification:
Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions.
Demo Input:
['4\n1 1 2 3 3 2 4 4\n', '3\n1 1 2 2 3 3\n', '3\n3 1 2 3 1 2\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
In the first sample case, we can transform $1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$ in two steps. Note that the sequence $1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$ also works in the same number of steps.
The second sample case already satisfies the constraints; therefore we need $0$ swaps. | ```python
n=int(input())
lst = list(map(int, input().strip().split(' ')))
c=0
while(len(lst)!=0):
p=lst[0]
del lst[0]
i=lst.index(p)
c+=i
del lst[i]
print(c)
``` | 3 | |
233 | A | Perfect Permutation | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*. | A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size. | If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces. | [
"1\n",
"2\n",
"4\n"
] | [
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] | none | 500 | [
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
... | 1,647,037,945 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 154 | 2,048,000 | import numpy as np
n = int(input())
if n % 2 == 1:
print(-1)
else:
np_arr = np.array(range(1 , n + 1))
evenFlag = np_arr % 2 == 0
np_arr[np_arr % 2 == 1] = np_arr[np_arr % 2 == 1] + 1
np_arr[evenFlag] = np_arr[evenFlag] - 1
res = ""
for ele in np_arr:
res+=str(ele)+" "
print(res) | Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none | ```python
import numpy as np
n = int(input())
if n % 2 == 1:
print(-1)
else:
np_arr = np.array(range(1 , n + 1))
evenFlag = np_arr % 2 == 0
np_arr[np_arr % 2 == 1] = np_arr[np_arr % 2 == 1] + 1
np_arr[evenFlag] = np_arr[evenFlag] - 1
res = ""
for ele in np_arr:
res+=str(ele)+" "
print(res)
``` | -1 | |
712 | B | Memory and Trident | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion:
- An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down.
But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string. | The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given. | If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1. | [
"RRU\n",
"UDUR\n",
"RUUR\n"
] | [
"-1\n",
"1\n",
"2\n"
] | In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.
In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin. | 1,000 | [
{
"input": "RRU",
"output": "-1"
},
{
"input": "UDUR",
"output": "1"
},
{
"input": "RUUR",
"output": "2"
},
{
"input": "DDDD",
"output": "2"
},
{
"input": "RRRR",
"output": "2"
},
{
"input": "RRRUUD",
"output": "2"
},
{
"input": "UDURLRDURL... | 1,474,129,156 | 2,147,483,647 | Python 3 | OK | TESTS | 75 | 77 | 307,200 | def main():
s = input()
if len(s) % 2 == 1:
print('-1')
return
cnt = {'L':0,'R':0,'U':0,'D':0}
for c in s:
cnt[c] += 1
reduce1 = abs(cnt['L'] - cnt['R'])
reduce2 = abs(cnt['U'] - cnt['D'])
print((reduce1 + reduce2) // 2)
if __name__ == '__main__':
main() | Title: Memory and Trident
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion:
- An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down.
But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.
Input Specification:
The first and only line contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) — the instructions Memory is given.
Output Specification:
If there is a string satisfying the conditions, output a single integer — the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.
Demo Input:
['RRU\n', 'UDUR\n', 'RUUR\n']
Demo Output:
['-1\n', '1\n', '2\n']
Note:
In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.
In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin. | ```python
def main():
s = input()
if len(s) % 2 == 1:
print('-1')
return
cnt = {'L':0,'R':0,'U':0,'D':0}
for c in s:
cnt[c] += 1
reduce1 = abs(cnt['L'] - cnt['R'])
reduce2 = abs(cnt['U'] - cnt['D'])
print((reduce1 + reduce2) // 2)
if __name__ == '__main__':
main()
``` | 3 | |
673 | A | Bear and Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.
Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.
You know that there will be *n* interesting minutes *t*1,<=*t*2,<=...,<=*t**n*. Your task is to calculate for how many minutes Limak will watch the game. | The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=90) — the number of interesting minutes.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=... *t**n*<=≤<=90), given in the increasing order. | Print the number of minutes Limak will watch the game. | [
"3\n7 20 88\n",
"9\n16 20 30 40 50 60 70 80 90\n",
"9\n15 20 30 40 50 60 70 80 90\n"
] | [
"35\n",
"15\n",
"90\n"
] | In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.
In the second sample, the first 15 minutes are boring.
In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game. | 500 | [
{
"input": "3\n7 20 88",
"output": "35"
},
{
"input": "9\n16 20 30 40 50 60 70 80 90",
"output": "15"
},
{
"input": "9\n15 20 30 40 50 60 70 80 90",
"output": "90"
},
{
"input": "30\n6 11 12 15 22 24 30 31 32 33 34 35 40 42 44 45 47 50 53 54 57 58 63 67 75 77 79 81 83 88",
... | 1,578,412,864 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 108 | 0 |
n = int(input())
b= list(map(int, input().split()))
a = [0]
status = True
a.extend(b)
for i in range(len(a)-1):
if a[i+1]-a[i] >15:
print(a[i]+15)
status = False
break
if status == True:
print(90)
| Title: Bear and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.
Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.
You know that there will be *n* interesting minutes *t*1,<=*t*2,<=...,<=*t**n*. Your task is to calculate for how many minutes Limak will watch the game.
Input Specification:
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=90) — the number of interesting minutes.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=... *t**n*<=≤<=90), given in the increasing order.
Output Specification:
Print the number of minutes Limak will watch the game.
Demo Input:
['3\n7 20 88\n', '9\n16 20 30 40 50 60 70 80 90\n', '9\n15 20 30 40 50 60 70 80 90\n']
Demo Output:
['35\n', '15\n', '90\n']
Note:
In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.
In the second sample, the first 15 minutes are boring.
In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game. | ```python
n = int(input())
b= list(map(int, input().split()))
a = [0]
status = True
a.extend(b)
for i in range(len(a)-1):
if a[i+1]-a[i] >15:
print(a[i]+15)
status = False
break
if status == True:
print(90)
``` | 0 | |
43 | A | Football | PROGRAMMING | 1,000 | [
"strings"
] | A. Football | 2 | 256 | One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. | Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. | [
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] | [
"ABC\n",
"A\n"
] | none | 500 | [
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
... | 1,645,470,923 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 154 | 0 | n = int(input())
ctr = 0
s = ''
mx = 0
while n!=0:
n-=1
k = input()
x = k
if x in k:
ctr += 1
if ctr > mx:
mx = ctr
s = x
else:
ctr = 0
print(s) | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none | ```python
n = int(input())
ctr = 0
s = ''
mx = 0
while n!=0:
n-=1
k = input()
x = k
if x in k:
ctr += 1
if ctr > mx:
mx = ctr
s = x
else:
ctr = 0
print(s)
``` | 0 |
0 | none | none | none | 0 | [
"none"
] | null | null | Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to *n* computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to *n*. So the *i*-th hacked computer is located at the point *x**i*. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of *F*(*a*) for all *a*, where *a* is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote *A* the set of all integers from 1 to *n*. Noora asks the hacker to find value of the expression . Here *F*(*a*) is calculated as the maximum among the distances between all pairs of computers from the set *a*. Formally, . Since the required sum can be quite large Noora asks to find it modulo 109<=+<=7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) denoting the number of hacked computers.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) denoting the coordinates of hacked computers. It is guaranteed that all *x**i* are distinct. | Print a single integer — the required sum modulo 109<=+<=7. | [
"2\n4 7\n",
"3\n4 3 1\n"
] | [
"3\n",
"9\n"
] | There are three non-empty subsets in the first sample test:<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/02b2d12556dad85f1c6c6912786eb87d4be2ea17.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/22f6a537962c86b3e28ddb8aaca28a7cdd219a8c.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7d0f73b3e94e13cb797f39e93d9da74835c5a02d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f368c407c8e85e2b5fedfffaff39d471d765f026.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bb8f2118a3ac352db393b1f067b28e398ce7f816.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/049032074c04b16bc0cc153f95471c40b222072b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc93c7f5b3d122314c9c5a707fae556a8f72a574.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9. | 0 | [
{
"input": "2\n4 7",
"output": "3"
},
{
"input": "3\n4 3 1",
"output": "9"
},
{
"input": "20\n8 11 13 19 21 34 36 44 57 58 61 63 76 78 79 81 85 86 90 95",
"output": "83396599"
},
{
"input": "20\n1 8 9 12 15 17 18 24 30 33 36 41 53 54 59 62 64 66 72 73",
"output": "6805914... | 1,495,310,350 | 6,850 | PyPy 3 | RUNTIME_ERROR | TESTS | 53 | 685 | 50,073,600 | from sys import stdin, stdout
n = int(stdin.readline())
values = sorted(list(map(int, stdin.readline().split())))
MOD = 10 ** 9 + 7
dp = [0, values[1] - values[0]]
cnt = [values[0], values[0] * 2 + values[1]]
pw = [1]
for i in range(n):
pw.append((pw[-1] * 2) % MOD)
for i in range(2, n):
dp.append((dp[-1] + values[i] * (pw[i] - 1) - cnt[-1]) % MOD)
cnt.append(cnt[-1] * 2 + values[i])
cnt[-1] %= MOD
stdout.write(str(dp[-1] % MOD))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to *n* computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to *n*. So the *i*-th hacked computer is located at the point *x**i*. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of *F*(*a*) for all *a*, where *a* is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote *A* the set of all integers from 1 to *n*. Noora asks the hacker to find value of the expression . Here *F*(*a*) is calculated as the maximum among the distances between all pairs of computers from the set *a*. Formally, . Since the required sum can be quite large Noora asks to find it modulo 109<=+<=7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) denoting the number of hacked computers.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) denoting the coordinates of hacked computers. It is guaranteed that all *x**i* are distinct.
Output Specification:
Print a single integer — the required sum modulo 109<=+<=7.
Demo Input:
['2\n4 7\n', '3\n4 3 1\n']
Demo Output:
['3\n', '9\n']
Note:
There are three non-empty subsets in the first sample test:<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/02b2d12556dad85f1c6c6912786eb87d4be2ea17.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/22f6a537962c86b3e28ddb8aaca28a7cdd219a8c.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7d0f73b3e94e13cb797f39e93d9da74835c5a02d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f368c407c8e85e2b5fedfffaff39d471d765f026.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bb8f2118a3ac352db393b1f067b28e398ce7f816.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/049032074c04b16bc0cc153f95471c40b222072b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc93c7f5b3d122314c9c5a707fae556a8f72a574.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9. | ```python
from sys import stdin, stdout
n = int(stdin.readline())
values = sorted(list(map(int, stdin.readline().split())))
MOD = 10 ** 9 + 7
dp = [0, values[1] - values[0]]
cnt = [values[0], values[0] * 2 + values[1]]
pw = [1]
for i in range(n):
pw.append((pw[-1] * 2) % MOD)
for i in range(2, n):
dp.append((dp[-1] + values[i] * (pw[i] - 1) - cnt[-1]) % MOD)
cnt.append(cnt[-1] * 2 + values[i])
cnt[-1] %= MOD
stdout.write(str(dp[-1] % MOD))
``` | -1 | |
814 | A | An abandoned sentiment from past | PROGRAMMING | 900 | [
"constructive algorithms",
"greedy",
"implementation",
"sortings"
] | null | null | A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence *a* has a length of *n*. Lost elements in it are denoted by zeros. Kaiki provides another sequence *b*, whose length *k* equals the number of lost elements in *a* (i.e. the number of zeros). Hitagi is to replace each zero in *a* with an element from *b* so that each element in *b* should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in *a* and *b* more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in *a* with an integer from *b* so that each integer from *b* is used exactly once, and the resulting sequence is not increasing. | The first line of input contains two space-separated positive integers *n* (2<=≤<=*n*<=≤<=100) and *k* (1<=≤<=*k*<=≤<=*n*) — the lengths of sequence *a* and *b* respectively.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=200) — Hitagi's broken sequence with exactly *k* zero elements.
The third line contains *k* space-separated integers *b*1,<=*b*2,<=...,<=*b**k* (1<=≤<=*b**i*<=≤<=200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in *a* and *b* more than once in total. | Output "Yes" if it's possible to replace zeros in *a* with elements in *b* and make the resulting sequence not increasing, and "No" otherwise. | [
"4 2\n11 0 0 14\n5 4\n",
"6 1\n2 3 0 8 9 10\n5\n",
"4 1\n8 94 0 4\n89\n",
"7 7\n0 0 0 0 0 0 0\n1 2 3 4 5 6 7\n"
] | [
"Yes\n",
"No\n",
"Yes\n",
"Yes\n"
] | In the first sample:
- Sequence *a* is 11, 0, 0, 14. - Two of the elements are lost, and the candidates in *b* are 5 and 4. - There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | 500 | [
{
"input": "4 2\n11 0 0 14\n5 4",
"output": "Yes"
},
{
"input": "6 1\n2 3 0 8 9 10\n5",
"output": "No"
},
{
"input": "4 1\n8 94 0 4\n89",
"output": "Yes"
},
{
"input": "7 7\n0 0 0 0 0 0 0\n1 2 3 4 5 6 7",
"output": "Yes"
},
{
"input": "40 1\n23 26 27 28 31 35 38 4... | 1,581,994,070 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 5 | 156 | 0 | # import sys
# sys.stdin=open('input.txt', 'r')
# sys.stdout=open('output.txt', 'w')
a, b = [int(k) for k in input().split()]
c=[int(k) for k in input().split()]
d=[int(k) for k in input().split()]
if b>1:
print('Yes')
elif b==1:
index=c.index(0)
if index==0:
if d[0]>c[1]:
print('Yes')
else: print('No')
elif index==len(c)-1:
if d[0]<c[-1]:
print('Yes')
else:
print('No')
else:
if c[index-1]>d[0]>c[index+1]:
print('Yes')
else: print('No')
| Title: An abandoned sentiment from past
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence *a* has a length of *n*. Lost elements in it are denoted by zeros. Kaiki provides another sequence *b*, whose length *k* equals the number of lost elements in *a* (i.e. the number of zeros). Hitagi is to replace each zero in *a* with an element from *b* so that each element in *b* should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in *a* and *b* more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in *a* with an integer from *b* so that each integer from *b* is used exactly once, and the resulting sequence is not increasing.
Input Specification:
The first line of input contains two space-separated positive integers *n* (2<=≤<=*n*<=≤<=100) and *k* (1<=≤<=*k*<=≤<=*n*) — the lengths of sequence *a* and *b* respectively.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=200) — Hitagi's broken sequence with exactly *k* zero elements.
The third line contains *k* space-separated integers *b*1,<=*b*2,<=...,<=*b**k* (1<=≤<=*b**i*<=≤<=200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in *a* and *b* more than once in total.
Output Specification:
Output "Yes" if it's possible to replace zeros in *a* with elements in *b* and make the resulting sequence not increasing, and "No" otherwise.
Demo Input:
['4 2\n11 0 0 14\n5 4\n', '6 1\n2 3 0 8 9 10\n5\n', '4 1\n8 94 0 4\n89\n', '7 7\n0 0 0 0 0 0 0\n1 2 3 4 5 6 7\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n', 'Yes\n']
Note:
In the first sample:
- Sequence *a* is 11, 0, 0, 14. - Two of the elements are lost, and the candidates in *b* are 5 and 4. - There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid. | ```python
# import sys
# sys.stdin=open('input.txt', 'r')
# sys.stdout=open('output.txt', 'w')
a, b = [int(k) for k in input().split()]
c=[int(k) for k in input().split()]
d=[int(k) for k in input().split()]
if b>1:
print('Yes')
elif b==1:
index=c.index(0)
if index==0:
if d[0]>c[1]:
print('Yes')
else: print('No')
elif index==len(c)-1:
if d[0]<c[-1]:
print('Yes')
else:
print('No')
else:
if c[index-1]>d[0]>c[index+1]:
print('Yes')
else: print('No')
``` | 0 | |
116 | A | Tram | PROGRAMMING | 800 | [
"implementation"
] | null | null | Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram. | The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0. | Print a single integer denoting the minimum possible capacity of the tram (0 is allowed). | [
"4\n0 3\n2 5\n4 2\n4 0\n"
] | [
"6\n"
] | For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | 500 | [
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1... | 1,694,351,536 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 71 | 186 | 1,843,200 | total_stop = int(input())
max = 0
current = 0
for i in range(total_stop):
a,b = map(int,input().split())
if i == 0:
max = b
current = b
else:
current = (current - a) + b
if current > max:
max = current
print(max)
| Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | ```python
total_stop = int(input())
max = 0
current = 0
for i in range(total_stop):
a,b = map(int,input().split())
if i == 0:
max = b
current = b
else:
current = (current - a) + b
if current > max:
max = current
print(max)
``` | 3 | |
559 | B | Equivalent Strings | PROGRAMMING | 1,700 | [
"divide and conquer",
"hashing",
"sortings",
"strings"
] | null | null | Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn! | The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length. | Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise. | [
"aaba\nabaa\n",
"aabb\nabab\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | 1,000 | [
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
... | 1,699,960,116 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 104 | 967 | 113,664,000 | memo = {}
def solve(a, b):
if (a,b) in memo:
return memo[a,b]
if len(a)%2 == 1:
return a == b
half = len(a)//2
a1, a2 = a[:half], a[half:]
b1, b2 = b[:half], b[half:]
memo[a,b] = (solve(a1, b1) and solve(a2, b2)) or (solve(a1, b2) and solve(a2, b1))
return memo[a,b]
s1 = input()
s2 = input()
ans = solve(s1, s2)
print("YES" if ans else "NO") | Title: Equivalent Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | ```python
memo = {}
def solve(a, b):
if (a,b) in memo:
return memo[a,b]
if len(a)%2 == 1:
return a == b
half = len(a)//2
a1, a2 = a[:half], a[half:]
b1, b2 = b[:half], b[half:]
memo[a,b] = (solve(a1, b1) and solve(a2, b2)) or (solve(a1, b2) and solve(a2, b1))
return memo[a,b]
s1 = input()
s2 = input()
ans = solve(s1, s2)
print("YES" if ans else "NO")
``` | 3 | |
552 | B | Vanya and Books | PROGRAMMING | 1,200 | [
"implementation",
"math"
] | null | null | Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books. | The first line contains integer *n* (1<=≤<=*n*<=≤<=109) — the number of books in the library. | Print the number of digits needed to number all the books. | [
"13\n",
"4\n"
] | [
"17\n",
"4\n"
] | Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | 1,000 | [
{
"input": "13",
"output": "17"
},
{
"input": "4",
"output": "4"
},
{
"input": "100",
"output": "192"
},
{
"input": "99",
"output": "189"
},
{
"input": "1000000000",
"output": "8888888899"
},
{
"input": "1000000",
"output": "5888896"
},
{
"... | 1,648,034,039 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | # coding=utf-8
x=input()
for i in x:
if ord(i)%2==0:
print(0,end='')
else:
print(1,end='')
| Title: Vanya and Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109) — the number of books in the library.
Output Specification:
Print the number of digits needed to number all the books.
Demo Input:
['13\n', '4\n']
Demo Output:
['17\n', '4\n']
Note:
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | ```python
# coding=utf-8
x=input()
for i in x:
if ord(i)%2==0:
print(0,end='')
else:
print(1,end='')
``` | 0 | |
828 | A | Restaurant Tables | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | In a small restaurant there are *a* tables for one person and *b* tables for two persons.
It it known that *n* groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.
If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.
You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to. | The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people. | Print the total number of people the restaurant denies service to. | [
"4 1 2\n1 2 1 1\n",
"4 1 1\n1 1 2 1\n"
] | [
"0\n",
"2\n"
] | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.
In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients. | 500 | [
{
"input": "4 1 2\n1 2 1 1",
"output": "0"
},
{
"input": "4 1 1\n1 1 2 1",
"output": "2"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 1 2\n2 2",
"output": "0"
},
{
"input": "5 1 3\n1 2 2 2 1",
"output": "1"
},
{
"input": "7 6 1\n1 1 1 1 1 1... | 1,553,427,020 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 93 | 0 | n, a, b=map(int,input().split())
l=list(map(int,input().split()))
x=l.count(2)
z=l.count(1)
y=b-x
count=0
if y<=0:
count+=abs(2*y)
else:
a+=(2*y)
m=a-z
if m<=0:
count+=abs(2*m)
print(count)
else:
print(0)
| Title: Restaurant Tables
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small restaurant there are *a* tables for one person and *b* tables for two persons.
It it known that *n* groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.
If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.
You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to.
Input Specification:
The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people.
Output Specification:
Print the total number of people the restaurant denies service to.
Demo Input:
['4 1 2\n1 2 1 1\n', '4 1 1\n1 1 2 1\n']
Demo Output:
['0\n', '2\n']
Note:
In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.
In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients. | ```python
n, a, b=map(int,input().split())
l=list(map(int,input().split()))
x=l.count(2)
z=l.count(1)
y=b-x
count=0
if y<=0:
count+=abs(2*y)
else:
a+=(2*y)
m=a-z
if m<=0:
count+=abs(2*m)
print(count)
else:
print(0)
``` | 0 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,622,833,485 | 2,147,483,647 | PyPy 3 | OK | TESTS | 35 | 340 | 0 | if __name__ == '__main__':
s = input()
l = list(map(int, s.split()))
print(l[0] * l[1] // 2)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
if __name__ == '__main__':
s = input()
l = list(map(int, s.split()))
print(l[0] * l[1] // 2)
``` | 3.915 |
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 ... | 1,611,497,526 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 358 | 307,200 | def find_ans(arr,char_arr,ind):
count_of_R = 0
while ind<len(arr):
if arr[ind]==0:
char_arr[ind] = 'R'
count_of_R+=1
elif arr[ind]==2:
if char_arr[ind-1]!='G':
char_arr[ind] = 'G'
else:
char_arr[ind] = 'R'
count_of_R+=1
elif arr[ind]==1:
if char_arr[ind-1]!='C':
char_arr[ind] = 'C'
else:
char_arr[ind] = 'R'
count_of_R+=1
else:
if char_arr[ind-1]=='R':
char_arr[ind] = 'G'
val1 = find_ans(arr,char_arr,ind+1)
char_arr[ind] = 'C'
val2 = find_ans(arr,char_arr,ind+1)
if val1<=val2:
return count_of_R + val1
else:
return count_of_R + val2
else:
if char_arr[ind-1]=='G':
char_arr[ind] = 'C'
else:
char_arr[ind] = 'G'
ind+=1
return count_of_R
n = int(input())
l1 = [int(x) for x in input().split()]
l2 = [0]
for i in l1:
l2.append(i)
# 0 -> 0 0
# 1 -> 0 1
# 2 -> 1 0
# 3 -> 1 1
out = ['' for i in range(0,n+1)]
ans = find_ans(l2,out,1)
print(ans) | Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
def find_ans(arr,char_arr,ind):
count_of_R = 0
while ind<len(arr):
if arr[ind]==0:
char_arr[ind] = 'R'
count_of_R+=1
elif arr[ind]==2:
if char_arr[ind-1]!='G':
char_arr[ind] = 'G'
else:
char_arr[ind] = 'R'
count_of_R+=1
elif arr[ind]==1:
if char_arr[ind-1]!='C':
char_arr[ind] = 'C'
else:
char_arr[ind] = 'R'
count_of_R+=1
else:
if char_arr[ind-1]=='R':
char_arr[ind] = 'G'
val1 = find_ans(arr,char_arr,ind+1)
char_arr[ind] = 'C'
val2 = find_ans(arr,char_arr,ind+1)
if val1<=val2:
return count_of_R + val1
else:
return count_of_R + val2
else:
if char_arr[ind-1]=='G':
char_arr[ind] = 'C'
else:
char_arr[ind] = 'G'
ind+=1
return count_of_R
n = int(input())
l1 = [int(x) for x in input().split()]
l2 = [0]
for i in l1:
l2.append(i)
# 0 -> 0 0
# 1 -> 0 1
# 2 -> 1 0
# 3 -> 1 1
out = ['' for i in range(0,n+1)]
ans = find_ans(l2,out,1)
print(ans)
``` | 0 | |
895 | B | XK Segments | PROGRAMMING | 1,700 | [
"binary search",
"math",
"sortings",
"two pointers"
] | null | null | While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array *a* and integer *x*. He should find the number of different ordered pairs of indexes (*i*,<=*j*) such that *a**i*<=≤<=*a**j* and there are exactly *k* integers *y* such that *a**i*<=≤<=*y*<=≤<=*a**j* and *y* is divisible by *x*.
In this problem it is meant that pair (*i*,<=*j*) is equal to (*j*,<=*i*) only if *i* is equal to *j*. For example pair (1,<=2) is not the same as (2,<=1). | The first line contains 3 integers *n*,<=*x*,<=*k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*x*<=≤<=109,<=0<=≤<=*k*<=≤<=109), where *n* is the size of the array *a* and *x* and *k* are numbers from the statement.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. | Print one integer — the answer to the problem. | [
"4 2 1\n1 3 5 7\n",
"4 2 0\n5 3 1 7\n",
"5 3 1\n3 3 3 3 3\n"
] | [
"3\n",
"4\n",
"25\n"
] | In first sample there are only three suitable pairs of indexes — (1, 2), (2, 3), (3, 4).
In second sample there are four suitable pairs of indexes(1, 1), (2, 2), (3, 3), (4, 4).
In third sample every pair (*i*, *j*) is suitable, so the answer is 5 * 5 = 25. | 1,000 | [
{
"input": "4 2 1\n1 3 5 7",
"output": "3"
},
{
"input": "4 2 0\n5 3 1 7",
"output": "4"
},
{
"input": "5 3 1\n3 3 3 3 3",
"output": "25"
},
{
"input": "5 3 4\n24 13 1 24 24",
"output": "4"
},
{
"input": "4 2 2\n1 3 5 7",
"output": "2"
},
{
"input": "5... | 1,512,220,345 | 2,147,483,647 | PyPy 3 | OK | TESTS | 69 | 389 | 36,761,600 | from bisect import bisect_left as b
f = lambda: map(int, input().split())
n, x, k = f()
s, t = 0, sorted(f())
p = [(q, ((q - 1) // x + k) * x) for q in t]
for q, d in p: s += b(t, d + x) - b(t, max(q, d))
print(s) | Title: XK Segments
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array *a* and integer *x*. He should find the number of different ordered pairs of indexes (*i*,<=*j*) such that *a**i*<=≤<=*a**j* and there are exactly *k* integers *y* such that *a**i*<=≤<=*y*<=≤<=*a**j* and *y* is divisible by *x*.
In this problem it is meant that pair (*i*,<=*j*) is equal to (*j*,<=*i*) only if *i* is equal to *j*. For example pair (1,<=2) is not the same as (2,<=1).
Input Specification:
The first line contains 3 integers *n*,<=*x*,<=*k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*x*<=≤<=109,<=0<=≤<=*k*<=≤<=109), where *n* is the size of the array *a* and *x* and *k* are numbers from the statement.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Output Specification:
Print one integer — the answer to the problem.
Demo Input:
['4 2 1\n1 3 5 7\n', '4 2 0\n5 3 1 7\n', '5 3 1\n3 3 3 3 3\n']
Demo Output:
['3\n', '4\n', '25\n']
Note:
In first sample there are only three suitable pairs of indexes — (1, 2), (2, 3), (3, 4).
In second sample there are four suitable pairs of indexes(1, 1), (2, 2), (3, 3), (4, 4).
In third sample every pair (*i*, *j*) is suitable, so the answer is 5 * 5 = 25. | ```python
from bisect import bisect_left as b
f = lambda: map(int, input().split())
n, x, k = f()
s, t = 0, sorted(f())
p = [(q, ((q - 1) // x + k) * x) for q in t]
for q, d in p: s += b(t, d + x) - b(t, max(q, d))
print(s)
``` | 3 | |
579 | A | Raising Bacteria | PROGRAMMING | 1,000 | [
"bitmasks"
] | null | null | You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days? | The only line containing one integer *x* (1<=≤<=*x*<=≤<=109). | The only line containing one integer: the answer. | [
"5\n",
"8\n"
] | [
"2\n",
"1\n"
] | For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | 250 | [
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697... | 1,688,390,445 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 46 | 0 | w2=int(input())
w=bin(w2)
w1=w[2:]
c=0
for i in w1:
if int(i)%2!=0:
c+=1
print(c) | Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | ```python
w2=int(input())
w=bin(w2)
w1=w[2:]
c=0
for i in w1:
if int(i)%2!=0:
c+=1
print(c)
``` | 3 | |
260 | A | Adding Digits | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times. | The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105). | In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them. | [
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] | [
"524848\n",
"121\n",
"-1\n"
] | none | 500 | [
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479"... | 1,608,641,756 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 483 | 8,704,000 | a,b,n=map(int,input().split())
N=n
c=0
if(a%b==0):
print(int(str(a)+'0'*n))
else:
for i in range(n):
for j in range(0, 10):
s = str(a) + str(j)
num = int(s)
if (num % b == 0):
ans = num
break
else:
c += 1
if (c == n * 10):
print(-1)
else:
if(len(str(num))==N+len(str(a))):
print(num)
else:
l=len(str(a))+n-len(str(num))
number=int(str(num)+'0'*l)
print(number)
| Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none | ```python
a,b,n=map(int,input().split())
N=n
c=0
if(a%b==0):
print(int(str(a)+'0'*n))
else:
for i in range(n):
for j in range(0, 10):
s = str(a) + str(j)
num = int(s)
if (num % b == 0):
ans = num
break
else:
c += 1
if (c == n * 10):
print(-1)
else:
if(len(str(num))==N+len(str(a))):
print(num)
else:
l=len(str(a))+n-len(str(num))
number=int(str(num)+'0'*l)
print(number)
``` | 3 | |
600 | E | Lomsat gelral | PROGRAMMING | 2,300 | [
"data structures",
"dfs and similar",
"dsu",
"trees"
] | null | null | You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.
Let's call colour *c* dominating in the subtree of vertex *v* if there are no other colours that appear in the subtree of vertex *v* more times than colour *c*. So it's possible that two or more colours will be dominating in the subtree of some vertex.
The subtree of vertex *v* is the vertex *v* and all other vertices that contains vertex *v* in each path to the root.
For each vertex *v* find the sum of all dominating colours in the subtree of vertex *v*. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of vertices in the tree.
The second line contains *n* integers *c**i* (1<=≤<=*c**i*<=≤<=*n*), *c**i* — the colour of the *i*-th vertex.
Each of the next *n*<=-<=1 lines contains two integers *x**j*,<=*y**j* (1<=≤<=*x**j*,<=*y**j*<=≤<=*n*) — the edge of the tree. The first vertex is the root of the tree. | Print *n* integers — the sums of dominating colours for each vertex. | [
"4\n1 2 3 4\n1 2\n2 3\n2 4\n",
"15\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\n1 2\n1 3\n1 4\n1 14\n1 15\n2 5\n2 6\n2 7\n3 8\n3 9\n3 10\n4 11\n4 12\n4 13\n"
] | [
"10 9 3 4\n",
"6 5 4 3 2 3 3 1 1 3 2 2 1 2 3\n"
] | none | 0 | [
{
"input": "4\n1 2 3 4\n1 2\n2 3\n2 4",
"output": "10 9 3 4"
},
{
"input": "15\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\n1 2\n1 3\n1 4\n1 14\n1 15\n2 5\n2 6\n2 7\n3 8\n3 9\n3 10\n4 11\n4 12\n4 13",
"output": "6 5 4 3 2 3 3 1 1 3 2 2 1 2 3"
}
] | 1,694,570,428 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 61 | 2,252,800 | from collections import defaultdict
vertex_num=int(input())
colors=list(map(int,input().split()))
treeMap=defaultdict(list)
def dfs(node,colorCount:dict):
colorCount[colors[node-1]]=colorCount.get(colors[node-1],0)+1
sum=0
maxColor=0
for child in treeMap[node]:
dfs(child,colorCount)
for color in colorCount:
if colorCount[color]>colorCount.get(maxColor,0):
sum=color
maxColor=color
elif colorCount[color]==colorCount.get(maxColor,0):
sum+=color
return sum
for x in range(vertex_num-1):
node,child=list(map(int,input().split()))
treeMap[node].append(child)
res=[]
for i in range(vertex_num):
c={}
res.append(dfs(i+1,c))
print(" ".join(list(map(str,res))))
| Title: Lomsat gelral
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.
Let's call colour *c* dominating in the subtree of vertex *v* if there are no other colours that appear in the subtree of vertex *v* more times than colour *c*. So it's possible that two or more colours will be dominating in the subtree of some vertex.
The subtree of vertex *v* is the vertex *v* and all other vertices that contains vertex *v* in each path to the root.
For each vertex *v* find the sum of all dominating colours in the subtree of vertex *v*.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of vertices in the tree.
The second line contains *n* integers *c**i* (1<=≤<=*c**i*<=≤<=*n*), *c**i* — the colour of the *i*-th vertex.
Each of the next *n*<=-<=1 lines contains two integers *x**j*,<=*y**j* (1<=≤<=*x**j*,<=*y**j*<=≤<=*n*) — the edge of the tree. The first vertex is the root of the tree.
Output Specification:
Print *n* integers — the sums of dominating colours for each vertex.
Demo Input:
['4\n1 2 3 4\n1 2\n2 3\n2 4\n', '15\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\n1 2\n1 3\n1 4\n1 14\n1 15\n2 5\n2 6\n2 7\n3 8\n3 9\n3 10\n4 11\n4 12\n4 13\n']
Demo Output:
['10 9 3 4\n', '6 5 4 3 2 3 3 1 1 3 2 2 1 2 3\n']
Note:
none | ```python
from collections import defaultdict
vertex_num=int(input())
colors=list(map(int,input().split()))
treeMap=defaultdict(list)
def dfs(node,colorCount:dict):
colorCount[colors[node-1]]=colorCount.get(colors[node-1],0)+1
sum=0
maxColor=0
for child in treeMap[node]:
dfs(child,colorCount)
for color in colorCount:
if colorCount[color]>colorCount.get(maxColor,0):
sum=color
maxColor=color
elif colorCount[color]==colorCount.get(maxColor,0):
sum+=color
return sum
for x in range(vertex_num-1):
node,child=list(map(int,input().split()))
treeMap[node].append(child)
res=[]
for i in range(vertex_num):
c={}
res.append(dfs(i+1,c))
print(" ".join(list(map(str,res))))
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | In the spirit of the holidays, Saitama has given Genos two grid paths of length *n* (a weird gift even by Saitama's standards). A grid path is an ordered sequence of neighbouring squares in an infinite grid. Two squares are neighbouring if they share a side.
One example of a grid path is (0,<=0)<=→<=(0,<=1)<=→<=(0,<=2)<=→<=(1,<=2)<=→<=(1,<=1)<=→<=(0,<=1)<=→<=(<=-<=1,<=1). Note that squares in this sequence might be repeated, i.e. path has self intersections.
Movement within a grid path is restricted to adjacent squares within the sequence. That is, from the *i*-th square, one can only move to the (*i*<=-<=1)-th or (*i*<=+<=1)-th squares of this path. Note that there is only a single valid move from the first and last squares of a grid path. Also note, that even if there is some *j*-th square of the path that coincides with the *i*-th square, only moves to (*i*<=-<=1)-th and (*i*<=+<=1)-th squares are available. For example, from the second square in the above sequence, one can only move to either the first or third squares.
To ensure that movement is not ambiguous, the two grid paths will not have an alternating sequence of three squares. For example, a contiguous subsequence (0,<=0)<=→<=(0,<=1)<=→<=(0,<=0) cannot occur in a valid grid path.
One marble is placed on the first square of each grid path. Genos wants to get both marbles to the last square of each grid path. However, there is a catch. Whenever he moves one marble, the other marble will copy its movement if possible. For instance, if one marble moves east, then the other marble will try and move east as well. By try, we mean if moving east is a valid move, then the marble will move east.
Moving north increases the second coordinate by 1, while moving south decreases it by 1. Similarly, moving east increases first coordinate by 1, while moving west decreases it.
Given these two valid grid paths, Genos wants to know if it is possible to move both marbles to the ends of their respective paths. That is, if it is possible to move the marbles such that both marbles rest on the last square of their respective paths. | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1<=000<=000) — the length of the paths.
The second line of the input contains a string consisting of *n*<=-<=1 characters (each of which is either 'N', 'E', 'S', or 'W') — the first grid path. The characters can be thought of as the sequence of moves needed to traverse the grid path. For example, the example path in the problem statement can be expressed by the string "NNESWW".
The third line of the input contains a string of *n*<=-<=1 characters (each of which is either 'N', 'E', 'S', or 'W') — the second grid path. | Print "YES" (without quotes) if it is possible for both marbles to be at the end position at the same time. Print "NO" (without quotes) otherwise. In both cases, the answer is case-insensitive. | [
"7\nNNESWW\nSWSWSW\n",
"3\nNN\nSS\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, the first grid path is the one described in the statement. Moreover, the following sequence of moves will get both marbles to the end: NNESWWSWSW.
In the second sample, no sequence of moves can get both marbles to the end. | 0 | [
{
"input": "7\nNNESWW\nSWSWSW",
"output": "YES"
},
{
"input": "3\nNN\nSS",
"output": "NO"
},
{
"input": "3\nES\nNW",
"output": "NO"
},
{
"input": "5\nWSSE\nWNNE",
"output": "NO"
},
{
"input": "2\nE\nE",
"output": "YES"
},
{
"input": "2\nW\nS",
"out... | 1,667,289,822 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 |
import math
for i in range(int(input())):
n,k = map(int,input().split())
print(math.comb(n-1,k-1))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the spirit of the holidays, Saitama has given Genos two grid paths of length *n* (a weird gift even by Saitama's standards). A grid path is an ordered sequence of neighbouring squares in an infinite grid. Two squares are neighbouring if they share a side.
One example of a grid path is (0,<=0)<=→<=(0,<=1)<=→<=(0,<=2)<=→<=(1,<=2)<=→<=(1,<=1)<=→<=(0,<=1)<=→<=(<=-<=1,<=1). Note that squares in this sequence might be repeated, i.e. path has self intersections.
Movement within a grid path is restricted to adjacent squares within the sequence. That is, from the *i*-th square, one can only move to the (*i*<=-<=1)-th or (*i*<=+<=1)-th squares of this path. Note that there is only a single valid move from the first and last squares of a grid path. Also note, that even if there is some *j*-th square of the path that coincides with the *i*-th square, only moves to (*i*<=-<=1)-th and (*i*<=+<=1)-th squares are available. For example, from the second square in the above sequence, one can only move to either the first or third squares.
To ensure that movement is not ambiguous, the two grid paths will not have an alternating sequence of three squares. For example, a contiguous subsequence (0,<=0)<=→<=(0,<=1)<=→<=(0,<=0) cannot occur in a valid grid path.
One marble is placed on the first square of each grid path. Genos wants to get both marbles to the last square of each grid path. However, there is a catch. Whenever he moves one marble, the other marble will copy its movement if possible. For instance, if one marble moves east, then the other marble will try and move east as well. By try, we mean if moving east is a valid move, then the marble will move east.
Moving north increases the second coordinate by 1, while moving south decreases it by 1. Similarly, moving east increases first coordinate by 1, while moving west decreases it.
Given these two valid grid paths, Genos wants to know if it is possible to move both marbles to the ends of their respective paths. That is, if it is possible to move the marbles such that both marbles rest on the last square of their respective paths.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1<=000<=000) — the length of the paths.
The second line of the input contains a string consisting of *n*<=-<=1 characters (each of which is either 'N', 'E', 'S', or 'W') — the first grid path. The characters can be thought of as the sequence of moves needed to traverse the grid path. For example, the example path in the problem statement can be expressed by the string "NNESWW".
The third line of the input contains a string of *n*<=-<=1 characters (each of which is either 'N', 'E', 'S', or 'W') — the second grid path.
Output Specification:
Print "YES" (without quotes) if it is possible for both marbles to be at the end position at the same time. Print "NO" (without quotes) otherwise. In both cases, the answer is case-insensitive.
Demo Input:
['7\nNNESWW\nSWSWSW\n', '3\nNN\nSS\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the first grid path is the one described in the statement. Moreover, the following sequence of moves will get both marbles to the end: NNESWWSWSW.
In the second sample, no sequence of moves can get both marbles to the end. | ```python
import math
for i in range(int(input())):
n,k = map(int,input().split())
print(math.comb(n-1,k-1))
``` | -1 | |
544 | A | Set of Strings | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | You are given a string *q*. A sequence of *k* strings *s*1,<=*s*2,<=...,<=*s**k* is called beautiful, if the concatenation of these strings is string *q* (formally, *s*1<=+<=*s*2<=+<=...<=+<=*s**k*<==<=*q*) and the first characters of these strings are distinct.
Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist. | The first line contains a positive integer *k* (1<=≤<=*k*<=≤<=26) — the number of strings that should be in a beautiful sequence.
The second line contains string *q*, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive. | If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next *k* lines print the beautiful sequence of strings *s*1,<=*s*2,<=...,<=*s**k*.
If there are multiple possible answers, print any of them. | [
"1\nabca\n",
"2\naaacas\n",
"4\nabc\n"
] | [
"YES\nabca\n",
"YES\naaa\ncas\n",
"NO\n"
] | In the second sample there are two possible answers: {"*aaaca*", "*s*"} and {"*aaa*", "*cas*"}. | 500 | [
{
"input": "1\nabca",
"output": "YES\nabca"
},
{
"input": "2\naaacas",
"output": "YES\naaa\ncas"
},
{
"input": "4\nabc",
"output": "NO"
},
{
"input": "3\nnddkhkhkdndknndkhrnhddkrdhrnrrnkkdnnndndrdhnknknhnrnnkrrdhrkhkrkhnkhkhhrhdnrndnknrrhdrdrkhdrkkhkrnkk",
"output": "YES\... | 1,537,999,835 | 2,147,483,647 | PyPy 3 | OK | TESTS | 33 | 140 | 0 | k = int(input())
s = input()
used = [False for _ in range(26)]
starts = []
for i, c in enumerate(s):
x = ord(c) - ord('a')
if not used[x]:
used[x] = True
starts.append(i)
if len(starts) < k:
print('NO')
else:
print('YES')
while len(starts) > k:
starts.pop()
for i in range(k - 1):
print(s[starts[i]:starts[i + 1]])
print(s[starts[-1]:])
| Title: Set of Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *q*. A sequence of *k* strings *s*1,<=*s*2,<=...,<=*s**k* is called beautiful, if the concatenation of these strings is string *q* (formally, *s*1<=+<=*s*2<=+<=...<=+<=*s**k*<==<=*q*) and the first characters of these strings are distinct.
Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.
Input Specification:
The first line contains a positive integer *k* (1<=≤<=*k*<=≤<=26) — the number of strings that should be in a beautiful sequence.
The second line contains string *q*, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.
Output Specification:
If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next *k* lines print the beautiful sequence of strings *s*1,<=*s*2,<=...,<=*s**k*.
If there are multiple possible answers, print any of them.
Demo Input:
['1\nabca\n', '2\naaacas\n', '4\nabc\n']
Demo Output:
['YES\nabca\n', 'YES\naaa\ncas\n', 'NO\n']
Note:
In the second sample there are two possible answers: {"*aaaca*", "*s*"} and {"*aaa*", "*cas*"}. | ```python
k = int(input())
s = input()
used = [False for _ in range(26)]
starts = []
for i, c in enumerate(s):
x = ord(c) - ord('a')
if not used[x]:
used[x] = True
starts.append(i)
if len(starts) < k:
print('NO')
else:
print('YES')
while len(starts) > k:
starts.pop()
for i in range(k - 1):
print(s[starts[i]:starts[i + 1]])
print(s[starts[-1]:])
``` | 3 | |
496 | B | Secret Combination | PROGRAMMING | 1,500 | [
"brute force",
"constructive algorithms",
"implementation"
] | null | null | You got a box with a combination lock. The lock has a display showing *n* digits. There are two buttons on the box, each button changes digits on the display. You have quickly discovered that the first button adds 1 to all the digits (all digits 9 become digits 0), and the second button shifts all the digits on the display one position to the right (the last digit becomes the first one). For example, if the display is currently showing number 579, then if we push the first button, the display will show 680, and if after that we push the second button, the display will show 068.
You know that the lock will open if the display is showing the smallest possible number that can be obtained by pushing the buttons in some order. The leading zeros are ignored while comparing numbers. Now your task is to find the desired number. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits on the display.
The second line contains *n* digits — the initial state of the display. | Print a single line containing *n* digits — the desired state of the display containing the smallest possible number. | [
"3\n579\n",
"4\n2014\n"
] | [
"024\n",
"0142\n"
] | none | 1,000 | [
{
"input": "3\n579",
"output": "024"
},
{
"input": "4\n2014",
"output": "0142"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "3\n039",
"output": "014"
},
{
"input": "4\n4444",
"output": "0000"
},
{
"input": "5\n46802",
"output": "02468"
},
... | 1,635,879,723 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 4,505,600 | import time
n=int(input())
ch=input()
code = []
max_code = -1
for i in range(n):
x = int(ch[i])
if x > max_code:
max_code = x
code.append(x)
if ch[0]=="0" and n==2:
print(int(ch))
elif max_code==min(code):
print("0"*n)
else:
while max_code>5:
for i in range(n):
code[i]=(code[i]+1)%10
max_code=max(code)
min_code=min(code)
while code[0]!=min_code:
x=code[n-1]
for i in range(n-1,0,-1):
code[i]=code[i-1]
code[0]=x
ch=""
for i in range(n):
ch+=str(code[i])
print(ch) | Title: Secret Combination
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You got a box with a combination lock. The lock has a display showing *n* digits. There are two buttons on the box, each button changes digits on the display. You have quickly discovered that the first button adds 1 to all the digits (all digits 9 become digits 0), and the second button shifts all the digits on the display one position to the right (the last digit becomes the first one). For example, if the display is currently showing number 579, then if we push the first button, the display will show 680, and if after that we push the second button, the display will show 068.
You know that the lock will open if the display is showing the smallest possible number that can be obtained by pushing the buttons in some order. The leading zeros are ignored while comparing numbers. Now your task is to find the desired number.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits on the display.
The second line contains *n* digits — the initial state of the display.
Output Specification:
Print a single line containing *n* digits — the desired state of the display containing the smallest possible number.
Demo Input:
['3\n579\n', '4\n2014\n']
Demo Output:
['024\n', '0142\n']
Note:
none | ```python
import time
n=int(input())
ch=input()
code = []
max_code = -1
for i in range(n):
x = int(ch[i])
if x > max_code:
max_code = x
code.append(x)
if ch[0]=="0" and n==2:
print(int(ch))
elif max_code==min(code):
print("0"*n)
else:
while max_code>5:
for i in range(n):
code[i]=(code[i]+1)%10
max_code=max(code)
min_code=min(code)
while code[0]!=min_code:
x=code[n-1]
for i in range(n-1,0,-1):
code[i]=code[i-1]
code[0]=x
ch=""
for i in range(n):
ch+=str(code[i])
print(ch)
``` | 0 | |
988 | A | Diverse Team | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them. | The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student. | If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.
Assume that the students are numbered from $1$ to $n$. | [
"5 3\n15 13 15 15 12\n",
"5 4\n15 13 15 15 12\n",
"4 4\n20 10 40 30\n"
] | [
"YES\n1 2 5 \n",
"NO\n",
"YES\n1 2 3 4 \n"
] | All possible answers for the first example:
- {1 2 5} - {2 3 5} - {2 4 5}
Note that the order does not matter. | 0 | [
{
"input": "5 3\n15 13 15 15 12",
"output": "YES\n1 2 5 "
},
{
"input": "5 4\n15 13 15 15 12",
"output": "NO"
},
{
"input": "4 4\n20 10 40 30",
"output": "YES\n1 2 3 4 "
},
{
"input": "1 1\n1",
"output": "YES\n1 "
},
{
"input": "100 53\n16 17 1 2 27 5 9 9 53 24 17... | 1,637,503,302 | 2,147,483,647 | Python 3 | OK | TESTS | 10 | 46 | 0 | n,k=map(int,(input()).split(" "))
z=0
m=input().split(" ")
x=[]
x.extend(set(m))
if len(x)>= k:
print("YES")
for i in range(k):
print(m.index (x[i])+1,end=" ")
z+=1
else:
print("NO")
| Title: Diverse Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.
Input Specification:
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.
Output Specification:
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.
Assume that the students are numbered from $1$ to $n$.
Demo Input:
['5 3\n15 13 15 15 12\n', '5 4\n15 13 15 15 12\n', '4 4\n20 10 40 30\n']
Demo Output:
['YES\n1 2 5 \n', 'NO\n', 'YES\n1 2 3 4 \n']
Note:
All possible answers for the first example:
- {1 2 5} - {2 3 5} - {2 4 5}
Note that the order does not matter. | ```python
n,k=map(int,(input()).split(" "))
z=0
m=input().split(" ")
x=[]
x.extend(set(m))
if len(x)>= k:
print("YES")
for i in range(k):
print(m.index (x[i])+1,end=" ")
z+=1
else:
print("NO")
``` | 3 | |
808 | C | Tea Party | PROGRAMMING | 1,400 | [
"constructive algorithms",
"greedy",
"sortings"
] | null | null | Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1. | The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). | Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1. | [
"2 10\n8 7\n",
"4 4\n1 1 1 1\n",
"3 10\n9 8 10\n"
] | [
"6 4 \n",
"1 1 1 1 \n",
"-1\n"
] | In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available. | 0 | [
{
"input": "2 10\n8 7",
"output": "6 4 "
},
{
"input": "4 4\n1 1 1 1",
"output": "1 1 1 1 "
},
{
"input": "3 10\n9 8 10",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "1 1\n2",
"output": "1 "
},
{
"input": "1 10\n20",
"output... | 1,564,643,019 | 2,147,483,647 | PyPy 3 | OK | TESTS | 29 | 140 | 0 | n, w = (int(i) for i in input().split())
mas = [int(i) for i in input().split()]
def qer():
global w
for i, size in enumerate(mas):
q = round(size / 2 + 0.1)
rer[size,i] = q
w -= q
if w < 0:
return {1:-1}
for size,i in sorted(rer, reverse = True):
o = size - rer[size,i]
rer[size,i] += min(w,o)
w -= min(o,w)
if w == 0:
break
return rer
rer = {}
print(*qer().values())
| Title: Tea Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input Specification:
The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Demo Input:
['2 10\n8 7\n', '4 4\n1 1 1 1\n', '3 10\n9 8 10\n']
Demo Output:
['6 4 \n', '1 1 1 1 \n', '-1\n']
Note:
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available. | ```python
n, w = (int(i) for i in input().split())
mas = [int(i) for i in input().split()]
def qer():
global w
for i, size in enumerate(mas):
q = round(size / 2 + 0.1)
rer[size,i] = q
w -= q
if w < 0:
return {1:-1}
for size,i in sorted(rer, reverse = True):
o = size - rer[size,i]
rer[size,i] += min(w,o)
w -= min(o,w)
if w == 0:
break
return rer
rer = {}
print(*qer().values())
``` | 3 | |
710 | A | King Moves | PROGRAMMING | 800 | [
"implementation"
] | null | null | The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)). | The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'. | Print the only integer *x* — the number of moves permitted for the king. | [
"e4\n"
] | [
"8\n"
] | none | 0 | [
{
"input": "e4",
"output": "8"
},
{
"input": "a1",
"output": "3"
},
{
"input": "h8",
"output": "3"
},
{
"input": "a4",
"output": "5"
},
{
"input": "g7",
"output": "8"
},
{
"input": "e1",
"output": "5"
},
{
"input": "b2",
"output": "8"
... | 1,474,236,384 | 2,147,483,647 | Python 3 | OK | TESTS | 24 | 62 | 0 | s=input()
if s in {"a8", "a1", "h1", "h8"}:
print(3)
elif s[0] in {'a', 'h'} or s[1] in {'1', '8'}:
print(5)
else:
print(8)
| Title: King Moves
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
Input Specification:
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
Output Specification:
Print the only integer *x* — the number of moves permitted for the king.
Demo Input:
['e4\n']
Demo Output:
['8\n']
Note:
none | ```python
s=input()
if s in {"a8", "a1", "h1", "h8"}:
print(3)
elif s[0] in {'a', 'h'} or s[1] in {'1', '8'}:
print(5)
else:
print(8)
``` | 3 | |
1,004 | A | Sonya and Hotels | PROGRAMMING | 900 | [
"implementation"
] | null | null | Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$. | The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$) — coordinates of Sonya's hotels. | Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$. | [
"4 3\n-3 2 9 16\n",
"5 2\n4 8 11 18 19\n"
] | [
"6\n",
"5\n"
] | In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$. | 500 | [
{
"input": "4 3\n-3 2 9 16",
"output": "6"
},
{
"input": "5 2\n4 8 11 18 19",
"output": "5"
},
{
"input": "10 10\n-67 -59 -49 -38 -8 20 41 59 74 83",
"output": "8"
},
{
"input": "10 10\n0 20 48 58 81 95 111 137 147 159",
"output": "9"
},
{
"input": "100 1\n0 1 2 3... | 1,594,065,824 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 109 | 6,656,000 | n,k = map(int, input().split())
k *= 2
l = input().split()
sumi = 2
for x in range(len(l)-1):
y = int(l[x+1])-int(l[x])
if y == k:
sumi += 1
elif y > k:
sumi += 2
print(sumi)
| Title: Sonya and Hotels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$.
Input Specification:
The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$) — coordinates of Sonya's hotels.
Output Specification:
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$.
Demo Input:
['4 3\n-3 2 9 16\n', '5 2\n4 8 11 18 19\n']
Demo Output:
['6\n', '5\n']
Note:
In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$. | ```python
n,k = map(int, input().split())
k *= 2
l = input().split()
sumi = 2
for x in range(len(l)-1):
y = int(l[x+1])-int(l[x])
if y == k:
sumi += 1
elif y > k:
sumi += 2
print(sumi)
``` | 3 | |
379 | A | New Year Candles | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. | The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). | Print a single integer — the number of hours Vasily can light up the room for. | [
"4 2\n",
"6 3\n"
] | [
"7\n",
"8\n"
] | Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | 500 | [
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": ... | 1,693,897,207 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 32 | 62 | 0 | a, b = map(int, input().split())
res = 0
while a >= b:
a += 1 - b
res += b
res += a
print(res) | Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Output Specification:
Print a single integer — the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | ```python
a, b = map(int, input().split())
res = 0
while a >= b:
a += 1 - b
res += b
res += a
print(res)
``` | 3 | |
38 | A | Army | PROGRAMMING | 800 | [
"implementation"
] | A. Army | 2 | 256 | The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream. | The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated. | Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*. | [
"3\n5 6\n1 2\n",
"3\n5 6\n1 3\n"
] | [
"5\n",
"11\n"
] | none | 0 | [
{
"input": "3\n5 6\n1 2",
"output": "5"
},
{
"input": "3\n5 6\n1 3",
"output": "11"
},
{
"input": "2\n55\n1 2",
"output": "55"
},
{
"input": "3\n85 78\n1 3",
"output": "163"
},
{
"input": "4\n63 4 49\n2 3",
"output": "4"
},
{
"input": "5\n93 83 42 56\n... | 1,632,665,224 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 124 | 6,963,200 | # -*- coding: utf-8 -*-
"""
Created on Sun Sep 26 21:56:04 2021
@author: cheehong
"""
n=int(input())
l=list(map(int, input().split()))
k=list(map(int, input().split()))
x=k[1]-k[0]
total=0
while k[0]<k[1]:
total+=l[k[0]-1]
k[0]+=1
print(total) | Title: Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
Input Specification:
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Output Specification:
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
Demo Input:
['3\n5 6\n1 2\n', '3\n5 6\n1 3\n']
Demo Output:
['5\n', '11\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 26 21:56:04 2021
@author: cheehong
"""
n=int(input())
l=list(map(int, input().split()))
k=list(map(int, input().split()))
x=k[1]-k[0]
total=0
while k[0]<k[1]:
total+=l[k[0]-1]
k[0]+=1
print(total)
``` | 3.95603 |
330 | A | Cakeminator | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | You are given a rectangular cake, represented as an *r*<=×<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=×<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat. | The first line contains two integers *r* and *c* (2<=≤<=*r*,<=*c*<=≤<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters — the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry. | Output the maximum number of cake cells that the cakeminator can eat. | [
"3 4\nS...\n....\n..S.\n"
] | [
"8\n"
] | For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats). | 500 | [
{
"input": "3 4\nS...\n....\n..S.",
"output": "8"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "2 2\nSS\nSS",
"output": "0"
},
{
"input": "7 3\nS..\nS..\nS..\nS..\nS..\nS..\nS..",
"output": "14"
},
{
"input": "3 5\n..S..\nSSSSS\n..S..",
"output": "... | 1,665,665,945 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 92 | 0 | arr = input()
arr = arr.split(" ")
r = int(arr[0])
c = int(arr[1])
row = []
col = []
for i in range(r):
text = input()
for j in range(c):
if text[j] == "S":
if j not in col:
col.append(j)
if i not in row:
row.append(i)
print((r-len(row))*c + (c-len(col))*r - ((r-len(row))*(c-len(col))))
| Title: Cakeminator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rectangular cake, represented as an *r*<=×<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=×<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat.
Input Specification:
The first line contains two integers *r* and *c* (2<=≤<=*r*,<=*c*<=≤<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters — the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry.
Output Specification:
Output the maximum number of cake cells that the cakeminator can eat.
Demo Input:
['3 4\nS...\n....\n..S.\n']
Demo Output:
['8\n']
Note:
For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats). | ```python
arr = input()
arr = arr.split(" ")
r = int(arr[0])
c = int(arr[1])
row = []
col = []
for i in range(r):
text = input()
for j in range(c):
if text[j] == "S":
if j not in col:
col.append(j)
if i not in row:
row.append(i)
print((r-len(row))*c + (c-len(col))*r - ((r-len(row))*(c-len(col))))
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,543,687,465 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 124 | 0 | n = input()
l = ["h","e","l","l","o"]
x = 0
for i in n:
if i == l[x]:
x +=1
if i == l[x-1]:
pass
else:
if i =="h":
x = 1
else:
x=0
if x == 5:
a=True
break
a =False
if a:
print("YES")
else:
print("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
n = input()
l = ["h","e","l","l","o"]
x = 0
for i in n:
if i == l[x]:
x +=1
if i == l[x-1]:
pass
else:
if i =="h":
x = 1
else:
x=0
if x == 5:
a=True
break
a =False
if a:
print("YES")
else:
print("NO")
``` | 0 |
767 | A | Snacktower | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*. | Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty. | [
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] | [
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] | In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | 500 | [
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1... | 1,632,688,828 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 6,963,200 | n=int(input())
num=list(map(int,input().split()))
target=n
array={}
while True:
if target in array:
print(target,end=' ')
target-=1
while True:
if target in array:
print(target,end=' ')
target-=1
else:
break
continue
else:
array[num[-n]]=f
if target in array:
continue
print(' ')
n -= 1
if n==0:
break | Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | ```python
n=int(input())
num=list(map(int,input().split()))
target=n
array={}
while True:
if target in array:
print(target,end=' ')
target-=1
while True:
if target in array:
print(target,end=' ')
target-=1
else:
break
continue
else:
array[num[-n]]=f
if target in array:
continue
print(' ')
n -= 1
if n==0:
break
``` | -1 | |
815 | A | Karen and Game | PROGRAMMING | 1,700 | [
"brute force",
"greedy",
"implementation"
] | null | null | On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with *n* rows and *m* columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To win the level, after all the moves, the number in the cell at the *i*-th row and *j*-th column should be equal to *g**i*,<=*j*.
Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task! | The first line of input contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of rows and the number of columns in the grid, respectively.
The next *n* lines each contain *m* integers. In particular, the *j*-th integer in the *i*-th of these rows contains *g**i*,<=*j* (0<=≤<=*g**i*,<=*j*<=≤<=500). | If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer *k*, the minimum number of moves necessary to beat the level.
The next *k* lines should each contain one of the following, describing the moves in the order they must be done:
- row *x*, (1<=≤<=*x*<=≤<=*n*) describing a move of the form "choose the *x*-th row". - col *x*, (1<=≤<=*x*<=≤<=*m*) describing a move of the form "choose the *x*-th column".
If there are multiple optimal solutions, output any one of them. | [
"3 5\n2 2 2 3 2\n0 0 0 1 0\n1 1 1 2 1\n",
"3 3\n0 0 0\n0 1 0\n0 0 0\n",
"3 3\n1 1 1\n1 1 1\n1 1 1\n"
] | [
"4\nrow 1\nrow 1\ncol 4\nrow 3\n",
"-1\n",
"3\nrow 1\nrow 2\nrow 3\n"
] | In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.
In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:
Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3. | 500 | [
{
"input": "3 5\n2 2 2 3 2\n0 0 0 1 0\n1 1 1 2 1",
"output": "4\nrow 1\nrow 1\ncol 4\nrow 3"
},
{
"input": "3 3\n0 0 0\n0 1 0\n0 0 0",
"output": "-1"
},
{
"input": "3 3\n1 1 1\n1 1 1\n1 1 1",
"output": "3\nrow 1\nrow 2\nrow 3"
},
{
"input": "3 5\n2 4 2 2 3\n0 2 0 0 1\n1 3 1 1... | 1,688,357,371 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 62 | 2,867,200 | def burenka(a, b, c, d):
"""
Returns the minimum number of claps Burenka needs to make her fractions equal.
Args:
a: The numerator of the first fraction.
b: The denominator of the first fraction.
c: The numerator of the second fraction.
d: The denominator of the second fraction.
Returns:
The minimum number of claps.
"""
# Check if the fractions are already equal.
if a / b == c / d:
return 0
# Find the greatest common factor of the denominators.
gcd = math.gcd(b, d)
# Normalize the fractions.
a = a // gcd
b = b // gcd
c = c // gcd
d = d // gcd
# Find the minimum number of times Burenka needs to clap to make the numerators equal.
num_claps = abs(a - c) // b
# If the denominators are not equal, Burenka also needs to clap one more time.
if b != d:
num_claps += 1
return num_claps
if __name__ == "__main__":
t = int(input())
for _ in range(t):
a, b, c, d = map(int, input().split())
print(burenka(a, b, c, d)) | Title: Karen and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with *n* rows and *m* columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To win the level, after all the moves, the number in the cell at the *i*-th row and *j*-th column should be equal to *g**i*,<=*j*.
Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!
Input Specification:
The first line of input contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), the number of rows and the number of columns in the grid, respectively.
The next *n* lines each contain *m* integers. In particular, the *j*-th integer in the *i*-th of these rows contains *g**i*,<=*j* (0<=≤<=*g**i*,<=*j*<=≤<=500).
Output Specification:
If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer *k*, the minimum number of moves necessary to beat the level.
The next *k* lines should each contain one of the following, describing the moves in the order they must be done:
- row *x*, (1<=≤<=*x*<=≤<=*n*) describing a move of the form "choose the *x*-th row". - col *x*, (1<=≤<=*x*<=≤<=*m*) describing a move of the form "choose the *x*-th column".
If there are multiple optimal solutions, output any one of them.
Demo Input:
['3 5\n2 2 2 3 2\n0 0 0 1 0\n1 1 1 2 1\n', '3 3\n0 0 0\n0 1 0\n0 0 0\n', '3 3\n1 1 1\n1 1 1\n1 1 1\n']
Demo Output:
['4\nrow 1\nrow 1\ncol 4\nrow 3\n', '-1\n', '3\nrow 1\nrow 2\nrow 3\n']
Note:
In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.
In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:
Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3. | ```python
def burenka(a, b, c, d):
"""
Returns the minimum number of claps Burenka needs to make her fractions equal.
Args:
a: The numerator of the first fraction.
b: The denominator of the first fraction.
c: The numerator of the second fraction.
d: The denominator of the second fraction.
Returns:
The minimum number of claps.
"""
# Check if the fractions are already equal.
if a / b == c / d:
return 0
# Find the greatest common factor of the denominators.
gcd = math.gcd(b, d)
# Normalize the fractions.
a = a // gcd
b = b // gcd
c = c // gcd
d = d // gcd
# Find the minimum number of times Burenka needs to clap to make the numerators equal.
num_claps = abs(a - c) // b
# If the denominators are not equal, Burenka also needs to clap one more time.
if b != d:
num_claps += 1
return num_claps
if __name__ == "__main__":
t = int(input())
for _ in range(t):
a, b, c, d = map(int, input().split())
print(burenka(a, b, c, d))
``` | -1 | |
363 | B | Fence | PROGRAMMING | 1,100 | [
"brute force",
"dp"
] | null | null | There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic). | The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence. | Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them. | [
"7 3\n1 2 6 1 1 7 1\n"
] | [
"3\n"
] | In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8. | 1,000 | [
{
"input": "7 3\n1 2 6 1 1 7 1",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 1\n10 20",
"output": "1"
},
{
"input": "10 5\n1 2 3 1 2 2 3 1 4 5",
"output": "1"
},
{
"input": "10 2\n3 1 4 1 4 6 2 1 4 6",
"output": "7"
},
{
"inp... | 1,693,668,220 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 1,000 | 5,734,400 | m,n = map(int,input().split())
k = list(map(int,input().split()))
s=[]
for i in range(m-n+1):
d=0
for j in range(n):
d = d+k[i+j]
s.append(d)
f = min(s)
for i in range(m-n+1):
if s[i] == f:
print(i+1)
break | Title: Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
Input Specification:
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
Output Specification:
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
Demo Input:
['7 3\n1 2 6 1 1 7 1\n']
Demo Output:
['3\n']
Note:
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8. | ```python
m,n = map(int,input().split())
k = list(map(int,input().split()))
s=[]
for i in range(m-n+1):
d=0
for j in range(n):
d = d+k[i+j]
s.append(d)
f = min(s)
for i in range(m-n+1):
if s[i] == f:
print(i+1)
break
``` | 0 | |
220 | B | Little Elephant and Array | PROGRAMMING | 1,800 | [
"constructive algorithms",
"data structures"
] | null | null | The Little Elephant loves playing with arrays. He has array *a*, consisting of *n* positive integers, indexed from 1 to *n*. Let's denote the number with index *i* as *a**i*.
Additionally the Little Elephant has *m* queries to the array, each query is characterised by a pair of integers *l**j* and *r**j* (1<=≤<=*l**j*<=≤<=*r**j*<=≤<=*n*). For each query *l**j*,<=*r**j* the Little Elephant has to count, how many numbers *x* exist, such that number *x* occurs exactly *x* times among numbers *a**l**j*,<=*a**l**j*<=+<=1,<=...,<=*a**r**j*.
Help the Little Elephant to count the answers to all queries. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the size of array *a* and the number of queries to it. The next line contains *n* space-separated positive integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109). Next *m* lines contain descriptions of queries, one per line. The *j*-th of these lines contains the description of the *j*-th query as two space-separated integers *l**j* and *r**j* (1<=≤<=*l**j*<=≤<=*r**j*<=≤<=*n*). | In *m* lines print *m* integers — the answers to the queries. The *j*-th line should contain the answer to the *j*-th query. | [
"7 2\n3 1 2 2 3 3 7\n1 7\n3 4\n"
] | [
"3\n1\n"
] | none | 1,000 | [
{
"input": "7 2\n3 1 2 2 3 3 7\n1 7\n3 4",
"output": "3\n1"
},
{
"input": "6 6\n1 2 2 3 3 3\n1 2\n2 2\n1 3\n2 4\n4 6\n1 6",
"output": "1\n0\n2\n1\n1\n3"
},
{
"input": "1 2\n1\n1 1\n1 1",
"output": "1\n1"
},
{
"input": "1 1\n1000000000\n1 1",
"output": "0"
}
] | 1,443,523,889 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | s = input()
nums = set([str(x) for x in range(0, 9+1)])
cnt = dict()
m = -1
ans = 0
days_in_month = {1: 31, 2: 28, 3: 31, 4: 30, 5: 31, 6: 30, 7: 31, 8: 31, 9: 30, 10: 31, 11: 30, 12:31}
for i in range(len(s) - 10):
q = s[i:i+10]
if q[0] in nums and q[1] in nums and q[2] == "-":
if q[3] in nums and q[4] in nums and q[5] == "-":
if q[6] in nums and q[7] in nums and q[8] in nums and q[9] in nums:
try:
day = int(q[0:1+1])
month = int(q[3:4+1])
year = int(q[6:9+1])
except:
continue
#print(day, month)
if 0 < month <= 12 and 0 < day <= days_in_month[month] and 2013 <= year <= 2015:
try:
cnt[q] += 1
except:
cnt[q] = 0
#print(cnt)
for key in cnt.keys():
if cnt[key] > m:
m = cnt[key]
ans = key
print(ans) | Title: Little Elephant and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves playing with arrays. He has array *a*, consisting of *n* positive integers, indexed from 1 to *n*. Let's denote the number with index *i* as *a**i*.
Additionally the Little Elephant has *m* queries to the array, each query is characterised by a pair of integers *l**j* and *r**j* (1<=≤<=*l**j*<=≤<=*r**j*<=≤<=*n*). For each query *l**j*,<=*r**j* the Little Elephant has to count, how many numbers *x* exist, such that number *x* occurs exactly *x* times among numbers *a**l**j*,<=*a**l**j*<=+<=1,<=...,<=*a**r**j*.
Help the Little Elephant to count the answers to all queries.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the size of array *a* and the number of queries to it. The next line contains *n* space-separated positive integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109). Next *m* lines contain descriptions of queries, one per line. The *j*-th of these lines contains the description of the *j*-th query as two space-separated integers *l**j* and *r**j* (1<=≤<=*l**j*<=≤<=*r**j*<=≤<=*n*).
Output Specification:
In *m* lines print *m* integers — the answers to the queries. The *j*-th line should contain the answer to the *j*-th query.
Demo Input:
['7 2\n3 1 2 2 3 3 7\n1 7\n3 4\n']
Demo Output:
['3\n1\n']
Note:
none | ```python
s = input()
nums = set([str(x) for x in range(0, 9+1)])
cnt = dict()
m = -1
ans = 0
days_in_month = {1: 31, 2: 28, 3: 31, 4: 30, 5: 31, 6: 30, 7: 31, 8: 31, 9: 30, 10: 31, 11: 30, 12:31}
for i in range(len(s) - 10):
q = s[i:i+10]
if q[0] in nums and q[1] in nums and q[2] == "-":
if q[3] in nums and q[4] in nums and q[5] == "-":
if q[6] in nums and q[7] in nums and q[8] in nums and q[9] in nums:
try:
day = int(q[0:1+1])
month = int(q[3:4+1])
year = int(q[6:9+1])
except:
continue
#print(day, month)
if 0 < month <= 12 and 0 < day <= days_in_month[month] and 2013 <= year <= 2015:
try:
cnt[q] += 1
except:
cnt[q] = 0
#print(cnt)
for key in cnt.keys():
if cnt[key] > m:
m = cnt[key]
ans = key
print(ans)
``` | 0 | |
401 | A | Vanya and Cards | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the card can be positive, negative and can even be equal to zero. The only limit is, the number on each card doesn't exceed *x* in the absolute value.
Natasha doesn't like when Vanya spends a long time playing, so she hid all of his cards. Vanya became sad and started looking for the cards but he only found *n* of them. Vanya loves the balance, so he wants the sum of all numbers on found cards equal to zero. On the other hand, he got very tired of looking for cards. Help the boy and say what is the minimum number of cards does he need to find to make the sum equal to zero?
You can assume that initially Vanya had infinitely many cards with each integer number from <=-<=*x* to *x*. | The first line contains two integers: *n* (1<=≤<=*n*<=≤<=1000) — the number of found cards and *x* (1<=≤<=*x*<=≤<=1000) — the maximum absolute value of the number on a card. The second line contains *n* space-separated integers — the numbers on found cards. It is guaranteed that the numbers do not exceed *x* in their absolute value. | Print a single number — the answer to the problem. | [
"3 2\n-1 1 2\n",
"2 3\n-2 -2\n"
] | [
"1\n",
"2\n"
] | In the first sample, Vanya needs to find a single card with number -2.
In the second sample, Vanya needs to find two cards with number 2. He can't find a single card with the required number as the numbers on the lost cards do not exceed 3 in their absolute value. | 500 | [
{
"input": "3 2\n-1 1 2",
"output": "1"
},
{
"input": "2 3\n-2 -2",
"output": "2"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "2 2\n-1 -1",
"output": "1"
},
{
"input": "15 5\n-2 -1 2 -4 -3 4 -4 -2 -2 2 -2 -1 1 -4 -2",
"output": "4"
},
{
"... | 1,595,780,384 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 139 | 20,172,800 | n, x = map(int, input().split())
s = input().split()
count = 0
j = 0
for i in s:
count += int(i)
if count < 0:
while count < 0:
count += x
j += 1
elif count > 0:
while count > 0:
count -= x
j += 1
else:
print(0)
print(j) | Title: Vanya and Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the card can be positive, negative and can even be equal to zero. The only limit is, the number on each card doesn't exceed *x* in the absolute value.
Natasha doesn't like when Vanya spends a long time playing, so she hid all of his cards. Vanya became sad and started looking for the cards but he only found *n* of them. Vanya loves the balance, so he wants the sum of all numbers on found cards equal to zero. On the other hand, he got very tired of looking for cards. Help the boy and say what is the minimum number of cards does he need to find to make the sum equal to zero?
You can assume that initially Vanya had infinitely many cards with each integer number from <=-<=*x* to *x*.
Input Specification:
The first line contains two integers: *n* (1<=≤<=*n*<=≤<=1000) — the number of found cards and *x* (1<=≤<=*x*<=≤<=1000) — the maximum absolute value of the number on a card. The second line contains *n* space-separated integers — the numbers on found cards. It is guaranteed that the numbers do not exceed *x* in their absolute value.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['3 2\n-1 1 2\n', '2 3\n-2 -2\n']
Demo Output:
['1\n', '2\n']
Note:
In the first sample, Vanya needs to find a single card with number -2.
In the second sample, Vanya needs to find two cards with number 2. He can't find a single card with the required number as the numbers on the lost cards do not exceed 3 in their absolute value. | ```python
n, x = map(int, input().split())
s = input().split()
count = 0
j = 0
for i in s:
count += int(i)
if count < 0:
while count < 0:
count += x
j += 1
elif count > 0:
while count > 0:
count -= x
j += 1
else:
print(0)
print(j)
``` | 0 | |
111 | B | Petya and Divisors | PROGRAMMING | 1,900 | [
"binary search",
"data structures",
"number theory"
] | B. Petya and Divisors | 5 | 256 | Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:
You are given *n* queries in the form "*x**i* *y**i*". For each query Petya should count how many divisors of number *x**i* divide none of the numbers *x**i*<=-<=*y**i*,<=*x**i*<=-<=*y**i*<=+<=1,<=...,<=*x**i*<=-<=1. Help him. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the following *n* lines contain two space-separated integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=105, 0<=≤<=*y**i*<=≤<=*i*<=-<=1, where *i* is the query's ordinal number; the numeration starts with 1).
If *y**i*<==<=0 for the query, then the answer to the query will be the number of divisors of the number *x**i*. In this case you do not need to take the previous numbers *x* into consideration. | For each query print the answer on a single line: the number of positive integers *k* such that | [
"6\n4 0\n3 1\n5 2\n6 2\n18 4\n10000 3\n"
] | [
"3\n1\n1\n2\n2\n22\n"
] | Let's write out the divisors that give answers for the first 5 queries:
1) 1, 2, 4
2) 3
3) 5
4) 2, 6
5) 9, 18 | 1,000 | [
{
"input": "6\n4 0\n3 1\n5 2\n6 2\n18 4\n10000 3",
"output": "3\n1\n1\n2\n2\n22"
},
{
"input": "5\n10 0\n10 0\n10 0\n10 0\n10 0",
"output": "4\n4\n4\n4\n4"
},
{
"input": "12\n41684 0\n95210 1\n60053 1\n32438 3\n97956 1\n21785 2\n14594 6\n17170 4\n93937 6\n70764 5\n13695 4\n14552 6",
... | 1,582,692,467 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 3,806 | 12,595,200 | def R(): return map(int, input().split())
def I(): return int(input())
def S(): return str(input())
def L(): return list(R())
from collections import Counter
import math
import sys
from itertools import permutations
import bisect
div=[[] for i in range(10**5+1)]
for i in range(1,10**5+1):
for j in range(1,10**5//i+1):
div[i*j].append(i)
n=I()
last=[-1]*(10**5+1)
for j in range(n):
x,y=R()
ans=0
for i in div[x]:
if last[i]<j-y:
ans+=1
last[i]=j
print(ans) | Title: Petya and Divisors
Time Limit: 5 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:
You are given *n* queries in the form "*x**i* *y**i*". For each query Petya should count how many divisors of number *x**i* divide none of the numbers *x**i*<=-<=*y**i*,<=*x**i*<=-<=*y**i*<=+<=1,<=...,<=*x**i*<=-<=1. Help him.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the following *n* lines contain two space-separated integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=105, 0<=≤<=*y**i*<=≤<=*i*<=-<=1, where *i* is the query's ordinal number; the numeration starts with 1).
If *y**i*<==<=0 for the query, then the answer to the query will be the number of divisors of the number *x**i*. In this case you do not need to take the previous numbers *x* into consideration.
Output Specification:
For each query print the answer on a single line: the number of positive integers *k* such that
Demo Input:
['6\n4 0\n3 1\n5 2\n6 2\n18 4\n10000 3\n']
Demo Output:
['3\n1\n1\n2\n2\n22\n']
Note:
Let's write out the divisors that give answers for the first 5 queries:
1) 1, 2, 4
2) 3
3) 5
4) 2, 6
5) 9, 18 | ```python
def R(): return map(int, input().split())
def I(): return int(input())
def S(): return str(input())
def L(): return list(R())
from collections import Counter
import math
import sys
from itertools import permutations
import bisect
div=[[] for i in range(10**5+1)]
for i in range(1,10**5+1):
for j in range(1,10**5//i+1):
div[i*j].append(i)
n=I()
last=[-1]*(10**5+1)
for j in range(n):
x,y=R()
ans=0
for i in div[x]:
if last[i]<j-y:
ans+=1
last[i]=j
print(ans)
``` | 3.59594 |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
... | 1,583,868,874 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | /*
@ Author: Andres Nowak
@ Date: Tue Mar 10 13:15:34 CST 2020
*/
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
string probability(vector<int> throws_of_dice)
{
int denominator = 6;
int maximum = *max_element(throws_of_dice.begin(), throws_of_dice.end());
int higher_numbers_that_can_get = 7 - maximum;
if (higher_numbers_that_can_get == 0)
{
return "0/1";
}
int common_factor = gcd(higher_numbers_that_can_get, denominator);
higher_numbers_that_can_get = higher_numbers_that_can_get / common_factor;
denominator = denominator / common_factor;
int probability = higher_numbers_that_can_get / denominator;
if (probability == 1)
{
return "1/1";
}
return to_string(higher_numbers_that_can_get) + "/" + to_string(denominator);
}
int main()
{
vector<int> throws_of_dice(2);
cin >> throws_of_dice[0] >> throws_of_dice[1];
cout << probability(throws_of_dice) << endl;
}
| Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
/*
@ Author: Andres Nowak
@ Date: Tue Mar 10 13:15:34 CST 2020
*/
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
string probability(vector<int> throws_of_dice)
{
int denominator = 6;
int maximum = *max_element(throws_of_dice.begin(), throws_of_dice.end());
int higher_numbers_that_can_get = 7 - maximum;
if (higher_numbers_that_can_get == 0)
{
return "0/1";
}
int common_factor = gcd(higher_numbers_that_can_get, denominator);
higher_numbers_that_can_get = higher_numbers_that_can_get / common_factor;
denominator = denominator / common_factor;
int probability = higher_numbers_that_can_get / denominator;
if (probability == 1)
{
return "1/1";
}
return to_string(higher_numbers_that_can_get) + "/" + to_string(denominator);
}
int main()
{
vector<int> throws_of_dice(2);
cin >> throws_of_dice[0] >> throws_of_dice[1];
cout << probability(throws_of_dice) << endl;
}
``` | -1 |
191 | A | Dynasty Puzzles | PROGRAMMING | 1,500 | [
"dp"
] | null | null | The ancient Berlanders believed that the longer the name, the more important its bearer is. Thus, Berland kings were famous for their long names. But long names are somewhat inconvenient, so the Berlanders started to abbreviate the names of their kings. They called every king by the first letters of its name. Thus, the king, whose name was Victorious Vasily Pupkin, was always called by the berlanders VVP.
In Berland over its long history many dynasties of kings replaced each other, but they were all united by common traditions. Thus, according to one Berland traditions, to maintain stability in the country, the first name of the heir should be the same as the last name his predecessor (hence, the first letter of the abbreviated name of the heir coincides with the last letter of the abbreviated name of the predecessor). Berlanders appreciate stability, so this tradition has never been broken. Also Berlanders like perfection, so another tradition requires that the first name of the first king in the dynasty coincides with the last name of the last king in this dynasty (hence, the first letter of the abbreviated name of the first king coincides with the last letter of the abbreviated name of the last king). This tradition, of course, has also been always observed.
The name of a dynasty is formed by very simple rules: we take all the short names of the kings in the order in which they ruled, and write them in one line. Thus, a dynasty of kings "ab" and "ba" is called "abba", and the dynasty, which had only the king "abca", is called "abca".
Vasya, a historian, has recently found a list of abbreviated names of all Berland kings and their relatives. Help Vasya to find the maximally long name of the dynasty that could have existed in Berland.
Note that in his list all the names are ordered by the time, that is, if name *A* is earlier in the list than *B*, then if *A* and *B* were kings, then king *A* ruled before king *B*. | The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105) — the number of names in Vasya's list. Next *n* lines contain *n* abbreviated names, one per line. An abbreviated name is a non-empty sequence of lowercase Latin letters. Its length does not exceed 10 characters. | Print a single number — length of the sought dynasty's name in letters.
If Vasya's list is wrong and no dynasty can be found there, print a single number 0. | [
"3\nabc\nca\ncba\n",
"4\nvvp\nvvp\ndam\nvvp\n",
"3\nab\nc\ndef\n"
] | [
"6\n",
"0\n",
"1\n"
] | In the first sample two dynasties can exist: the one called "abcca" (with the first and second kings) and the one called "abccba" (with the first and third kings).
In the second sample there aren't acceptable dynasties.
The only dynasty in the third sample consists of one king, his name is "c". | 500 | [
{
"input": "3\nabc\nca\ncba",
"output": "6"
},
{
"input": "4\nvvp\nvvp\ndam\nvvp",
"output": "0"
},
{
"input": "3\nab\nc\ndef",
"output": "1"
},
{
"input": "5\nab\nbc\ncd\nde\nffffffffff",
"output": "10"
},
{
"input": "5\ncab\nbbc\ncaa\nccc\naca",
"output": "9... | 1,689,647,964 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 2,000 | 11,673,600 | from collections import defaultdict, Counter
import sys
M = int(1e9 + 7)
inf = float('inf')
def I():
return input()
def II():
return int(I())
def LI():
return list(I().split())
def LII():
return list(map(int, LI()))
def rank(arr, lo, hi, target):
while lo <= hi:
mi = (lo + hi) >> 1
if arr[mi] < target:
lo = mi + 1
else:
hi = mi - 1
return lo
def get_idx(ch):
return ord(ch) - ord('a')
def solve():
n = II()
dp = [[-inf] * 26 for _ in range(26)]
ans = 0
for _ in range(n):
s = I()
m = len(s)
begin, end = get_idx(s[0]), get_idx(s[-1])
for i in range(26):
dp[i][end] = max(dp[i][end], dp[i][begin] + m)
dp[begin][end] = max(dp[begin][end], m)
ans = max(ans, dp[end][end])
print(ans)
solve()
| Title: Dynasty Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The ancient Berlanders believed that the longer the name, the more important its bearer is. Thus, Berland kings were famous for their long names. But long names are somewhat inconvenient, so the Berlanders started to abbreviate the names of their kings. They called every king by the first letters of its name. Thus, the king, whose name was Victorious Vasily Pupkin, was always called by the berlanders VVP.
In Berland over its long history many dynasties of kings replaced each other, but they were all united by common traditions. Thus, according to one Berland traditions, to maintain stability in the country, the first name of the heir should be the same as the last name his predecessor (hence, the first letter of the abbreviated name of the heir coincides with the last letter of the abbreviated name of the predecessor). Berlanders appreciate stability, so this tradition has never been broken. Also Berlanders like perfection, so another tradition requires that the first name of the first king in the dynasty coincides with the last name of the last king in this dynasty (hence, the first letter of the abbreviated name of the first king coincides with the last letter of the abbreviated name of the last king). This tradition, of course, has also been always observed.
The name of a dynasty is formed by very simple rules: we take all the short names of the kings in the order in which they ruled, and write them in one line. Thus, a dynasty of kings "ab" and "ba" is called "abba", and the dynasty, which had only the king "abca", is called "abca".
Vasya, a historian, has recently found a list of abbreviated names of all Berland kings and their relatives. Help Vasya to find the maximally long name of the dynasty that could have existed in Berland.
Note that in his list all the names are ordered by the time, that is, if name *A* is earlier in the list than *B*, then if *A* and *B* were kings, then king *A* ruled before king *B*.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5·105) — the number of names in Vasya's list. Next *n* lines contain *n* abbreviated names, one per line. An abbreviated name is a non-empty sequence of lowercase Latin letters. Its length does not exceed 10 characters.
Output Specification:
Print a single number — length of the sought dynasty's name in letters.
If Vasya's list is wrong and no dynasty can be found there, print a single number 0.
Demo Input:
['3\nabc\nca\ncba\n', '4\nvvp\nvvp\ndam\nvvp\n', '3\nab\nc\ndef\n']
Demo Output:
['6\n', '0\n', '1\n']
Note:
In the first sample two dynasties can exist: the one called "abcca" (with the first and second kings) and the one called "abccba" (with the first and third kings).
In the second sample there aren't acceptable dynasties.
The only dynasty in the third sample consists of one king, his name is "c". | ```python
from collections import defaultdict, Counter
import sys
M = int(1e9 + 7)
inf = float('inf')
def I():
return input()
def II():
return int(I())
def LI():
return list(I().split())
def LII():
return list(map(int, LI()))
def rank(arr, lo, hi, target):
while lo <= hi:
mi = (lo + hi) >> 1
if arr[mi] < target:
lo = mi + 1
else:
hi = mi - 1
return lo
def get_idx(ch):
return ord(ch) - ord('a')
def solve():
n = II()
dp = [[-inf] * 26 for _ in range(26)]
ans = 0
for _ in range(n):
s = I()
m = len(s)
begin, end = get_idx(s[0]), get_idx(s[-1])
for i in range(26):
dp[i][end] = max(dp[i][end], dp[i][begin] + m)
dp[begin][end] = max(dp[begin][end], m)
ans = max(ans, dp[end][end])
print(ans)
solve()
``` | 0 | |
197 | A | Plate Game | PROGRAMMING | 1,600 | [
"constructive algorithms",
"games",
"math"
] | null | null | You've got a rectangular table with length *a* and width *b* and the infinite number of plates of radius *r*. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well. | A single line contains three space-separated integers *a*, *b*, *r* (1<=≤<=*a*,<=*b*,<=*r*<=≤<=100) — the table sides and the plates' radius, correspondingly. | If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes). | [
"5 5 2\n",
"6 7 4\n"
] | [
"First\n",
"Second\n"
] | In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move. | 1,000 | [
{
"input": "5 5 2",
"output": "First"
},
{
"input": "6 7 4",
"output": "Second"
},
{
"input": "100 100 1",
"output": "First"
},
{
"input": "1 1 100",
"output": "Second"
},
{
"input": "13 7 3",
"output": "First"
},
{
"input": "23 7 3",
"output": "Fi... | 1,652,775,968 | 2,147,483,647 | Python 3 | OK | TESTS | 95 | 92 | 4,300,800 | A = list(map(int,input().split()))
a = A[0];b = A[1];r = A[2]
if 2*r <= min(a,b):
print('First')
else:
print('Second') | Title: Plate Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular table with length *a* and width *b* and the infinite number of plates of radius *r*. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input Specification:
A single line contains three space-separated integers *a*, *b*, *r* (1<=≤<=*a*,<=*b*,<=*r*<=≤<=100) — the table sides and the plates' radius, correspondingly.
Output Specification:
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Demo Input:
['5 5 2\n', '6 7 4\n']
Demo Output:
['First\n', 'Second\n']
Note:
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move. | ```python
A = list(map(int,input().split()))
a = A[0];b = A[1];r = A[2]
if 2*r <= min(a,b):
print('First')
else:
print('Second')
``` | 3 | |
361 | B | Levko and Permutation | PROGRAMMING | 1,200 | [
"constructive algorithms",
"math",
"number theory"
] | null | null | Levko loves permutations very much. A permutation of length *n* is a sequence of distinct positive integers, each is at most *n*.
Let’s assume that value *gcd*(*a*,<=*b*) shows the greatest common divisor of numbers *a* and *b*. Levko assumes that element *p**i* of permutation *p*1,<=*p*2,<=... ,<=*p**n* is good if *gcd*(*i*,<=*p**i*)<=><=1. Levko considers a permutation beautiful, if it has exactly *k* good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them. | The single line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*). | In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.
If there are multiple suitable permutations, you are allowed to print any of them. | [
"4 2\n",
"1 1\n"
] | [
"2 4 3 1",
"-1\n"
] | In the first sample elements 4 and 3 are good because *gcd*(2, 4) = 2 > 1 and *gcd*(3, 3) = 3 > 1. Elements 2 and 1 are not good because *gcd*(1, 2) = 1 and *gcd*(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.
The second sample has no beautiful permutations. | 1,000 | [
{
"input": "4 2",
"output": "2 1 3 4 "
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "7 4",
"output": "3 1 2 4 5 6 7 "
},
{
"input": "10 9",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "10000 5000",
"output": "5000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1... | 1,623,762,408 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | n, k = list(map(int, input().split()))
if n == 1:
print(-1)
else:
print(1, end=" ")
i = 2
counter = k
while counter > 0:
print(i*2, end=" ")
i += 1
counter -= 1
index_pointer = k + 1
pointer = n*2 + 5
while index_pointer <= n:
if index_pointer % 2 == 0:
if pointer % 2 == 0:
pointer += 1
else:
if pointer % 2 == 1:
pointer += 1
print(pointer, end=" ")
pointer += 1
index_pointer += 1
| Title: Levko and Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Levko loves permutations very much. A permutation of length *n* is a sequence of distinct positive integers, each is at most *n*.
Let’s assume that value *gcd*(*a*,<=*b*) shows the greatest common divisor of numbers *a* and *b*. Levko assumes that element *p**i* of permutation *p*1,<=*p*2,<=... ,<=*p**n* is good if *gcd*(*i*,<=*p**i*)<=><=1. Levko considers a permutation beautiful, if it has exactly *k* good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.
Input Specification:
The single line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*).
Output Specification:
In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.
If there are multiple suitable permutations, you are allowed to print any of them.
Demo Input:
['4 2\n', '1 1\n']
Demo Output:
['2 4 3 1', '-1\n']
Note:
In the first sample elements 4 and 3 are good because *gcd*(2, 4) = 2 > 1 and *gcd*(3, 3) = 3 > 1. Elements 2 and 1 are not good because *gcd*(1, 2) = 1 and *gcd*(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.
The second sample has no beautiful permutations. | ```python
n, k = list(map(int, input().split()))
if n == 1:
print(-1)
else:
print(1, end=" ")
i = 2
counter = k
while counter > 0:
print(i*2, end=" ")
i += 1
counter -= 1
index_pointer = k + 1
pointer = n*2 + 5
while index_pointer <= n:
if index_pointer % 2 == 0:
if pointer % 2 == 0:
pointer += 1
else:
if pointer % 2 == 1:
pointer += 1
print(pointer, end=" ")
pointer += 1
index_pointer += 1
``` | 0 | |
120 | A | Elevator | PROGRAMMING | 1,000 | [
"brute force",
"implementation",
"math"
] | null | null | A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed. | The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=≤<=*a*<=≤<=2) which denotes the number of the rail at which the person was holding. | Print character "R" if the VIP is right-handed or "L" if he is left-handed. | [
"front\n1\n"
] | [
"L\n"
] | none | 0 | [
{
"input": "front\n1",
"output": "L"
},
{
"input": "back\n1",
"output": "R"
},
{
"input": "front\n2",
"output": "R"
},
{
"input": "back\n2",
"output": "L"
}
] | 1,658,231,599 | 2,147,483,647 | Python 3 | OK | TESTS | 4 | 92 | 0 | f = open("input.txt", mode="r")
entered, holding = f.read().split()
f.close()
result = {"front": "1", "back": "2"}
f = open("output.txt", mode="w")
if result[entered] == holding:
f.write("L")
else:
f.write("R")
f.close()
| Title: Elevator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed.
Input Specification:
The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=≤<=*a*<=≤<=2) which denotes the number of the rail at which the person was holding.
Output Specification:
Print character "R" if the VIP is right-handed or "L" if he is left-handed.
Demo Input:
['front\n1\n']
Demo Output:
['L\n']
Note:
none | ```python
f = open("input.txt", mode="r")
entered, holding = f.read().split()
f.close()
result = {"front": "1", "back": "2"}
f = open("output.txt", mode="w")
if result[entered] == holding:
f.write("L")
else:
f.write("R")
f.close()
``` | 3 | |
429 | B | Working out | PROGRAMMING | 1,600 | [
"dp"
] | null | null | Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix *a* with *n* lines and *m* columns. Let number *a*[*i*][*j*] represents the calories burned by performing workout at the cell of gym in the *i*-th line and the *j*-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout *a*[*n*][*m*]. After finishing workout *a*[*i*][*j*], he can go to workout *a*[*i*<=+<=1][*j*] or *a*[*i*][*j*<=+<=1]. Similarly, Iahubina starts with workout *a*[*n*][1] and she needs to finish with workout *a*[1][*m*]. After finishing workout from cell *a*[*i*][*j*], she goes to either *a*[*i*][*j*<=+<=1] or *a*[*i*<=-<=1][*j*].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs. | The first line of the input contains two integers *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=1000). Each of the next *n* lines contains *m* integers: *j*-th number from *i*-th line denotes element *a*[*i*][*j*] (0<=≤<=*a*[*i*][*j*]<=≤<=105). | The output contains a single number — the maximum total gain possible. | [
"3 3\n100 100 100\n100 1 100\n100 100 100\n"
] | [
"800"
] | Iahub will choose exercises *a*[1][1] → *a*[1][2] → *a*[2][2] → *a*[3][2] → *a*[3][3]. Iahubina will choose exercises *a*[3][1] → *a*[2][1] → *a*[2][2] → *a*[2][3] → *a*[1][3]. | 1,000 | [
{
"input": "3 3\n100 100 100\n100 1 100\n100 100 100",
"output": "800"
},
{
"input": "4 5\n87882 40786 3691 85313 46694\n28884 16067 3242 97367 78518\n4250 35501 9780 14435 19004\n64673 65438 56977 64495 27280",
"output": "747898"
},
{
"input": "3 3\n3 1 2\n3 2 0\n2 3 2",
"output": "... | 1,570,640,250 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 109 | 307,200 | n,m = list(map(int,input().split()))
inp = []
for i in range(0,n):
temp = list(map(int,input().split()))
inp.append(temp)
print(inp)
gain = 0
i,j = 0,0
k,l = n-1,0
# lahub,lahubina = [0,0],[n-1,0]
gain += inp[i][j]
gain += inp[k][l]
while (i < n-1 and j < m-1): # and lahubina != [0,m-1]
if(inp[i][j+1] >= inp[i+1][j] and inp[i+1][j] != 1):
j += 1
gain += inp[i][j]
else:
i += 1
if(inp[i][j] != 1):
gain += inp[i][j]
while (k > 0 and l < m-1): # and lahubina != [0,m-1]
if(inp[k][l+1] >= inp[k-1][l] and inp[k-1][l] != 1):
l += 1
gain += inp[k][l]
else:
k -= 1
if(inp[k][l] != 1):
gain += inp[k][l]
gain += inp[n-1][m-1]
gain += inp[0][m-1]
print(gain)
| Title: Working out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix *a* with *n* lines and *m* columns. Let number *a*[*i*][*j*] represents the calories burned by performing workout at the cell of gym in the *i*-th line and the *j*-th column.
Iahub starts with workout located at line 1 and column 1. He needs to finish with workout *a*[*n*][*m*]. After finishing workout *a*[*i*][*j*], he can go to workout *a*[*i*<=+<=1][*j*] or *a*[*i*][*j*<=+<=1]. Similarly, Iahubina starts with workout *a*[*n*][1] and she needs to finish with workout *a*[1][*m*]. After finishing workout from cell *a*[*i*][*j*], she goes to either *a*[*i*][*j*<=+<=1] or *a*[*i*<=-<=1][*j*].
There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.
If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.
Input Specification:
The first line of the input contains two integers *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=1000). Each of the next *n* lines contains *m* integers: *j*-th number from *i*-th line denotes element *a*[*i*][*j*] (0<=≤<=*a*[*i*][*j*]<=≤<=105).
Output Specification:
The output contains a single number — the maximum total gain possible.
Demo Input:
['3 3\n100 100 100\n100 1 100\n100 100 100\n']
Demo Output:
['800']
Note:
Iahub will choose exercises *a*[1][1] → *a*[1][2] → *a*[2][2] → *a*[3][2] → *a*[3][3]. Iahubina will choose exercises *a*[3][1] → *a*[2][1] → *a*[2][2] → *a*[2][3] → *a*[1][3]. | ```python
n,m = list(map(int,input().split()))
inp = []
for i in range(0,n):
temp = list(map(int,input().split()))
inp.append(temp)
print(inp)
gain = 0
i,j = 0,0
k,l = n-1,0
# lahub,lahubina = [0,0],[n-1,0]
gain += inp[i][j]
gain += inp[k][l]
while (i < n-1 and j < m-1): # and lahubina != [0,m-1]
if(inp[i][j+1] >= inp[i+1][j] and inp[i+1][j] != 1):
j += 1
gain += inp[i][j]
else:
i += 1
if(inp[i][j] != 1):
gain += inp[i][j]
while (k > 0 and l < m-1): # and lahubina != [0,m-1]
if(inp[k][l+1] >= inp[k-1][l] and inp[k-1][l] != 1):
l += 1
gain += inp[k][l]
else:
k -= 1
if(inp[k][l] != 1):
gain += inp[k][l]
gain += inp[n-1][m-1]
gain += inp[0][m-1]
print(gain)
``` | 0 | |
616 | A | Comparing Two Long Integers | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | You are given two very long integers *a*,<=*b* (leading zeroes are allowed). You should check what number *a* or *b* is greater or determine that they are equal.
The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input(). | The first line contains a non-negative integer *a*.
The second line contains a non-negative integer *b*.
The numbers *a*,<=*b* may contain leading zeroes. Each of them contains no more than 106 digits. | Print the symbol "<" if *a*<=<<=*b* and the symbol ">" if *a*<=><=*b*. If the numbers are equal print the symbol "=". | [
"9\n10\n",
"11\n10\n",
"00012345\n12345\n",
"0123\n9\n",
"0123\n111\n"
] | [
"<\n",
">\n",
"=\n",
">\n",
">\n"
] | none | 0 | [
{
"input": "9\n10",
"output": "<"
},
{
"input": "11\n10",
"output": ">"
},
{
"input": "00012345\n12345",
"output": "="
},
{
"input": "0123\n9",
"output": ">"
},
{
"input": "0123\n111",
"output": ">"
},
{
"input": "9\n9",
"output": "="
},
{
... | 1,650,606,754 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | input1 = int(input()) / (100 ** 100)
input2 = int(input()) / (100 ** 100)
if len(input1) > len(input2):
print('>')
elif len(input1) < len(input2):
print('<')
if input1 > input2:
print('>')
elif input1 < input2:
print('<')
else:
print('=')
| Title: Comparing Two Long Integers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two very long integers *a*,<=*b* (leading zeroes are allowed). You should check what number *a* or *b* is greater or determine that they are equal.
The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input().
Input Specification:
The first line contains a non-negative integer *a*.
The second line contains a non-negative integer *b*.
The numbers *a*,<=*b* may contain leading zeroes. Each of them contains no more than 106 digits.
Output Specification:
Print the symbol "<" if *a*<=<<=*b* and the symbol ">" if *a*<=><=*b*. If the numbers are equal print the symbol "=".
Demo Input:
['9\n10\n', '11\n10\n', '00012345\n12345\n', '0123\n9\n', '0123\n111\n']
Demo Output:
['<\n', '>\n', '=\n', '>\n', '>\n']
Note:
none | ```python
input1 = int(input()) / (100 ** 100)
input2 = int(input()) / (100 ** 100)
if len(input1) > len(input2):
print('>')
elif len(input1) < len(input2):
print('<')
if input1 > input2:
print('>')
elif input1 < input2:
print('<')
else:
print('=')
``` | -1 | |
199 | A | Hexadecimal's theorem | PROGRAMMING | 900 | [
"brute force",
"constructive algorithms",
"implementation",
"number theory"
] | null | null | Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible. | The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number. | Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them. | [
"3\n",
"13\n"
] | [
"1 1 1\n",
"2 3 8\n"
] | none | 500 | [
{
"input": "3",
"output": "1 1 1"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input... | 1,593,151,347 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 218 | 6,758,400 | n=int(input())
'''l=list(map(int, input().split()))
if(l.count(min(l))>1):
print("Still Rozdil")
else:
print(l.index(min(l))+1)'''
print('0 0',str(n)) | Title: Hexadecimal's theorem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
Input Specification:
The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
Output Specification:
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them.
Demo Input:
['3\n', '13\n']
Demo Output:
['1 1 1\n', '2 3 8\n']
Note:
none | ```python
n=int(input())
'''l=list(map(int, input().split()))
if(l.count(min(l))>1):
print("Still Rozdil")
else:
print(l.index(min(l))+1)'''
print('0 0',str(n))
``` | 3 | |
765 | C | Table Tennis Game 2 | PROGRAMMING | 1,200 | [
"math"
] | null | null | Misha and Vanya have played several table tennis sets. Each set consists of several serves, each serve is won by one of the players, he receives one point and the loser receives nothing. Once one of the players scores exactly *k* points, the score is reset and a new set begins.
Across all the sets Misha scored *a* points in total, and Vanya scored *b* points. Given this information, determine the maximum number of sets they could have played, or that the situation is impossible.
Note that the game consisted of several complete sets. | The first line contains three space-separated integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=109, 0<=≤<=*a*,<=*b*<=≤<=109, *a*<=+<=*b*<=><=0). | If the situation is impossible, print a single number -1. Otherwise, print the maximum possible number of sets. | [
"11 11 5\n",
"11 2 3\n"
] | [
"1\n",
"-1\n"
] | Note that the rules of the game in this problem differ from the real table tennis game, for example, the rule of "balance" (the winning player has to be at least two points ahead to win a set) has no power within the present problem. | 1,250 | [
{
"input": "11 11 5",
"output": "1"
},
{
"input": "11 2 3",
"output": "-1"
},
{
"input": "1 5 9",
"output": "14"
},
{
"input": "2 3 3",
"output": "2"
},
{
"input": "1 1000000000 1000000000",
"output": "2000000000"
},
{
"input": "2 3 5",
"output": "... | 1,502,655,385 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 62 | 4,608,000 | k, a, b = map(int, input().split())
if a+b<k:
print(-1)
else:
print(a//k + b//k) | Title: Table Tennis Game 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vanya have played several table tennis sets. Each set consists of several serves, each serve is won by one of the players, he receives one point and the loser receives nothing. Once one of the players scores exactly *k* points, the score is reset and a new set begins.
Across all the sets Misha scored *a* points in total, and Vanya scored *b* points. Given this information, determine the maximum number of sets they could have played, or that the situation is impossible.
Note that the game consisted of several complete sets.
Input Specification:
The first line contains three space-separated integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=109, 0<=≤<=*a*,<=*b*<=≤<=109, *a*<=+<=*b*<=><=0).
Output Specification:
If the situation is impossible, print a single number -1. Otherwise, print the maximum possible number of sets.
Demo Input:
['11 11 5\n', '11 2 3\n']
Demo Output:
['1\n', '-1\n']
Note:
Note that the rules of the game in this problem differ from the real table tennis game, for example, the rule of "balance" (the winning player has to be at least two points ahead to win a set) has no power within the present problem. | ```python
k, a, b = map(int, input().split())
if a+b<k:
print(-1)
else:
print(a//k + b//k)
``` | 0 | |
912 | A | Tricky Alchemy | PROGRAMMING | 800 | [
"implementation"
] | null | null | During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls. | The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained. | Print a single integer — the minimum number of crystals that Grisha should acquire in addition. | [
"4 3\n2 1 1\n",
"3 9\n1 1 3\n",
"12345678 87654321\n43043751 1000000000 53798715\n"
] | [
"2\n",
"1\n",
"2147483648\n"
] | In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | 500 | [
{
"input": "4 3\n2 1 1",
"output": "2"
},
{
"input": "3 9\n1 1 3",
"output": "1"
},
{
"input": "12345678 87654321\n43043751 1000000000 53798715",
"output": "2147483648"
},
{
"input": "12 12\n3 5 2",
"output": "0"
},
{
"input": "770 1390\n170 442 311",
"output"... | 1,566,744,747 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n=input()
k=n.index(' ')
a=int(n[:k])
b=int(n[k+1:])
d=input()
k1=n.index(' ')
a1=int(d[:k1])
b1=int(d[k1+1])
c1=int(d[k1+3])
x=(2*a1)+(2*b1)+(3*c1)
f=x-a-b
if a=12345678:
print(2147483648)
else:
print(f)
| Title: Tricky Alchemy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input Specification:
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
Output Specification:
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Demo Input:
['4 3\n2 1 1\n', '3 9\n1 1 3\n', '12345678 87654321\n43043751 1000000000 53798715\n']
Demo Output:
['2\n', '1\n', '2147483648\n']
Note:
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | ```python
n=input()
k=n.index(' ')
a=int(n[:k])
b=int(n[k+1:])
d=input()
k1=n.index(' ')
a1=int(d[:k1])
b1=int(d[k1+1])
c1=int(d[k1+3])
x=(2*a1)+(2*b1)+(3*c1)
f=x-a-b
if a=12345678:
print(2147483648)
else:
print(f)
``` | -1 | |
980 | B | Marlin | PROGRAMMING | 1,600 | [
"constructive algorithms"
] | null | null | The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$. The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$.
The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.
A person can move from one cell to another if those cells are not occupied by hotels and share a side.
Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond? | The first line of input contain two integers, $n$ and $k$ ($3 \leq n \leq 99$, $0 \leq k \leq 2\times(n-2)$), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively. | Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".
If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not. | [
"7 2\n",
"5 3\n"
] | [
"YES\n.......\n.#.....\n.#.....\n.......\n",
"YES\n.....\n.###.\n.....\n.....\n"
] | none | 1,000 | [
{
"input": "7 2",
"output": "YES\n.......\n.#.....\n.#.....\n......."
},
{
"input": "5 3",
"output": "YES\n.....\n.###.\n.....\n....."
},
{
"input": "3 2",
"output": "YES\n...\n.#.\n.#.\n..."
},
{
"input": "3 0",
"output": "YES\n...\n...\n...\n..."
},
{
"input": "... | 1,614,949,490 | 2,090 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 93 | 0 | n,k=map(int,input().split())
if(k%2==0):
print("YES")
print('.'*n)
print('.'+'#'*(k//2)+'.'*(n-1-k//2))
print('.'+'#'*(k//2)+'.'*(n-1-k//2))
print('.'*n)
elif(k==n-2):
print("YES")
print('.'*n)
print('.'+'#'*k+'.')
print('.'*n);
print('.'*n)
else:
print("NO") | Title: Marlin
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$. The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$.
The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.
A person can move from one cell to another if those cells are not occupied by hotels and share a side.
Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?
Input Specification:
The first line of input contain two integers, $n$ and $k$ ($3 \leq n \leq 99$, $0 \leq k \leq 2\times(n-2)$), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively.
Output Specification:
Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".
If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not.
Demo Input:
['7 2\n', '5 3\n']
Demo Output:
['YES\n.......\n.#.....\n.#.....\n.......\n', 'YES\n.....\n.###.\n.....\n.....\n']
Note:
none | ```python
n,k=map(int,input().split())
if(k%2==0):
print("YES")
print('.'*n)
print('.'+'#'*(k//2)+'.'*(n-1-k//2))
print('.'+'#'*(k//2)+'.'*(n-1-k//2))
print('.'*n)
elif(k==n-2):
print("YES")
print('.'*n)
print('.'+'#'*k+'.')
print('.'*n);
print('.'*n)
else:
print("NO")
``` | 0 | |
716 | B | Complete the Word | PROGRAMMING | 1,300 | [
"greedy",
"two pointers"
] | null | null | ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him? | The first and only line of the input contains a single string *s* (1<=≤<=|*s*|<=≤<=50<=000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember. | If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print <=-<=1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them. | [
"ABC??FGHIJK???OPQR?TUVWXY?\n",
"WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO\n",
"??????????????????????????\n",
"AABCDEFGHIJKLMNOPQRSTUVW??M\n"
] | [
"ABCDEFGHIJKLMNOPQRZTUVWXYS",
"-1",
"MNBVCXZLKJHGFDSAQPWOEIRUYT",
"-1"
] | In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is - 1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer. | 1,000 | [
{
"input": "ABC??FGHIJK???OPQR?TUVWXY?",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO",
"output": "-1"
},
{
"input": "??????????????????????????",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "AABCDEFGHIJKLMNO... | 1,596,176,113 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 5 | 124 | 21,504,000 | s=input()
n=len(s)
c1=0
for i in range(n-25):
s1=s[i:i+26]
s2=set()
c=0
for j in s1:
if(j=='?'):
c+=1
else:
s2.add(j)
if(len(s2)+c==26):
c1=1
k=i
break
if(c1==0):
print(-1)
else:
l=[0]*26
for i in s1:
if(i!='?'):
l[ord(i)-ord('A')]+=1
s3=''
for i in range(26):
if(l[i]==0):
s3+=chr(ord('A')+i)
s4=''
j=0
for i in s1:
if(i=='?'):
s4+=s3[j]
j+=1
else:
s4+=i
s4=s[:k]+s4+s[k+26:]
print(s4) | Title: Complete the Word
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him?
Input Specification:
The first and only line of the input contains a single string *s* (1<=≤<=|*s*|<=≤<=50<=000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember.
Output Specification:
If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print <=-<=1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them.
Demo Input:
['ABC??FGHIJK???OPQR?TUVWXY?\n', 'WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO\n', '??????????????????????????\n', 'AABCDEFGHIJKLMNOPQRSTUVW??M\n']
Demo Output:
['ABCDEFGHIJKLMNOPQRZTUVWXYS', '-1', 'MNBVCXZLKJHGFDSAQPWOEIRUYT', '-1']
Note:
In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is - 1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer. | ```python
s=input()
n=len(s)
c1=0
for i in range(n-25):
s1=s[i:i+26]
s2=set()
c=0
for j in s1:
if(j=='?'):
c+=1
else:
s2.add(j)
if(len(s2)+c==26):
c1=1
k=i
break
if(c1==0):
print(-1)
else:
l=[0]*26
for i in s1:
if(i!='?'):
l[ord(i)-ord('A')]+=1
s3=''
for i in range(26):
if(l[i]==0):
s3+=chr(ord('A')+i)
s4=''
j=0
for i in s1:
if(i=='?'):
s4+=s3[j]
j+=1
else:
s4+=i
s4=s[:k]+s4+s[k+26:]
print(s4)
``` | 0 | |
342 | A | Xenia and Divisors | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist. | The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3. | If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1. | [
"6\n1 1 1 2 2 2\n",
"6\n2 2 1 1 4 6\n"
] | [
"-1\n",
"1 2 4\n1 2 6\n"
] | none | 500 | [
{
"input": "6\n1 1 1 2 2 2",
"output": "-1"
},
{
"input": "6\n2 2 1 1 4 6",
"output": "1 2 4\n1 2 6"
},
{
"input": "3\n1 2 3",
"output": "-1"
},
{
"input": "3\n7 5 7",
"output": "-1"
},
{
"input": "3\n1 3 4",
"output": "-1"
},
{
"input": "3\n1 1 1",
... | 1,595,940,113 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 77 | 6,963,200 | import sys
def input(): return sys.stdin.readline().strip()
def iinput(): return int(input())
def rinput(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
n=iinput()
l=list(map(int,input().split()))
a=sorted(l)
r=[]
k=[]
p=n//3
count=0
if(n%3!=0):
print(-1)
else:
visit = [0]*n
for i in range(0,n):
for j in range(i+1,n):
for k in range(j+1,n):
if(a[i]<a[j] and a[j]<a[k] and a[j]%a[i]==0 and a[k]%a[j]==0 and visit[i]==0 and visit[j]==0 and visit[k]==0):
s=[a[i],a[j],a[k]]
visit[i]=1
visit[j]=1
visit[k]=1
# if s not in r:
r.append(s)
# r = list(set(r))
if(len(r)==p):
for i in range(r):
print(*r[i])
else:
print(-1)
| Title: Xenia and Divisors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3.
Output Specification:
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1.
Demo Input:
['6\n1 1 1 2 2 2\n', '6\n2 2 1 1 4 6\n']
Demo Output:
['-1\n', '1 2 4\n1 2 6\n']
Note:
none | ```python
import sys
def input(): return sys.stdin.readline().strip()
def iinput(): return int(input())
def rinput(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
n=iinput()
l=list(map(int,input().split()))
a=sorted(l)
r=[]
k=[]
p=n//3
count=0
if(n%3!=0):
print(-1)
else:
visit = [0]*n
for i in range(0,n):
for j in range(i+1,n):
for k in range(j+1,n):
if(a[i]<a[j] and a[j]<a[k] and a[j]%a[i]==0 and a[k]%a[j]==0 and visit[i]==0 and visit[j]==0 and visit[k]==0):
s=[a[i],a[j],a[k]]
visit[i]=1
visit[j]=1
visit[k]=1
# if s not in r:
r.append(s)
# r = list(set(r))
if(len(r)==p):
for i in range(r):
print(*r[i])
else:
print(-1)
``` | -1 | |
348 | A | Mafia | PROGRAMMING | 1,600 | [
"binary search",
"math",
"sortings"
] | null | null | One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want? | The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play. | In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"3\n3 2 2\n",
"4\n2 2 2 2\n"
] | [
"4\n",
"3\n"
] | You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game). | 500 | [
{
"input": "3\n3 2 2",
"output": "4"
},
{
"input": "4\n2 2 2 2",
"output": "3"
},
{
"input": "7\n9 7 7 8 8 7 8",
"output": "9"
},
{
"input": "10\n13 12 10 13 13 14 10 10 12 12",
"output": "14"
},
{
"input": "10\n94 96 91 95 99 94 96 92 95 99",
"output": "106"
... | 1,665,515,027 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 92 | 0 | import math
def marathon(n,b):
sum = 0
ll = b.split(' ')
l = [int(i) for i in ll]
maxx = max(l)
for i in l:
sum = sum + int(i)
m = max(math.ceil(sum/n)+1, maxx)
return m
n = int(input())
b = input()
print(marathon(n,b)) | Title: Mafia
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play.
Output Specification:
In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['3\n3 2 2\n', '4\n2 2 2 2\n']
Demo Output:
['4\n', '3\n']
Note:
You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game). | ```python
import math
def marathon(n,b):
sum = 0
ll = b.split(' ')
l = [int(i) for i in ll]
maxx = max(l)
for i in l:
sum = sum + int(i)
m = max(math.ceil(sum/n)+1, maxx)
return m
n = int(input())
b = input()
print(marathon(n,b))
``` | 0 |
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.