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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
699 | B | One Bomb | PROGRAMMING | 1,400 | [
"implementation"
] | null | null | You are given a description of a depot. It is a rectangular checkered field of *n*<=×<=*m* size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").
You have one bomb. If you lay the bomb at the cell (*x*,<=*y*), then after triggering it will wipe out all walls in the row *x* and all walls in the column *y*.
You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall. | The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the depot field.
The next *n* lines contain *m* symbols "." and "*" each — the description of the field. *j*-th symbol in *i*-th of them stands for cell (*i*,<=*j*). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall. | If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).
Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them. | [
"3 4\n.*..\n....\n.*..\n",
"3 3\n..*\n.*.\n*..\n",
"6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..\n"
] | [
"YES\n1 2\n",
"NO\n",
"YES\n3 3\n"
] | none | 1,000 | [
{
"input": "3 4\n.*..\n....\n.*..",
"output": "YES\n1 2"
},
{
"input": "3 3\n..*\n.*.\n*..",
"output": "NO"
},
{
"input": "6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..",
"output": "YES\n3 3"
},
{
"input": "1 10\n**********",
"output": "YES\n1 1"
},
{
"input": "10... | 1,557,574,189 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 1,000 | 10,854,400 | s=[int(n) for n in input().split()]
z=[]
m=0
for n in range(s[0]):
z.append([n for n in input()])
m+=z[n].count('*')
t=[]
for n in range(len(z[0])):
t.append([])
for k in range(len(z)):
t[n].append(z[k][n])
a=0
l=1
#print(t)
for n in range(len(z)):
for k in range(len(z[n])):
a=z[n].count('*')
b=t[k].count('*')
if (a+b-1)==m:
l=0
print('YES')
print(n+1,k+1)
break
if l==0:
break
else:
l=1
continue
if l==1:
print('NO')
| Title: One Bomb
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a description of a depot. It is a rectangular checkered field of *n*<=×<=*m* size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").
You have one bomb. If you lay the bomb at the cell (*x*,<=*y*), then after triggering it will wipe out all walls in the row *x* and all walls in the column *y*.
You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the depot field.
The next *n* lines contain *m* symbols "." and "*" each — the description of the field. *j*-th symbol in *i*-th of them stands for cell (*i*,<=*j*). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.
Output Specification:
If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).
Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.
Demo Input:
['3 4\n.*..\n....\n.*..\n', '3 3\n..*\n.*.\n*..\n', '6 5\n..*..\n..*..\n*****\n..*..\n..*..\n..*..\n']
Demo Output:
['YES\n1 2\n', 'NO\n', 'YES\n3 3\n']
Note:
none | ```python
s=[int(n) for n in input().split()]
z=[]
m=0
for n in range(s[0]):
z.append([n for n in input()])
m+=z[n].count('*')
t=[]
for n in range(len(z[0])):
t.append([])
for k in range(len(z)):
t[n].append(z[k][n])
a=0
l=1
#print(t)
for n in range(len(z)):
for k in range(len(z[n])):
a=z[n].count('*')
b=t[k].count('*')
if (a+b-1)==m:
l=0
print('YES')
print(n+1,k+1)
break
if l==0:
break
else:
l=1
continue
if l==1:
print('NO')
``` | 0 | |
669 | A | Little Artem and Presents | PROGRAMMING | 800 | [
"math"
] | null | null | Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha? | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday. | Print the maximum possible number of times Artem can give presents to Masha. | [
"1\n",
"2\n",
"3\n",
"4\n"
] | [
"1\n",
"1\n",
"2\n",
"3\n"
] | In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again. | 500 | [
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "3"
},
{
"input": "100",
"output": "67"
},
{
"input": "101",
"output": "67"
},
{
"input": "102",
"output": "68"... | 1,566,237,816 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 124 | 0 | n=int(input())
n=2*n
n=n+1
print(n//3) | Title: Little Artem and Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
Output Specification:
Print the maximum possible number of times Artem can give presents to Masha.
Demo Input:
['1\n', '2\n', '3\n', '4\n']
Demo Output:
['1\n', '1\n', '2\n', '3\n']
Note:
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again. | ```python
n=int(input())
n=2*n
n=n+1
print(n//3)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question. | The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants | Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | 0 | [
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
... | 1,512,589,300 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 62 | 5,529,600 | n = [int(i) for i in input().split()]
p = ['012','013','014','015','123','124','125','234','235','345','023','024','025','034','035','045','134','135','145','245']
try:
for i in p:
if sum([n[int(j)] for j in i]) == sum(n)/2:
print('YES')
3/0
print('NO')
except ZeroDivisionError:
pass | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | ```python
n = [int(i) for i in input().split()]
p = ['012','013','014','015','123','124','125','234','235','345','023','024','025','034','035','045','134','135','145','245']
try:
for i in p:
if sum([n[int(j)] for j in i]) == sum(n)/2:
print('YES')
3/0
print('NO')
except ZeroDivisionError:
pass
``` | 3 | |
915 | C | Permute Digits | PROGRAMMING | 1,700 | [
"dp",
"greedy"
] | null | null | You are given two positive integer numbers *a* and *b*. Permute (change order) of the digits of *a* to construct maximal number not exceeding *b*. No number in input and/or output can start with the digit 0.
It is allowed to leave *a* as it is. | The first line contains integer *a* (1<=≤<=*a*<=≤<=1018). The second line contains integer *b* (1<=≤<=*b*<=≤<=1018). Numbers don't have leading zeroes. It is guaranteed that answer exists. | Print the maximum possible number that is a permutation of digits of *a* and is not greater than *b*. The answer can't have any leading zeroes. It is guaranteed that the answer exists.
The number in the output should have exactly the same length as number *a*. It should be a permutation of digits of *a*. | [
"123\n222\n",
"3921\n10000\n",
"4940\n5000\n"
] | [
"213\n",
"9321\n",
"4940\n"
] | none | 0 | [
{
"input": "123\n222",
"output": "213"
},
{
"input": "3921\n10000",
"output": "9321"
},
{
"input": "4940\n5000",
"output": "4940"
},
{
"input": "23923472834\n23589234723",
"output": "23498743322"
},
{
"input": "102391019\n491010301",
"output": "399211100"
},... | 1,687,777,051 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | import sys
def solve(a,b):
return a
a = input()
b = input()
print(solve(a,b)) | Title: Permute Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two positive integer numbers *a* and *b*. Permute (change order) of the digits of *a* to construct maximal number not exceeding *b*. No number in input and/or output can start with the digit 0.
It is allowed to leave *a* as it is.
Input Specification:
The first line contains integer *a* (1<=≤<=*a*<=≤<=1018). The second line contains integer *b* (1<=≤<=*b*<=≤<=1018). Numbers don't have leading zeroes. It is guaranteed that answer exists.
Output Specification:
Print the maximum possible number that is a permutation of digits of *a* and is not greater than *b*. The answer can't have any leading zeroes. It is guaranteed that the answer exists.
The number in the output should have exactly the same length as number *a*. It should be a permutation of digits of *a*.
Demo Input:
['123\n222\n', '3921\n10000\n', '4940\n5000\n']
Demo Output:
['213\n', '9321\n', '4940\n']
Note:
none | ```python
import sys
def solve(a,b):
return a
a = input()
b = input()
print(solve(a,b))
``` | 0 | |
23 | B | Party | PROGRAMMING | 1,600 | [
"constructive algorithms",
"graphs",
"math"
] | B. Party | 2 | 256 | *n* people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2,<=3,<=...,<=*n*<=-<=1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end? | The first input line contains one number *t* — amount of tests (1<=≤<=*t*<=≤<=105). Each of the following *t* lines contains one integer number *n* (1<=≤<=*n*<=≤<=105). | For each test output in a separate line one number — the maximum amount of people that could stay in the end. | [
"1\n3\n"
] | [
"1\n"
] | none | 0 | [
{
"input": "1\n3",
"output": "1"
}
] | 1,625,663,348 | 2,147,483,647 | Python 3 | OK | TESTS | 2 | 1,526 | 0 | test_case = int(input());
while test_case > 0:
people = int(input());
if people>2:
people=people-2;
else:
people=0;
print(people);
test_case -= 1 | Title: Party
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2,<=3,<=...,<=*n*<=-<=1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input Specification:
The first input line contains one number *t* — amount of tests (1<=≤<=*t*<=≤<=105). Each of the following *t* lines contains one integer number *n* (1<=≤<=*n*<=≤<=105).
Output Specification:
For each test output in a separate line one number — the maximum amount of people that could stay in the end.
Demo Input:
['1\n3\n']
Demo Output:
['1\n']
Note:
none | ```python
test_case = int(input());
while test_case > 0:
people = int(input());
if people>2:
people=people-2;
else:
people=0;
print(people);
test_case -= 1
``` | 3.6185 |
748 | C | Santa Claus and Robot | PROGRAMMING | 1,400 | [
"constructive algorithms",
"math"
] | null | null | Santa Claus has Robot which lives on the infinite grid and can move along its lines. He can also, having a sequence of *m* points *p*1,<=*p*2,<=...,<=*p**m* with integer coordinates, do the following: denote its initial location by *p*0. First, the robot will move from *p*0 to *p*1 along one of the shortest paths between them (please notice that since the robot moves only along the grid lines, there can be several shortest paths). Then, after it reaches *p*1, it'll move to *p*2, again, choosing one of the shortest ways, then to *p*3, and so on, until he has visited all points in the given order. Some of the points in the sequence may coincide, in that case Robot will visit that point several times according to the sequence order.
While Santa was away, someone gave a sequence of points to Robot. This sequence is now lost, but Robot saved the protocol of its unit movements. Please, find the minimum possible length of the sequence. | The first line of input contains the only positive integer *n* (1<=≤<=*n*<=≤<=2·105) which equals the number of unit segments the robot traveled. The second line contains the movements protocol, which consists of *n* letters, each being equal either L, or R, or U, or D. *k*-th letter stands for the direction which Robot traveled the *k*-th unit segment in: L means that it moved to the left, R — to the right, U — to the top and D — to the bottom. Have a look at the illustrations for better explanation. | The only line of input should contain the minimum possible length of the sequence. | [
"4\nRURD\n",
"6\nRRULDD\n",
"26\nRRRULURURUULULLLDLDDRDRDLD\n",
"3\nRLL\n",
"4\nLRLR\n"
] | [
"2\n",
"2\n",
"7\n",
"2\n",
"4\n"
] | The illustrations to the first three tests are given below.
<img class="tex-graphics" src="https://espresso.codeforces.com/832fb8f97a482be815e0f87edde26c9791a0d330.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/119a8ba68772b2c2bf76f2acdc58027f6c5cde1f.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/c7b4534f24cbad48148bcba24bc44f37bf7a2dbf.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The last example illustrates that each point in the sequence should be counted as many times as it is presented in the sequence. | 1,500 | [
{
"input": "4\nRURD",
"output": "2"
},
{
"input": "6\nRRULDD",
"output": "2"
},
{
"input": "26\nRRRULURURUULULLLDLDDRDRDLD",
"output": "7"
},
{
"input": "3\nRLL",
"output": "2"
},
{
"input": "4\nLRLR",
"output": "4"
},
{
"input": "5\nLRDLR",
"outpu... | 1,482,660,900 | 4,200 | Python 3 | OK | TESTS | 32 | 826 | 5,222,400 | class Point:
x = 0
y = 0
def __init__(self, new_x, new_y):
self.x = new_x
self.y = new_y
def is_shortest(a, b, turn):
if a.x == b.x and a.y < b.y:
return not turn[0] and not turn[1] and not turn[3]
elif a.x > b.x and a.y < b.y:
return not turn[0] and not turn[3]
elif a.x > b.x and a.y == b.y:
return not turn[0] and not turn[2] and not turn[3]
elif a.x > b.x and a.y > b.y:
return not turn[0] and not turn[2]
elif a.x == b.x and a.y > b.y:
return not turn[0] and not turn[1] and not turn[2]
elif a.x < b.x and a.y > b.y:
return not turn[1] and not turn[2]
elif a.x < b.x and a.y == b.y:
return not turn[1] and not turn[2] and not turn[3]
elif a.x < b.x and a.y < b.y:
return not turn[1] and not turn[3]
else:
return not turn[0] and not turn[1] and not turn[2] and not turn[3]
#print(is_shortest(Point(2, 2), Point(3, 2), [1, 0, 0, 0]))
input()
point = 0
turn = [0] * 4
A = Point(0, 0)
B = Point(0, 0)
state = Point(0, 0)
s = input().strip()
i = 0
while i < len(s):
c = s[i]
B.x = state.x
B.y = state.y
if c == "R":
B.x += 1
turn[0] = 1
elif c == "L":
B.x -= 1
turn[1] = 1
elif c == "U":
B.y += 1
turn[2] = 1
else:
B.y -= 1
turn[3] = 1
#print("#")
#print(A.x, A.y)
#print(B.x, B.y)
if not is_shortest(A, B, turn):
A.x = state.x
A.y = state.y
#print(A.x, A.y)
point += 1
turn = [0] * 4
else:
state.x = B.x
state.y = B.y
i += 1
print(point + 1) | Title: Santa Claus and Robot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus has Robot which lives on the infinite grid and can move along its lines. He can also, having a sequence of *m* points *p*1,<=*p*2,<=...,<=*p**m* with integer coordinates, do the following: denote its initial location by *p*0. First, the robot will move from *p*0 to *p*1 along one of the shortest paths between them (please notice that since the robot moves only along the grid lines, there can be several shortest paths). Then, after it reaches *p*1, it'll move to *p*2, again, choosing one of the shortest ways, then to *p*3, and so on, until he has visited all points in the given order. Some of the points in the sequence may coincide, in that case Robot will visit that point several times according to the sequence order.
While Santa was away, someone gave a sequence of points to Robot. This sequence is now lost, but Robot saved the protocol of its unit movements. Please, find the minimum possible length of the sequence.
Input Specification:
The first line of input contains the only positive integer *n* (1<=≤<=*n*<=≤<=2·105) which equals the number of unit segments the robot traveled. The second line contains the movements protocol, which consists of *n* letters, each being equal either L, or R, or U, or D. *k*-th letter stands for the direction which Robot traveled the *k*-th unit segment in: L means that it moved to the left, R — to the right, U — to the top and D — to the bottom. Have a look at the illustrations for better explanation.
Output Specification:
The only line of input should contain the minimum possible length of the sequence.
Demo Input:
['4\nRURD\n', '6\nRRULDD\n', '26\nRRRULURURUULULLLDLDDRDRDLD\n', '3\nRLL\n', '4\nLRLR\n']
Demo Output:
['2\n', '2\n', '7\n', '2\n', '4\n']
Note:
The illustrations to the first three tests are given below.
<img class="tex-graphics" src="https://espresso.codeforces.com/832fb8f97a482be815e0f87edde26c9791a0d330.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/119a8ba68772b2c2bf76f2acdc58027f6c5cde1f.png" style="max-width: 100.0%;max-height: 100.0%;"/> <img class="tex-graphics" src="https://espresso.codeforces.com/c7b4534f24cbad48148bcba24bc44f37bf7a2dbf.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The last example illustrates that each point in the sequence should be counted as many times as it is presented in the sequence. | ```python
class Point:
x = 0
y = 0
def __init__(self, new_x, new_y):
self.x = new_x
self.y = new_y
def is_shortest(a, b, turn):
if a.x == b.x and a.y < b.y:
return not turn[0] and not turn[1] and not turn[3]
elif a.x > b.x and a.y < b.y:
return not turn[0] and not turn[3]
elif a.x > b.x and a.y == b.y:
return not turn[0] and not turn[2] and not turn[3]
elif a.x > b.x and a.y > b.y:
return not turn[0] and not turn[2]
elif a.x == b.x and a.y > b.y:
return not turn[0] and not turn[1] and not turn[2]
elif a.x < b.x and a.y > b.y:
return not turn[1] and not turn[2]
elif a.x < b.x and a.y == b.y:
return not turn[1] and not turn[2] and not turn[3]
elif a.x < b.x and a.y < b.y:
return not turn[1] and not turn[3]
else:
return not turn[0] and not turn[1] and not turn[2] and not turn[3]
#print(is_shortest(Point(2, 2), Point(3, 2), [1, 0, 0, 0]))
input()
point = 0
turn = [0] * 4
A = Point(0, 0)
B = Point(0, 0)
state = Point(0, 0)
s = input().strip()
i = 0
while i < len(s):
c = s[i]
B.x = state.x
B.y = state.y
if c == "R":
B.x += 1
turn[0] = 1
elif c == "L":
B.x -= 1
turn[1] = 1
elif c == "U":
B.y += 1
turn[2] = 1
else:
B.y -= 1
turn[3] = 1
#print("#")
#print(A.x, A.y)
#print(B.x, B.y)
if not is_shortest(A, B, turn):
A.x = state.x
A.y = state.y
#print(A.x, A.y)
point += 1
turn = [0] * 4
else:
state.x = B.x
state.y = B.y
i += 1
print(point + 1)
``` | 3 | |
500 | A | New Year Transportation | PROGRAMMING | 1,000 | [
"dfs and similar",
"graphs",
"implementation"
] | null | null | New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system. | The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World. | If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO". | [
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | 500 | [
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 ... | 1,674,648,566 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 1,945,600 | n = [int(m) for m in input().split()]
s = [int(c) for c in input().split()]
t = 0
for i in range(len(s)):
t += s[i]
if t > n[1]:
print("NO")
break
elif t == n[1]:
print("YES")
break
| Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | ```python
n = [int(m) for m in input().split()]
s = [int(c) for c in input().split()]
t = 0
for i in range(len(s)):
t += s[i]
if t > n[1]:
print("NO")
break
elif t == n[1]:
print("YES")
break
``` | 0 | |
429 | A | Xor-tree | PROGRAMMING | 1,300 | [
"dfs and similar",
"trees"
] | null | null | Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init**i*, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.
The goal of the game is to get each node *i* to have value *goal**i*, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*) meaning there is an edge between nodes *u**i* and *v**i*.
The next line contains *n* integer numbers, the *i*-th of them corresponds to *init**i* (*init**i* is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal**i* (*goal**i* is either 0 or 1). | In the first line output an integer number *cnt*, representing the minimal number of operations you perform. Each of the next *cnt* lines should contain an integer *x**i*, representing that you pick a node *x**i*. | [
"10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1\n"
] | [
"2\n4\n7\n"
] | none | 500 | [
{
"input": "10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1",
"output": "2\n4\n7"
},
{
"input": "15\n2 1\n3 2\n4 3\n5 4\n6 5\n7 6\n8 7\n9 8\n10 9\n11 10\n12 11\n13 12\n14 13\n15 14\n0 1 0 0 1 1 1 1 1 1 0 0 0 1 1\n1 1 1 1 0 0 1 1 0 1 0 0 1 1 0",
"output"... | 1,399,858,941 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 1,000 | 30,412,800 | import fileinput
odd_change = 0
even_change = 0
class Node:
def __init__(self, id, value):
self.id = id
self.children = []
self.value = value
self.odd_change = 0
self.even_change = 0
self.is_odd = True
def add_child(self, child):
# print child
self.children.append(child)
def try_flip(self, target):
rv = False
# print str(self.id) + " " + str(self.even_change) + " " + str(self.odd_change)
if not self.is_odd and self.even_change % 2 != 0:
self.flip_value()
elif self.is_odd and self.odd_change % 2 != 0:
self.flip_value()
if not (int(self.value) == int(target)):
self.flip_value()
if not self.is_odd:
self.even_change = self.even_change + 1
else:
self.odd_change = self.odd_change + 1
rv = True
for c in self.children:
c.odd_change = self.odd_change
c.even_change = self.even_change
return rv
def flip_value(self):
if int(self.value) == 1:
self.value = 0
else:
self.value = 1
def __repr__(self):
# return str(self.id) + " " + str(self.value) + " " + str(self.is_odd)
return str(self.value)
full_input = fileinput.input()
n = int(full_input[0])
nodes = {}
for i in range(n):
nodes[i + 1] = Node(i + 1, 0)
for i in range(n - 1):
edge = full_input[i + 1].rstrip().split(' ')
# print "here"
# print int(edge[0])
parent = int(edge[1])
child = int(edge[0])
nodes[parent].add_child(nodes[child])
if nodes[parent].is_odd:
nodes[child].is_odd = False
else:
nodes[child].is_odd = True
orig = full_input[n].rstrip().split(' ')
for i in range(n):
nodes[i + 1].value = orig[i]
# string = bytearray()
switches = []
cnt = 0
target = full_input[n+1].rstrip().split(' ')
for i in range(n):
# print nodes
# print str(nodes[i + 1]) + " " + str(target[i])
if nodes[i + 1].try_flip(target[i]):
# print i + 1
# print nodes
# print ""
value = str(i + 1)
# string.extend(value)
# string.extend("\n")
switches.append(i + 1)
cnt = cnt + 1
# print nodes
print(cnt)
# print switches
# print "\n".join(str(x) for x in switches)
# print(string)
for v in switches:
print(v)
| Title: Xor-tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init**i*, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.
The goal of the game is to get each node *i* to have value *goal**i*, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*) meaning there is an edge between nodes *u**i* and *v**i*.
The next line contains *n* integer numbers, the *i*-th of them corresponds to *init**i* (*init**i* is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal**i* (*goal**i* is either 0 or 1).
Output Specification:
In the first line output an integer number *cnt*, representing the minimal number of operations you perform. Each of the next *cnt* lines should contain an integer *x**i*, representing that you pick a node *x**i*.
Demo Input:
['10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1\n']
Demo Output:
['2\n4\n7\n']
Note:
none | ```python
import fileinput
odd_change = 0
even_change = 0
class Node:
def __init__(self, id, value):
self.id = id
self.children = []
self.value = value
self.odd_change = 0
self.even_change = 0
self.is_odd = True
def add_child(self, child):
# print child
self.children.append(child)
def try_flip(self, target):
rv = False
# print str(self.id) + " " + str(self.even_change) + " " + str(self.odd_change)
if not self.is_odd and self.even_change % 2 != 0:
self.flip_value()
elif self.is_odd and self.odd_change % 2 != 0:
self.flip_value()
if not (int(self.value) == int(target)):
self.flip_value()
if not self.is_odd:
self.even_change = self.even_change + 1
else:
self.odd_change = self.odd_change + 1
rv = True
for c in self.children:
c.odd_change = self.odd_change
c.even_change = self.even_change
return rv
def flip_value(self):
if int(self.value) == 1:
self.value = 0
else:
self.value = 1
def __repr__(self):
# return str(self.id) + " " + str(self.value) + " " + str(self.is_odd)
return str(self.value)
full_input = fileinput.input()
n = int(full_input[0])
nodes = {}
for i in range(n):
nodes[i + 1] = Node(i + 1, 0)
for i in range(n - 1):
edge = full_input[i + 1].rstrip().split(' ')
# print "here"
# print int(edge[0])
parent = int(edge[1])
child = int(edge[0])
nodes[parent].add_child(nodes[child])
if nodes[parent].is_odd:
nodes[child].is_odd = False
else:
nodes[child].is_odd = True
orig = full_input[n].rstrip().split(' ')
for i in range(n):
nodes[i + 1].value = orig[i]
# string = bytearray()
switches = []
cnt = 0
target = full_input[n+1].rstrip().split(' ')
for i in range(n):
# print nodes
# print str(nodes[i + 1]) + " " + str(target[i])
if nodes[i + 1].try_flip(target[i]):
# print i + 1
# print nodes
# print ""
value = str(i + 1)
# string.extend(value)
# string.extend("\n")
switches.append(i + 1)
cnt = cnt + 1
# print nodes
print(cnt)
# print switches
# print "\n".join(str(x) for x in switches)
# print(string)
for v in switches:
print(v)
``` | 0 | |
27 | A | Next Test | PROGRAMMING | 1,200 | [
"implementation",
"sortings"
] | A. Next Test | 2 | 256 | «Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests. | Output the required default value for the next test index. | [
"3\n1 7 2\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "3\n3 4 1",
"output": "2"
},
{
"input": "4\n6 4 3 5",
"output": "1"
},
{
"input": "5\n3 2 1 7 4",
"output": "5"
},
{
"input": "6\n4 1 2 5 3 7",
"output": "6"
},
... | 1,519,051,134 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 124 | 5,632,000 |
n = int(input())
a = list(map(int,input().split()))
d = [0]*3005
for i in a:
d[i-1] = 1
for i in range(1,3002):
if d[i]==0:
print (i+1)
break | Title: Next Test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
Output Specification:
Output the required default value for the next test index.
Demo Input:
['3\n1 7 2\n']
Demo Output:
['3\n']
Note:
none | ```python
n = int(input())
a = list(map(int,input().split()))
d = [0]*3005
for i in a:
d[i-1] = 1
for i in range(1,3002):
if d[i]==0:
print (i+1)
break
``` | 0 |
960 | A | Check the string | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes). | The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'. | Print "YES" or "NO", according to the condition. | [
"aaabccc\n",
"bbacc\n",
"aabc\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | 500 | [
{
"input": "aaabccc",
"output": "YES"
},
{
"input": "bbacc",
"output": "NO"
},
{
"input": "aabc",
"output": "YES"
},
{
"input": "aabbcc",
"output": "YES"
},
{
"input": "aaacccbb",
"output": "NO"
},
{
"input": "abc",
"output": "YES"
},
{
"in... | 1,694,053,623 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 16 | 31 | 0 | strn=str(input())
sortedstrn=sorted(strn)
sortedstrn=''.join(sorted(sortedstrn))
a,b,c=0,0,0
# print((sortedstrn))
if strn==str(sortedstrn):
a=strn.count('a')
b=strn.count('b')
c=strn.count('c')
# print(a,b,c)
if c==a+b or c==a or c==b:
print("Yes")
else:
print("No")
else:
print("No") | Title: Check the string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
Input Specification:
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
Output Specification:
Print "YES" or "NO", according to the condition.
Demo Input:
['aaabccc\n', 'bbacc\n', 'aabc\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | ```python
strn=str(input())
sortedstrn=sorted(strn)
sortedstrn=''.join(sorted(sortedstrn))
a,b,c=0,0,0
# print((sortedstrn))
if strn==str(sortedstrn):
a=strn.count('a')
b=strn.count('b')
c=strn.count('c')
# print(a,b,c)
if c==a+b or c==a or c==b:
print("Yes")
else:
print("No")
else:
print("No")
``` | 0 | |
270 | A | Fancy Fence | PROGRAMMING | 1,100 | [
"geometry",
"implementation",
"math"
] | null | null | Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? | The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. | For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. | [
"3\n30\n60\n90\n"
] | [
"NO\nYES\nYES\n"
] | In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | 500 | [
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,655,959,087 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 3 | 122 | 2,048,000 | t = int(input())
while t>0:
n= int(input())
check = False
for i in range(3, 361):
if n*i == (i-2)*180:
print("YES")
check = True
break
if check == False:
print("NO")
t-=1 | Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | ```python
t = int(input())
while t>0:
n= int(input())
check = False
for i in range(3, 361):
if n*i == (i-2)*180:
print("YES")
check = True
break
if check == False:
print("NO")
t-=1
``` | 3 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,679,399,701 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 46 | 0 | l = list(map(int, input().split()))
print(l[0]//l[2] + 1 + l[1]//l[2] + 1)
| Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
l = list(map(int, input().split()))
print(l[0]//l[2] + 1 + l[1]//l[2] + 1)
``` | 0 |
816 | A | Karen and Morning | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | null | null | Karen is getting ready for a new school day!
It is currently hh:mm, given in a 24-hour format. As you know, Karen loves palindromes, and she believes that it is good luck to wake up when the time is a palindrome.
What is the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome?
Remember that a palindrome is a string that reads the same forwards and backwards. For instance, 05:39 is not a palindrome, because 05:39 backwards is 93:50. On the other hand, 05:50 is a palindrome, because 05:50 backwards is 05:50. | The first and only line of input contains a single string in the format hh:mm (00<=≤<= hh <=≤<=23, 00<=≤<= mm <=≤<=59). | Output a single integer on a line by itself, the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome. | [
"05:39\n",
"13:31\n",
"23:59\n"
] | [
"11\n",
"0\n",
"1\n"
] | In the first test case, the minimum number of minutes Karen should sleep for is 11. She can wake up at 05:50, when the time is a palindrome.
In the second test case, Karen can wake up immediately, as the current time, 13:31, is already a palindrome.
In the third test case, the minimum number of minutes Karen should sleep for is 1 minute. She can wake up at 00:00, when the time is a palindrome. | 500 | [
{
"input": "05:39",
"output": "11"
},
{
"input": "13:31",
"output": "0"
},
{
"input": "23:59",
"output": "1"
},
{
"input": "13:32",
"output": "69"
},
{
"input": "14:40",
"output": "1"
},
{
"input": "14:00",
"output": "41"
},
{
"input": "05:... | 1,593,292,381 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 93 | 6,656,000 | h,m = input().split(":")
r = h[::-1]
ans = 0
if m < r:
ans = int(r) - int(m)
else:
if m > r:
ans = 60 - int(m) + int(r)
else:
ans = 0
print(ans) | Title: Karen and Morning
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Karen is getting ready for a new school day!
It is currently hh:mm, given in a 24-hour format. As you know, Karen loves palindromes, and she believes that it is good luck to wake up when the time is a palindrome.
What is the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome?
Remember that a palindrome is a string that reads the same forwards and backwards. For instance, 05:39 is not a palindrome, because 05:39 backwards is 93:50. On the other hand, 05:50 is a palindrome, because 05:50 backwards is 05:50.
Input Specification:
The first and only line of input contains a single string in the format hh:mm (00<=≤<= hh <=≤<=23, 00<=≤<= mm <=≤<=59).
Output Specification:
Output a single integer on a line by itself, the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome.
Demo Input:
['05:39\n', '13:31\n', '23:59\n']
Demo Output:
['11\n', '0\n', '1\n']
Note:
In the first test case, the minimum number of minutes Karen should sleep for is 11. She can wake up at 05:50, when the time is a palindrome.
In the second test case, Karen can wake up immediately, as the current time, 13:31, is already a palindrome.
In the third test case, the minimum number of minutes Karen should sleep for is 1 minute. She can wake up at 00:00, when the time is a palindrome. | ```python
h,m = input().split(":")
r = h[::-1]
ans = 0
if m < r:
ans = int(r) - int(m)
else:
if m > r:
ans = 60 - int(m) + int(r)
else:
ans = 0
print(ans)
``` | 0 | |
119 | A | Epic Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game. | The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. | If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). | [
"3 5 9\n",
"1 1 100\n"
] | [
"0",
"1"
] | The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | 500 | [
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"inpu... | 1,577,179,980 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 218 | 0 | import math
a,b,c=map(int,input().split());cnt=int(0);
while True:
if cnt==0:
cnt=1;
if math.gcd(a,c)<=c:
c-=math.gcd(a,c)
else:
print(1)
exit(0)
else:
cnt=0
if(math.gcd(b,c)<=c):
c-=math.gcd(b,c)
else:
print(0)
exit(0) | Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | ```python
import math
a,b,c=map(int,input().split());cnt=int(0);
while True:
if cnt==0:
cnt=1;
if math.gcd(a,c)<=c:
c-=math.gcd(a,c)
else:
print(1)
exit(0)
else:
cnt=0
if(math.gcd(b,c)<=c):
c-=math.gcd(b,c)
else:
print(0)
exit(0)
``` | 3 | |
877 | A | Alex and broken contest | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive. | The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem. | Print "YES", if problem is from this contest, and "NO" otherwise. | [
"Alex_and_broken_contest\n",
"NikitaAndString\n",
"Danil_and_Olya\n"
] | [
"NO",
"YES",
"NO"
] | none | 500 | [
{
"input": "Alex_and_broken_contest",
"output": "NO"
},
{
"input": "NikitaAndString",
"output": "YES"
},
{
"input": "Danil_and_Olya",
"output": "NO"
},
{
"input": "Slava____and_the_game",
"output": "YES"
},
{
"input": "Olya_and_energy_drinks",
"output": "YES"
... | 1,612,066,154 | 2,147,483,647 | PyPy 3 | OK | TESTS | 55 | 109 | 0 | s=input()
a="Danil"
b="Olya"
c="Ann"
d="Nikita"
e="Slava"
j=s.count(a)+s.count(b)+s.count(c)+s.count(d)+s.count(e)
if j==1:
print("YES")
else:
print("NO")
| Title: Alex and broken contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input Specification:
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem.
Output Specification:
Print "YES", if problem is from this contest, and "NO" otherwise.
Demo Input:
['Alex_and_broken_contest\n', 'NikitaAndString\n', 'Danil_and_Olya\n']
Demo Output:
['NO', 'YES', 'NO']
Note:
none | ```python
s=input()
a="Danil"
b="Olya"
c="Ann"
d="Nikita"
e="Slava"
j=s.count(a)+s.count(b)+s.count(c)+s.count(d)+s.count(e)
if j==1:
print("YES")
else:
print("NO")
``` | 3 | |
858 | A | k-rounding | PROGRAMMING | 1,100 | [
"brute force",
"math",
"number theory"
] | null | null | For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*. | The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8). | Print the *k*-rounding of *n*. | [
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] | [
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] | none | 750 | [
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output":... | 1,511,377,255 | 2,147,483,647 | PyPy 3 | OK | TESTS | 96 | 108 | 307,200 | n,k=map(int,input().split())
x=10**k
if(n>x):
big=n
small=x
else:
big=x
small=n
mod=-1
while(mod!=0):
mod=big%small
big=small
if(mod!=0):
small=mod
print(small*int(n/small)*int(x/small))
| Title: k-rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none | ```python
n,k=map(int,input().split())
x=10**k
if(n>x):
big=n
small=x
else:
big=x
small=n
mod=-1
while(mod!=0):
mod=big%small
big=small
if(mod!=0):
small=mod
print(small*int(n/small)*int(x/small))
``` | 3 | |
355 | A | Vasya and Digital Root | PROGRAMMING | 1,100 | [
"constructive algorithms",
"implementation"
] | null | null | Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that *S*(*n*) is the sum of digits of number *n*, for example, *S*(4098)<==<=4<=+<=0<=+<=9<=+<=8<==<=21. Then the digital root of number *n* equals to:
1. *dr*(*n*)<==<=*S*(*n*), if *S*(*n*)<=<<=10; 1. *dr*(*n*)<==<=*dr*(<=*S*(*n*)<=), if *S*(*n*)<=≥<=10.
For example, *dr*(4098)<=<==<=<=*dr*(21)<=<==<=<=3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that *dr*(*n*)<=<==<=<=*S*(<=*S*(<=*S*(<=*S*(*n*)<=)<=)<=) (*n*<=≤<=101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers *k* and *d*, find the number consisting of exactly *k* digits (the leading zeroes are not allowed), with digital root equal to *d*, or else state that such number does not exist. | The first line contains two integers *k* and *d* (1<=≤<=*k*<=≤<=1000; 0<=≤<=*d*<=≤<=9). | In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly *k* digits. We assume that number 0 doesn't contain any leading zeroes. | [
"4 4\n",
"5 1\n",
"1 0\n"
] | [
"5881\n",
"36172\n",
"0\n"
] | For the first test sample *dr*(5881) = *dr*(22) = 4.
For the second test sample *dr*(36172) = *dr*(19) = *dr*(10) = 1. | 500 | [
{
"input": "4 4",
"output": "5881"
},
{
"input": "5 1",
"output": "36172"
},
{
"input": "1 0",
"output": "0"
},
{
"input": "8 7",
"output": "49722154"
},
{
"input": "487 0",
"output": "No solution"
},
{
"input": "1000 5",
"output": "854193955406789... | 1,415,381,744 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 62 | 0 | import sys
import math
k, d = [int(x) for x in (sys.stdin.readline()).split()]
if(d == 0 and k == 1):
print(0)
exit()
elif(d == 0):
print("No solution")
exit()
t = []
for i in range(k - 1):
t.append(str(9))
t.append(str(d))
print("".join(t)) | Title: Vasya and Digital Root
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that *S*(*n*) is the sum of digits of number *n*, for example, *S*(4098)<==<=4<=+<=0<=+<=9<=+<=8<==<=21. Then the digital root of number *n* equals to:
1. *dr*(*n*)<==<=*S*(*n*), if *S*(*n*)<=<<=10; 1. *dr*(*n*)<==<=*dr*(<=*S*(*n*)<=), if *S*(*n*)<=≥<=10.
For example, *dr*(4098)<=<==<=<=*dr*(21)<=<==<=<=3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that *dr*(*n*)<=<==<=<=*S*(<=*S*(<=*S*(<=*S*(*n*)<=)<=)<=) (*n*<=≤<=101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers *k* and *d*, find the number consisting of exactly *k* digits (the leading zeroes are not allowed), with digital root equal to *d*, or else state that such number does not exist.
Input Specification:
The first line contains two integers *k* and *d* (1<=≤<=*k*<=≤<=1000; 0<=≤<=*d*<=≤<=9).
Output Specification:
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly *k* digits. We assume that number 0 doesn't contain any leading zeroes.
Demo Input:
['4 4\n', '5 1\n', '1 0\n']
Demo Output:
['5881\n', '36172\n', '0\n']
Note:
For the first test sample *dr*(5881) = *dr*(22) = 4.
For the second test sample *dr*(36172) = *dr*(19) = *dr*(10) = 1. | ```python
import sys
import math
k, d = [int(x) for x in (sys.stdin.readline()).split()]
if(d == 0 and k == 1):
print(0)
exit()
elif(d == 0):
print("No solution")
exit()
t = []
for i in range(k - 1):
t.append(str(9))
t.append(str(d))
print("".join(t))
``` | 3 | |
244 | B | Undoubtedly Lucky Numbers | PROGRAMMING | 1,600 | [
"bitmasks",
"brute force",
"dfs and similar"
] | null | null | Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits *x* and *y*. For example, if *x*<==<=4, and *y*<==<=7, then numbers 47, 744, 4 are lucky.
Let's call a positive integer *a* undoubtedly lucky, if there are such digits *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=9), that the decimal representation of number *a* (without leading zeroes) contains only digits *x* and *y*.
Polycarpus has integer *n*. He wants to know how many positive integers that do not exceed *n*, are undoubtedly lucky. Help him, count this number. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=109) — Polycarpus's number. | Print a single integer that says, how many positive integers that do not exceed *n* are undoubtedly lucky. | [
"10\n",
"123\n"
] | [
"10\n",
"113\n"
] | In the first test sample all numbers that do not exceed 10 are undoubtedly lucky.
In the second sample numbers 102, 103, 104, 105, 106, 107, 108, 109, 120, 123 are not undoubtedly lucky. | 1,000 | [
{
"input": "10",
"output": "10"
},
{
"input": "123",
"output": "113"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "1000",
"output": "352"
},
{
"input": "1000000000",
"output": "40744"
},
{
"input": "999999... | 1,683,990,738 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 530 | 0 | n=int(input())
g=len(str(n))
c=0
def sol(k,l):
global c
if k and int(k)>n:
return
if len(set(k))>2:
return
if l>g:
return
if k:
c+=1
for i in range(0 if k else 1,10):
sol(k+str(i),l+1)
sol("",0)
print(c) | Title: Undoubtedly Lucky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits *x* and *y*. For example, if *x*<==<=4, and *y*<==<=7, then numbers 47, 744, 4 are lucky.
Let's call a positive integer *a* undoubtedly lucky, if there are such digits *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=9), that the decimal representation of number *a* (without leading zeroes) contains only digits *x* and *y*.
Polycarpus has integer *n*. He wants to know how many positive integers that do not exceed *n*, are undoubtedly lucky. Help him, count this number.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=109) — Polycarpus's number.
Output Specification:
Print a single integer that says, how many positive integers that do not exceed *n* are undoubtedly lucky.
Demo Input:
['10\n', '123\n']
Demo Output:
['10\n', '113\n']
Note:
In the first test sample all numbers that do not exceed 10 are undoubtedly lucky.
In the second sample numbers 102, 103, 104, 105, 106, 107, 108, 109, 120, 123 are not undoubtedly lucky. | ```python
n=int(input())
g=len(str(n))
c=0
def sol(k,l):
global c
if k and int(k)>n:
return
if len(set(k))>2:
return
if l>g:
return
if k:
c+=1
for i in range(0 if k else 1,10):
sol(k+str(i),l+1)
sol("",0)
print(c)
``` | 3 | |
628 | B | New Skateboard | PROGRAMMING | 1,300 | [
"dp"
] | null | null | Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. | The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9. | Print integer *a* — the number of substrings of the string *s* that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | [
"124\n",
"04\n",
"5810438174\n"
] | [
"4\n",
"3\n",
"9\n"
] | none | 0 | [
{
"input": "124",
"output": "4"
},
{
"input": "04",
"output": "3"
},
{
"input": "5810438174",
"output": "9"
},
{
"input": "1",
"output": "0"
},
{
"input": "039",
"output": "1"
},
{
"input": "97247",
"output": "6"
},
{
"input": "5810438174",... | 1,613,754,121 | 2,147,483,647 | PyPy 3 | OK | TESTS | 22 | 202 | 10,854,400 | s = input()
s1 = list(s)
count = 0
count += s1.count('0')
count += s1.count('4')
count += s1.count('8')
for i in range(len(s)-1):
if(int(''.join(s[i:i+2]))%4==0):
count += i+1
print(count) | Title: New Skateboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input Specification:
The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9.
Output Specification:
Print integer *a* — the number of substrings of the string *s* that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Demo Input:
['124\n', '04\n', '5810438174\n']
Demo Output:
['4\n', '3\n', '9\n']
Note:
none | ```python
s = input()
s1 = list(s)
count = 0
count += s1.count('0')
count += s1.count('4')
count += s1.count('8')
for i in range(len(s)-1):
if(int(''.join(s[i:i+2]))%4==0):
count += i+1
print(count)
``` | 3 | |
596 | D | Wilbur and Trees | PROGRAMMING | 2,300 | [
"dp",
"math",
"probabilities",
"sortings"
] | null | null | Wilbur the pig really wants to be a beaver, so he decided today to pretend he is a beaver and bite at trees to cut them down.
There are *n* trees located at various positions on a line. Tree *i* is located at position *x**i*. All the given positions of the trees are distinct.
The trees are equal, i.e. each tree has height *h*. Due to the wind, when a tree is cut down, it either falls left with probability *p*, or falls right with probability 1<=-<=*p*. If a tree hits another tree while falling, that tree will fall in the same direction as the tree that hit it. A tree can hit another tree only if the distance between them is strictly less than *h*.
For example, imagine there are 4 trees located at positions 1, 3, 5 and 8, while *h*<==<=3 and the tree at position 1 falls right. It hits the tree at position 3 and it starts to fall too. In it's turn it hits the tree at position 5 and it also starts to fall. The distance between 8 and 5 is exactly 3, so the tree at position 8 will not fall.
As long as there are still trees standing, Wilbur will select either the leftmost standing tree with probability 0.5 or the rightmost standing tree with probability 0.5. Selected tree is then cut down. If there is only one tree remaining, Wilbur always selects it. As the ground is covered with grass, Wilbur wants to know the expected total length of the ground covered with fallen trees after he cuts them all down because he is concerned about his grass-eating cow friends. Please help Wilbur. | The first line of the input contains two integers, *n* (1<=≤<=*n*<=≤<=2000) and *h* (1<=≤<=*h*<=≤<=108) and a real number *p* (0<=≤<=*p*<=≤<=1), given with no more than six decimal places.
The second line of the input contains *n* integers, *x*1,<=*x*2,<=...,<=*x**n* (<=-<=108<=≤<=*x**i*<=≤<=108) in no particular order. | Print a single real number — the expected total length of the ground covered by trees when they have all fallen down. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
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 2 0.500000\n1 2\n",
"4 3 0.4\n4 3 1 2\n"
] | [
"3.250000000\n",
"6.631200000\n"
] | Consider the first example, we have 2 trees with height 2.
1. Both trees falls left. This can either happen with the right tree falling left first, which has <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d0f634778b5b369e93c5e4521921161fd08259cb.png" style="max-width: 100.0%;max-height: 100.0%;"/> probability (also knocking down the left tree), or the left tree can fall left and then the right tree can fall left, which has <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/641d3274da41b4a98609d2f4fad1fa35c9a8a0a1.png" style="max-width: 100.0%;max-height: 100.0%;"/> probability. Total probability is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ef2d54b84025327b541b1c00210b659d1ed21135.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
2. Both trees fall right. This is analogous to (1), so the probability of this happening is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/43481774386d410a1bf5213c8638987fd86ac23a.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
3. The left tree fall left and the right tree falls right. This is the only remaining scenario so it must have <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0166ed98b57be8fa0496443c205adee6e32c4f0b.png" style="max-width: 100.0%;max-height: 100.0%;"/> probability.
Cases 1 and 2 lead to a total of 3 units of ground covered, while case 3 leads to a total of 4 units of ground covered. Thus, the expected value is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/558a70d9a0e95f0b97db05aac6ef64a78d8e80fb.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 2,250 | [] | 1,689,633,920 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | print("_RANDOM_GUESS_1689633920.0478349")# 1689633920.047856 | Title: Wilbur and Trees
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Wilbur the pig really wants to be a beaver, so he decided today to pretend he is a beaver and bite at trees to cut them down.
There are *n* trees located at various positions on a line. Tree *i* is located at position *x**i*. All the given positions of the trees are distinct.
The trees are equal, i.e. each tree has height *h*. Due to the wind, when a tree is cut down, it either falls left with probability *p*, or falls right with probability 1<=-<=*p*. If a tree hits another tree while falling, that tree will fall in the same direction as the tree that hit it. A tree can hit another tree only if the distance between them is strictly less than *h*.
For example, imagine there are 4 trees located at positions 1, 3, 5 and 8, while *h*<==<=3 and the tree at position 1 falls right. It hits the tree at position 3 and it starts to fall too. In it's turn it hits the tree at position 5 and it also starts to fall. The distance between 8 and 5 is exactly 3, so the tree at position 8 will not fall.
As long as there are still trees standing, Wilbur will select either the leftmost standing tree with probability 0.5 or the rightmost standing tree with probability 0.5. Selected tree is then cut down. If there is only one tree remaining, Wilbur always selects it. As the ground is covered with grass, Wilbur wants to know the expected total length of the ground covered with fallen trees after he cuts them all down because he is concerned about his grass-eating cow friends. Please help Wilbur.
Input Specification:
The first line of the input contains two integers, *n* (1<=≤<=*n*<=≤<=2000) and *h* (1<=≤<=*h*<=≤<=108) and a real number *p* (0<=≤<=*p*<=≤<=1), given with no more than six decimal places.
The second line of the input contains *n* integers, *x*1,<=*x*2,<=...,<=*x**n* (<=-<=108<=≤<=*x**i*<=≤<=108) in no particular order.
Output Specification:
Print a single real number — the expected total length of the ground covered by trees when they have all fallen down. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
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 2 0.500000\n1 2\n', '4 3 0.4\n4 3 1 2\n']
Demo Output:
['3.250000000\n', '6.631200000\n']
Note:
Consider the first example, we have 2 trees with height 2.
1. Both trees falls left. This can either happen with the right tree falling left first, which has <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d0f634778b5b369e93c5e4521921161fd08259cb.png" style="max-width: 100.0%;max-height: 100.0%;"/> probability (also knocking down the left tree), or the left tree can fall left and then the right tree can fall left, which has <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/641d3274da41b4a98609d2f4fad1fa35c9a8a0a1.png" style="max-width: 100.0%;max-height: 100.0%;"/> probability. Total probability is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ef2d54b84025327b541b1c00210b659d1ed21135.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
2. Both trees fall right. This is analogous to (1), so the probability of this happening is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/43481774386d410a1bf5213c8638987fd86ac23a.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
3. The left tree fall left and the right tree falls right. This is the only remaining scenario so it must have <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0166ed98b57be8fa0496443c205adee6e32c4f0b.png" style="max-width: 100.0%;max-height: 100.0%;"/> probability.
Cases 1 and 2 lead to a total of 3 units of ground covered, while case 3 leads to a total of 4 units of ground covered. Thus, the expected value is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/558a70d9a0e95f0b97db05aac6ef64a78d8e80fb.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
print("_RANDOM_GUESS_1689633920.0478349")# 1689633920.047856
``` | 0 | |
711 | A | Bus to Udayland | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit? | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details. | If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them. | [
"6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n",
"5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n"
] | [
"YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"NO\n",
"YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n"
] | Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | 500 | [
{
"input": "6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX",
"output": "YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX"
},
{
"input": "4\nXO|OX\nXO|XX\nOX|OX\nXX|OX",
"output": "NO"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO"
... | 1,598,276,458 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 109 | 6,963,200 | n = int(input())
result = []
z = 0
for i in range(n):
place = input()
if place.count('OO') > 0:
if z == 0:
result.append(place.replace('OO', '++', 1))
z += 1
else:
result.append(place)
else:
result.append(place)
if z == 1:
print('YES')
for elem in result:
print(elem, end="\n")
else:
print('NO')
| Title: Bus to Udayland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
Output Specification:
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
Demo Input:
['6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', '4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n', '5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n']
Demo Output:
['YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', 'NO\n', 'YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n']
Note:
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | ```python
n = int(input())
result = []
z = 0
for i in range(n):
place = input()
if place.count('OO') > 0:
if z == 0:
result.append(place.replace('OO', '++', 1))
z += 1
else:
result.append(place)
else:
result.append(place)
if z == 1:
print('YES')
for elem in result:
print(elem, end="\n")
else:
print('NO')
``` | 3 | |
165 | A | Supercentral Point | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | [
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] | [
"2\n",
"1\n"
] | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | 500 | [
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n... | 1,604,874,159 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 280 | 307,200 | n = int(input())
lst = []
for i in range(n):
lst.append(list(map(int, input().split())))
counter = 0
for i in range(n):
up, low, left, right = False, False, False, False
for j in range(n):
if lst[i][0] > lst[j][0] and lst[i][1] == lst[j][1]: right = True
if lst[i][0] < lst[j][0] and lst[i][1] == lst[j][1]: left = True
if lst[i][0] == lst[j][0] and lst[i][1] < lst[j][1]: low = True
if lst[i][0] == lst[j][0] and lst[i][1] > lst[j][1]: up = True
if up and low and right and left:
counter += 1
break
print(counter) | Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | ```python
n = int(input())
lst = []
for i in range(n):
lst.append(list(map(int, input().split())))
counter = 0
for i in range(n):
up, low, left, right = False, False, False, False
for j in range(n):
if lst[i][0] > lst[j][0] and lst[i][1] == lst[j][1]: right = True
if lst[i][0] < lst[j][0] and lst[i][1] == lst[j][1]: left = True
if lst[i][0] == lst[j][0] and lst[i][1] < lst[j][1]: low = True
if lst[i][0] == lst[j][0] and lst[i][1] > lst[j][1]: up = True
if up and low and right and left:
counter += 1
break
print(counter)
``` | 3 | |
581 | A | Vasya the Hipster | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him? | The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got. | Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day. | [
"3 1\n",
"2 3\n",
"7 3\n"
] | [
"1 1\n",
"2 0\n",
"3 2\n"
] | In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day. | 500 | [
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input":... | 1,664,397,157 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 46 | 0 | strs = input()
a,b = tuple(strs.split())
a,b = int(a), int(b)
if a == b:
fashion = a
onecolor = 0
else:
if a > b:
fashion = b
onecolor = (a-b)//2
else:
fashion = a
onecolor = (b-a)//2
print("{} {}".format(fashion,onecolor)) | Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day. | ```python
strs = input()
a,b = tuple(strs.split())
a,b = int(a), int(b)
if a == b:
fashion = a
onecolor = 0
else:
if a > b:
fashion = b
onecolor = (a-b)//2
else:
fashion = a
onecolor = (b-a)//2
print("{} {}".format(fashion,onecolor))
``` | 3 | |
670 | A | Holidays | PROGRAMMING | 900 | [
"brute force",
"constructive algorithms",
"greedy",
"math"
] | null | null | On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars. | The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars. | Print two integers — the minimum possible and the maximum possible number of days off per year on Mars. | [
"14\n",
"2\n"
] | [
"4 4\n",
"0 2\n"
] | In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off. | 500 | [
{
"input": "14",
"output": "4 4"
},
{
"input": "2",
"output": "0 2"
},
{
"input": "1",
"output": "0 1"
},
{
"input": "3",
"output": "0 2"
},
{
"input": "4",
"output": "0 2"
},
{
"input": "5",
"output": "0 2"
},
{
"input": "6",
"output":... | 1,695,392,297 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 198 | 77 | 1,638,400 | def calculate_days_off(n, start_day):
days_off = 0
current_day = start_day
for i in range(1, n+1):
if current_day == 6 or current_day == 7: # Saturday or Sunday
days_off += 1
current_day = (current_day % 7) + 1
return days_off
n = int(input())
min_start_day = 1
max_start_day = 6
min_days_off = calculate_days_off(n, min_start_day)
max_days_off = calculate_days_off(n, max_start_day)
print(min_days_off, max_days_off)
| Title: Holidays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars.
Output Specification:
Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.
Demo Input:
['14\n', '2\n']
Demo Output:
['4 4\n', '0 2\n']
Note:
In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off. | ```python
def calculate_days_off(n, start_day):
days_off = 0
current_day = start_day
for i in range(1, n+1):
if current_day == 6 or current_day == 7: # Saturday or Sunday
days_off += 1
current_day = (current_day % 7) + 1
return days_off
n = int(input())
min_start_day = 1
max_start_day = 6
min_days_off = calculate_days_off(n, min_start_day)
max_days_off = calculate_days_off(n, max_start_day)
print(min_days_off, max_days_off)
``` | 3 | |
44 | E | Anfisa the Monkey | PROGRAMMING | 1,400 | [
"dp"
] | E. Anfisa the Monkey | 2 | 256 | Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into *k* lines not shorter than *a* and not longer than *b*, for the text to resemble human speech more. Help Anfisa. | The first line contains three integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=200, 1<=≤<=*a*<=≤<=*b*<=≤<=200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols. | Print *k* lines, each of which contains no less than *a* and no more than *b* symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes). | [
"3 2 5\nabrakadabra\n",
"4 1 2\nabrakadabra\n"
] | [
"ab\nrakad\nabra\n",
"No solution\n"
] | none | 0 | [
{
"input": "3 2 5\nabrakadabra",
"output": "abra\nkada\nbra"
},
{
"input": "4 1 2\nabrakadabra",
"output": "No solution"
},
{
"input": "3 1 2\nvgnfpo",
"output": "vg\nnf\npo"
},
{
"input": "5 3 4\nvrrdnhazvexzjfv",
"output": "vrr\ndnh\nazv\nexz\njfv"
},
{
"input":... | 1,572,263,553 | 2,147,483,647 | PyPy 3 | OK | TESTS | 32 | 280 | 0 | k,a,b = map(int, input().split())
s=input()
length = len(s)
if length < a*k or length > b*k:
print("No solution")
quit()
parts=[a]*k
length -= a*k
cur=0
while length > 0:
if length < b-a:
parts[cur] += length
length = 0
else:
parts[cur] += b-a
length -= (b-a)
cur += 1
for part in parts:
print(s[:part])
s = s[part:] | Title: Anfisa the Monkey
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into *k* lines not shorter than *a* and not longer than *b*, for the text to resemble human speech more. Help Anfisa.
Input Specification:
The first line contains three integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=200, 1<=≤<=*a*<=≤<=*b*<=≤<=200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols.
Output Specification:
Print *k* lines, each of which contains no less than *a* and no more than *b* symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes).
Demo Input:
['3 2 5\nabrakadabra\n', '4 1 2\nabrakadabra\n']
Demo Output:
['ab\nrakad\nabra\n', 'No solution\n']
Note:
none | ```python
k,a,b = map(int, input().split())
s=input()
length = len(s)
if length < a*k or length > b*k:
print("No solution")
quit()
parts=[a]*k
length -= a*k
cur=0
while length > 0:
if length < b-a:
parts[cur] += length
length = 0
else:
parts[cur] += b-a
length -= (b-a)
cur += 1
for part in parts:
print(s[:part])
s = s[part:]
``` | 3.93 |
296 | A | Yaroslav and Permutations | PROGRAMMING | 1,100 | [
"greedy",
"math"
] | null | null | Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements. | In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise. | [
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs. | 500 | [
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 53... | 1,627,738,166 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 124 | 6,963,200 | n = int(input())
array = list(map(int, input().split()))
hash_map = {}
for x in array:
if x not in hash_map:
hash_map[x] = 1
else:
hash_map[x] += 1
flag = True
for v in hash_map.values():
if v > (n + 1) / 2:
flag = False
break
if flag:
print("YES")
else:
print("NO")
| Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs. | ```python
n = int(input())
array = list(map(int, input().split()))
hash_map = {}
for x in array:
if x not in hash_map:
hash_map[x] = 1
else:
hash_map[x] += 1
flag = True
for v in hash_map.values():
if v > (n + 1) / 2:
flag = False
break
if flag:
print("YES")
else:
print("NO")
``` | 3 | |
950 | A | Left-handers, Right-handers and Ambidexters | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively. | The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training. | Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players. | [
"1 4 2\n",
"5 5 5\n",
"0 2 0\n"
] | [
"6\n",
"14\n",
"0\n"
] | In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand. | 500 | [
{
"input": "1 4 2",
"output": "6"
},
{
"input": "5 5 5",
"output": "14"
},
{
"input": "0 2 0",
"output": "0"
},
{
"input": "30 70 34",
"output": "128"
},
{
"input": "89 32 24",
"output": "112"
},
{
"input": "89 44 77",
"output": "210"
},
{
... | 1,536,071,866 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 108 | 0 | l, r, a = map(int,input().split())
if l > r:
r += a - (l - r)
if a != 0:
l += a//2
r += a//2
else:
l += a - (r - l)
if a != 0:
l += a//2
r += a//2
k = 0
while l > 0 and r > 0:
l -= 1
r -= 1
k += 2
print(k) | Title: Left-handers, Right-handers and Ambidexters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input Specification:
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output Specification:
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Demo Input:
['1 4 2\n', '5 5 5\n', '0 2 0\n']
Demo Output:
['6\n', '14\n', '0\n']
Note:
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand. | ```python
l, r, a = map(int,input().split())
if l > r:
r += a - (l - r)
if a != 0:
l += a//2
r += a//2
else:
l += a - (r - l)
if a != 0:
l += a//2
r += a//2
k = 0
while l > 0 and r > 0:
l -= 1
r -= 1
k += 2
print(k)
``` | 0 | |
195 | B | After Training | PROGRAMMING | 1,300 | [
"data structures",
"implementation",
"math"
] | null | null | After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has *n* balls and *m* baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to *m*, correspondingly. The balls are numbered with numbers from 1 to *n*.
Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which is minimum, where *i* is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.
For every ball print the number of the basket where it will go according to Valeric's scheme.
Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on. | The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of balls and baskets, correspondingly. | Print *n* numbers, one per line. The *i*-th line must contain the number of the basket for the *i*-th ball. | [
"4 3\n",
"3 1\n"
] | [
"2\n1\n3\n2\n",
"1\n1\n1\n"
] | none | 1,000 | [
{
"input": "4 3",
"output": "2\n1\n3\n2"
},
{
"input": "3 1",
"output": "1\n1\n1"
},
{
"input": "10 3",
"output": "2\n1\n3\n2\n1\n3\n2\n1\n3\n2"
},
{
"input": "6 5",
"output": "3\n2\n4\n1\n5\n3"
},
{
"input": "2 6",
"output": "3\n4"
},
{
"input": "5 2"... | 1,648,838,445 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 6,758,400 | def chooseBasket(x, baskets, m):
smaller = -1
smallerList = []
for i in range(m):
if len(baskets[i]) < smaller or smaller == -1:
smallerList = [i]
smaller = len(baskets[i])
elif len(baskets[i]) == smaller:
smallerList.append(i)
if len(smallerList) == 1:
baskets[smallerList[0]].append(x)
return
closest = -1
closestList = []
for small in smallerList:
aux = abs(0.5*(m+1) - (small+1))
if aux < closest or closest == -1:
closestList = [small]
closest = aux
elif aux == closest:
closestList.append(small)
if len(closestList) == 1:
baskets[closestList[0]].append(x)
return
smaller = -1
for i in range(len(closestList)):
if closestList[i] < smaller or smaller == -1:
smaller = closestList[i]
baskets[smaller].append(x)
n,m = map(int, input().split())
baskets = []
for i in range(m):
baskets.append([])
for i in range(n):
chooseBasket(i, baskets, m)
outputList = []
for i in range(m):
for ball in baskets[i]:
outputList.append([ball, i+1])
outputList = sorted(outputList, key=lambda el : el[0])
for x in outputList:
print(x[1])
| Title: After Training
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has *n* balls and *m* baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to *m*, correspondingly. The balls are numbered with numbers from 1 to *n*.
Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which is minimum, where *i* is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.
For every ball print the number of the basket where it will go according to Valeric's scheme.
Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on.
Input Specification:
The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of balls and baskets, correspondingly.
Output Specification:
Print *n* numbers, one per line. The *i*-th line must contain the number of the basket for the *i*-th ball.
Demo Input:
['4 3\n', '3 1\n']
Demo Output:
['2\n1\n3\n2\n', '1\n1\n1\n']
Note:
none | ```python
def chooseBasket(x, baskets, m):
smaller = -1
smallerList = []
for i in range(m):
if len(baskets[i]) < smaller or smaller == -1:
smallerList = [i]
smaller = len(baskets[i])
elif len(baskets[i]) == smaller:
smallerList.append(i)
if len(smallerList) == 1:
baskets[smallerList[0]].append(x)
return
closest = -1
closestList = []
for small in smallerList:
aux = abs(0.5*(m+1) - (small+1))
if aux < closest or closest == -1:
closestList = [small]
closest = aux
elif aux == closest:
closestList.append(small)
if len(closestList) == 1:
baskets[closestList[0]].append(x)
return
smaller = -1
for i in range(len(closestList)):
if closestList[i] < smaller or smaller == -1:
smaller = closestList[i]
baskets[smaller].append(x)
n,m = map(int, input().split())
baskets = []
for i in range(m):
baskets.append([])
for i in range(n):
chooseBasket(i, baskets, m)
outputList = []
for i in range(m):
for ball in baskets[i]:
outputList.append([ball, i+1])
outputList = sorted(outputList, key=lambda el : el[0])
for x in outputList:
print(x[1])
``` | 0 | |
570 | B | Simple Game | PROGRAMMING | 1,300 | [
"constructive algorithms",
"games",
"greedy",
"implementation",
"math"
] | null | null | One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive). | The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively. | Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them. | [
"3 1\n",
"4 3\n"
] | [
"2",
"2"
] | In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less. | 1,000 | [
{
"input": "3 1",
"output": "2"
},
{
"input": "4 3",
"output": "2"
},
{
"input": "5 5",
"output": "4"
},
{
"input": "10 5",
"output": "6"
},
{
"input": "20 13",
"output": "12"
},
{
"input": "51 1",
"output": "2"
},
{
"input": "100 50",
... | 1,527,497,775 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 77 | 0 | a = list(input().split())
n = int (a[0])
m = int (a[1])
if n%2==0:
if m<n/2:
print(int (m+1))
else:
print(int (m-1))
else:
if m<=n//2:
print(int (m+1))
else:
print(int (m-1)) | Title: Simple Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive).
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively.
Output Specification:
Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.
Demo Input:
['3 1\n', '4 3\n']
Demo Output:
['2', '2']
Note:
In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less. | ```python
a = list(input().split())
n = int (a[0])
m = int (a[1])
if n%2==0:
if m<n/2:
print(int (m+1))
else:
print(int (m-1))
else:
if m<=n//2:
print(int (m+1))
else:
print(int (m-1))
``` | 0 | |
80 | A | Panoramix's Prediction | PROGRAMMING | 800 | [
"brute force"
] | A. Panoramix's Prediction | 2 | 256 | A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix? | The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4. | Print YES, if *m* is the next prime number after *n*, or NO otherwise. | [
"3 5\n",
"7 11\n",
"7 9\n"
] | [
"YES",
"YES",
"NO"
] | none | 500 | [
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
... | 1,692,386,295 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 28 | 92 | 0 | v,v1=map(int,input().split());c=v+1;c1=v1
while c%2==0 or (c%3==0 and c!=3) or (c%5==0 and c!=5) or (c%7==0 and c1!=7):
c+=1
if c==c1 and c<=50:print("YES")
else:print("NO") | Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none | ```python
v,v1=map(int,input().split());c=v+1;c1=v1
while c%2==0 or (c%3==0 and c!=3) or (c%5==0 and c!=5) or (c%7==0 and c1!=7):
c+=1
if c==c1 and c<=50:print("YES")
else:print("NO")
``` | 0 |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 6... | 1,670,326,206 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | n=int(input())
d={}
for x in range(3):
d[x]=set(input().split())
print(int(*(d[0]-d[1])))
print(int(*(d[1]-d[2]))) | Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
n=int(input())
d={}
for x in range(3):
d[x]=set(input().split())
print(int(*(d[0]-d[1])))
print(int(*(d[1]-d[2])))
``` | 0 | |
908 | A | New Year and Counting Cards | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true. | The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit. | Print a single integer, the minimum number of cards you must turn over to verify your claim. | [
"ee\n",
"z\n",
"0ay1\n"
] | [
"2\n",
"0\n",
"2\n"
] | In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | 500 | [
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01... | 1,618,191,270 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 77 | 0 | string = str(input())
counter = 0
l = ['1','3','5','7','9','a','i','o','u','e']
for ch in string:
if ch in l:
counter += 1
print(counter)
| Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | ```python
string = str(input())
counter = 0
l = ['1','3','5','7','9','a','i','o','u','e']
for ch in string:
if ch in l:
counter += 1
print(counter)
``` | 3 | |
79 | B | Colorful Field | PROGRAMMING | 1,400 | [
"implementation",
"sortings"
] | B. Colorful Field | 2 | 256 | Fox Ciel saw a large field while she was on a bus. The field was a *n*<=×<=*m* rectangle divided into 1<=×<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=→<=...<=→<=(1,<=*m*)<=→<=(2,<=1)<=→<=...<=→<=(2,<=*m*)<=→<=...<=→<=(*n*,<=1)<=→<=...<=→<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells. | In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=≤<=*n*<=≤<=4·104,<=1<=≤<=*m*<=≤<=4·104,<=1<=≤<=*k*<=≤<=103,<=1<=≤<=*t*<=≤<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*n*,<=1<=≤<=*b*<=≤<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*j*<=≤<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*). | For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes. | [
"4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n"
] | [
"Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n"
] | The sample corresponds to the figure in the statement. | 1,000 | [
{
"input": "4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1",
"output": "Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots"
},
{
"input": "2 3 2 2\n1 1\n2 2\n2 1\n2 2",
"output": "Grapes\nWaste"
},
{
"input": "31 31 31 4\n4 9\n16 27\n11 29\n8 28\n11 2\n10 7\n22 6\n1 25\n14 8... | 1,628,612,638 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 434 | 6,963,200 | n,m,k,t=map(int,input().split())
plant=['Grapes','Carrots','Kiwis']
waste=[]
for i in range(0,k):
x,y=map(int,input().split())
waste.append((x-1)*m+(y-1))
for i in range(0,t):
x,y=map(int,input().split())
index=(x-1)*m+(y-1)
if index in waste:
print("Waste")
else:
num=index+1
for j in waste:
if j<index:
num=num-1
print(plant[num%3])
| Title: Colorful Field
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Fox Ciel saw a large field while she was on a bus. The field was a *n*<=×<=*m* rectangle divided into 1<=×<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=→<=...<=→<=(1,<=*m*)<=→<=(2,<=1)<=→<=...<=→<=(2,<=*m*)<=→<=...<=→<=(*n*,<=1)<=→<=...<=→<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells.
Input Specification:
In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=≤<=*n*<=≤<=4·104,<=1<=≤<=*m*<=≤<=4·104,<=1<=≤<=*k*<=≤<=103,<=1<=≤<=*t*<=≤<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*n*,<=1<=≤<=*b*<=≤<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*j*<=≤<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*).
Output Specification:
For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes.
Demo Input:
['4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n']
Demo Output:
['Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n']
Note:
The sample corresponds to the figure in the statement. | ```python
n,m,k,t=map(int,input().split())
plant=['Grapes','Carrots','Kiwis']
waste=[]
for i in range(0,k):
x,y=map(int,input().split())
waste.append((x-1)*m+(y-1))
for i in range(0,t):
x,y=map(int,input().split())
index=(x-1)*m+(y-1)
if index in waste:
print("Waste")
else:
num=index+1
for j in waste:
if j<index:
num=num-1
print(plant[num%3])
``` | 3.87853 |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,623,655,348 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 77 | 0 | s = ''
s1,s2 = input(), input()
for i,j in zip(s1,s2):
if i != j:
s += '1'
else:
s += '0'
print(s)
| Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
s = ''
s1,s2 = input(), input()
for i,j in zip(s1,s2):
if i != j:
s += '1'
else:
s += '0'
print(s)
``` | 3.98075 |
501 | B | Misha and Changing Handles | PROGRAMMING | 1,100 | [
"data structures",
"dsu",
"strings"
] | null | null | Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that. | The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone. | In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description. | [
"5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n"
] | [
"3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n"
] | none | 500 | [
{
"input": "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov",
"output": "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123"
},
{
"input": "1\nMisha Vasya",
"output": "1\nMisha Vasya"
},
{
"input": "10\na b\nb c\nc d\nd... | 1,625,535,858 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 77 | 0 | times = int(input())
old_to_new = {}
new_to_old = {}
for _ in range(times):
(old, new) = input().split(" ")
if old in new_to_old:
old_to_new[new_to_old[old]] = new
new_to_old[new] = new_to_old[old]
else:
old_to_new[old] = new
new_to_old[new] = old
print(len(old_to_new))
for k in old_to_new:
print(f"{k} {old_to_new[k]}") | Title: Misha and Changing Handles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input Specification:
The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.
Output Specification:
In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Demo Input:
['5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n']
Demo Output:
['3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n']
Note:
none | ```python
times = int(input())
old_to_new = {}
new_to_old = {}
for _ in range(times):
(old, new) = input().split(" ")
if old in new_to_old:
old_to_new[new_to_old[old]] = new
new_to_old[new] = new_to_old[old]
else:
old_to_new[old] = new
new_to_old[new] = old
print(len(old_to_new))
for k in old_to_new:
print(f"{k} {old_to_new[k]}")
``` | 3 | |
538 | B | Quasi Binary | PROGRAMMING | 1,400 | [
"constructive algorithms",
"dp",
"greedy",
"implementation"
] | null | null | A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106). | In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them. | [
"9\n",
"32\n"
] | [
"9\n1 1 1 1 1 1 1 1 1 \n",
"3\n10 11 11 \n"
] | none | 1,000 | [
{
"input": "9",
"output": "9\n1 1 1 1 1 1 1 1 1 "
},
{
"input": "32",
"output": "3\n10 11 11 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "415",
"output": "5\n1 101 101 101 111 "
},
{
"input": "10011",
"output": "1\n10011 "
},
{
"input": "10201... | 1,443,185,209 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 61 | 0 | #In the name of Allah
from sys import stdin, stdout
input = stdin.readline
a = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000]
n = int(input())
def f(n):
for i in range(64):
if a[i] > n:
i -= 1
break
return i
ans = []
while n:
ans.append(a[f(n)])
n -= ans[-1]
stdout.write(str(len(ans)) + "\n")
stdout.write(" ".join(map(str, ans)))
| Title: Quasi Binary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106).
Output Specification:
In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Demo Input:
['9\n', '32\n']
Demo Output:
['9\n1 1 1 1 1 1 1 1 1 \n', '3\n10 11 11 \n']
Note:
none | ```python
#In the name of Allah
from sys import stdin, stdout
input = stdin.readline
a = [1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000]
n = int(input())
def f(n):
for i in range(64):
if a[i] > n:
i -= 1
break
return i
ans = []
while n:
ans.append(a[f(n)])
n -= ans[-1]
stdout.write(str(len(ans)) + "\n")
stdout.write(" ".join(map(str, ans)))
``` | 0 | |
868 | B | Race Against Time | PROGRAMMING | 1,400 | [
"implementation"
] | null | null | Have you ever tried to explain to the coordinator, why it is eight hours to the contest and not a single problem has been prepared yet? Misha had. And this time he has a really strong excuse: he faced a space-time paradox! Space and time replaced each other.
The entire universe turned into an enormous clock face with three hands — hour, minute, and second. Time froze, and clocks now show the time *h* hours, *m* minutes, *s* seconds.
Last time Misha talked with the coordinator at *t*1 o'clock, so now he stands on the number *t*1 on the clock face. The contest should be ready by *t*2 o'clock. In the terms of paradox it means that Misha has to go to number *t*2 somehow. Note that he doesn't have to move forward only: in these circumstances time has no direction.
Clock hands are very long, and Misha cannot get round them. He also cannot step over as it leads to the collapse of space-time. That is, if hour clock points 12 and Misha stands at 11 then he cannot move to 1 along the top arc. He has to follow all the way round the clock center (of course, if there are no other hands on his way).
Given the hands' positions, *t*1, and *t*2, find if Misha can prepare the contest on time (or should we say on space?). That is, find if he can move from *t*1 to *t*2 by the clock face. | Five integers *h*, *m*, *s*, *t*1, *t*2 (1<=≤<=*h*<=≤<=12, 0<=≤<=*m*,<=*s*<=≤<=59, 1<=≤<=*t*1,<=*t*2<=≤<=12, *t*1<=≠<=*t*2).
Misha's position and the target time do not coincide with the position of any hand. | Print "YES" (quotes for clarity), if Misha can prepare the contest on time, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"12 30 45 3 11\n",
"12 0 1 12 1\n",
"3 47 0 4 9\n"
] | [
"NO\n",
"YES\n",
"YES\n"
] | The three examples are shown on the pictures below from left to right. The starting position of Misha is shown with green, the ending position is shown with pink. Note that the positions of the hands on the pictures are not exact, but are close to the exact and the answer is the same. | 500 | [
{
"input": "12 30 45 3 11",
"output": "NO"
},
{
"input": "12 0 1 12 1",
"output": "YES"
},
{
"input": "3 47 0 4 9",
"output": "YES"
},
{
"input": "10 22 59 6 10",
"output": "YES"
},
{
"input": "3 1 13 12 3",
"output": "NO"
},
{
"input": "11 19 28 9 10"... | 1,525,127,780 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 77 | 7,065,600 | x=list(map(int,input().split()))
x[0]=int((x[0]*(60/12))) % 60
m=x[1]
s=x[2]
x[3]=int((x[3]*(60/12))) % 60
x[4]=int((x[4]*(60/12))) % 60
z=[x[1],x[2],x[0]]
z.sort()
x1=[]
x2=[]
for i in range(1,3):
if z[i-1]==z[i]:
del(z[i])
break
if len(z)==3:
if x[3]>z[0] and x[3]<z[1]:
x1.append(1)
if x[3]>z[1] and x[3]<z[2]:
x1.append(2)
if x[3]==z[0]:
x1.append(1)
x1.append(3)
if x[3]==z[1]:
x1.append(2)
x1.append(1)
if x[3]==z[2]:
x1.append(2)
x1.append(3)
elif len(x1)==0:
x1.append(3)
if x[4]>z[0] and x[4]<z[1]:
x2.append(1)
if x[4]>z[1] and x[4]<z[2]:
x2.append(2)
if x[4]==z[0]:
x2.append(1)
x2.append(3)
if x[4]==z[1]:
x2.append(2)
x2.append(1)
if x[4]==z[2]:
x2.append(2)
x2.append(3)
elif len(x2)==0:
x2.append(3)
else:
if x[3]>z[0] and x[3]<z[1]:
x1.append(1)
if x[3]==z[0]:
x1.append(1)
x1.append(2)
if x[3]==z[1]:
x1.append(1)
x1.append(2)
elif len(x)==0:
x1.append(2)
if x[4]>z[0] and x[4]<z[1]:
x1.append(1)
if x[4]==z[0]:
x2.append(1)
x2.append(2)
if x[4]==z[1]:
x2.append(1)
x2.append(2)
elif len(x2)==0:
x2.append(2)
d=x1
f=len(x1)
s='NO'
if len(x1)<len(x2):
d=x2
f=len(x2)
x2=x1
for i in range(f):
if d[i]in x2:
s='YES'
print(s) | Title: Race Against Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Have you ever tried to explain to the coordinator, why it is eight hours to the contest and not a single problem has been prepared yet? Misha had. And this time he has a really strong excuse: he faced a space-time paradox! Space and time replaced each other.
The entire universe turned into an enormous clock face with three hands — hour, minute, and second. Time froze, and clocks now show the time *h* hours, *m* minutes, *s* seconds.
Last time Misha talked with the coordinator at *t*1 o'clock, so now he stands on the number *t*1 on the clock face. The contest should be ready by *t*2 o'clock. In the terms of paradox it means that Misha has to go to number *t*2 somehow. Note that he doesn't have to move forward only: in these circumstances time has no direction.
Clock hands are very long, and Misha cannot get round them. He also cannot step over as it leads to the collapse of space-time. That is, if hour clock points 12 and Misha stands at 11 then he cannot move to 1 along the top arc. He has to follow all the way round the clock center (of course, if there are no other hands on his way).
Given the hands' positions, *t*1, and *t*2, find if Misha can prepare the contest on time (or should we say on space?). That is, find if he can move from *t*1 to *t*2 by the clock face.
Input Specification:
Five integers *h*, *m*, *s*, *t*1, *t*2 (1<=≤<=*h*<=≤<=12, 0<=≤<=*m*,<=*s*<=≤<=59, 1<=≤<=*t*1,<=*t*2<=≤<=12, *t*1<=≠<=*t*2).
Misha's position and the target time do not coincide with the position of any hand.
Output Specification:
Print "YES" (quotes for clarity), if Misha can prepare the contest on time, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['12 30 45 3 11\n', '12 0 1 12 1\n', '3 47 0 4 9\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n']
Note:
The three examples are shown on the pictures below from left to right. The starting position of Misha is shown with green, the ending position is shown with pink. Note that the positions of the hands on the pictures are not exact, but are close to the exact and the answer is the same. | ```python
x=list(map(int,input().split()))
x[0]=int((x[0]*(60/12))) % 60
m=x[1]
s=x[2]
x[3]=int((x[3]*(60/12))) % 60
x[4]=int((x[4]*(60/12))) % 60
z=[x[1],x[2],x[0]]
z.sort()
x1=[]
x2=[]
for i in range(1,3):
if z[i-1]==z[i]:
del(z[i])
break
if len(z)==3:
if x[3]>z[0] and x[3]<z[1]:
x1.append(1)
if x[3]>z[1] and x[3]<z[2]:
x1.append(2)
if x[3]==z[0]:
x1.append(1)
x1.append(3)
if x[3]==z[1]:
x1.append(2)
x1.append(1)
if x[3]==z[2]:
x1.append(2)
x1.append(3)
elif len(x1)==0:
x1.append(3)
if x[4]>z[0] and x[4]<z[1]:
x2.append(1)
if x[4]>z[1] and x[4]<z[2]:
x2.append(2)
if x[4]==z[0]:
x2.append(1)
x2.append(3)
if x[4]==z[1]:
x2.append(2)
x2.append(1)
if x[4]==z[2]:
x2.append(2)
x2.append(3)
elif len(x2)==0:
x2.append(3)
else:
if x[3]>z[0] and x[3]<z[1]:
x1.append(1)
if x[3]==z[0]:
x1.append(1)
x1.append(2)
if x[3]==z[1]:
x1.append(1)
x1.append(2)
elif len(x)==0:
x1.append(2)
if x[4]>z[0] and x[4]<z[1]:
x1.append(1)
if x[4]==z[0]:
x2.append(1)
x2.append(2)
if x[4]==z[1]:
x2.append(1)
x2.append(2)
elif len(x2)==0:
x2.append(2)
d=x1
f=len(x1)
s='NO'
if len(x1)<len(x2):
d=x2
f=len(x2)
x2=x1
for i in range(f):
if d[i]in x2:
s='YES'
print(s)
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not. | In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly. | For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower). | [
"6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n"
] | [
"Yes\nYes\nYes\nNo\nNo\nYes\n"
] | First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3. | 0 | [
{
"input": "6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000",
"output": "Yes\nYes\nYes\nNo\nNo\nYes"
},
{
"input": "3\n1 1\n8 27\n1000 1331",
"output": "Yes\nNo\nNo"
},
{
"input": "1\n12004 18012002",
"output": "Yes"
},
{
"input": "1\n3331 11095561",
"output": "Yes... | 1,608,538,312 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 1,000 | 117,043,200 |
def CreateCubes(n):
cant=n**3+1
cubes={}
for x in range(1,n+1):
cubes[x**3] = x
return cubes
def TheMeaninglessGame():
n = int(input())
cubes=CreateCubes(1000000)
for x in range(n):
i=input().split()
a=int(i[0])
b=int(i[1])
r=cubes.get(a*b)
if not r:
print('No')
continue
X_1=a%r
X_2=b%r
bl=X_1==X_2==0
if bl:print('Yes')
else: print('No')
TheMeaninglessGame()
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input Specification:
In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly.
Output Specification:
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n']
Demo Output:
['Yes\nYes\nYes\nNo\nNo\nYes\n']
Note:
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3. | ```python
def CreateCubes(n):
cant=n**3+1
cubes={}
for x in range(1,n+1):
cubes[x**3] = x
return cubes
def TheMeaninglessGame():
n = int(input())
cubes=CreateCubes(1000000)
for x in range(n):
i=input().split()
a=int(i[0])
b=int(i[1])
r=cubes.get(a*b)
if not r:
print('No')
continue
X_1=a%r
X_2=b%r
bl=X_1==X_2==0
if bl:print('Yes')
else: print('No')
TheMeaninglessGame()
``` | 0 | |
774 | J | Stepan's Series | PROGRAMMING | 2,000 | [
"*special",
"dp"
] | null | null | Well, the series which Stepan watched for a very long time, ended. In total, the series had *n* episodes. For each of them, Stepan remembers either that he definitely has watched it, or that he definitely hasn't watched it, or he is unsure, has he watched this episode or not.
Stepan's dissatisfaction is the maximum number of consecutive series that Stepan did not watch.
Your task is to determine according to Stepan's memories if his dissatisfaction could be exactly equal to *k*. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=*n*) — the number of episodes in the series and the dissatisfaction which should be checked.
The second line contains the sequence which consists of *n* symbols "Y", "N" and "?". If the *i*-th symbol equals "Y", Stepan remembers that he has watched the episode number *i*. If the *i*-th symbol equals "N", Stepan remembers that he hasn't watched the epizode number *i*. If the *i*-th symbol equals "?", Stepan doesn't exactly remember if he has watched the episode number *i* or not. | If Stepan's dissatisfaction can be exactly equal to *k*, then print "YES" (without qoutes). Otherwise print "NO" (without qoutes). | [
"5 2\nNYNNY\n",
"6 1\n????NN\n"
] | [
"YES\n",
"NO\n"
] | In the first test Stepan remembers about all the episodes whether he has watched them or not. His dissatisfaction is 2, because he hasn't watch two episodes in a row — the episode number 3 and the episode number 4. The answer is "YES", because *k* = 2.
In the second test *k* = 1, Stepan's dissatisfaction is greater than or equal to 2 (because he remembers that he hasn't watch at least two episodes in a row — number 5 and number 6), even if he has watched the episodes from the first to the fourth, inclusive. | 0 | [
{
"input": "5 2\nNYNNY",
"output": "YES"
},
{
"input": "6 1\n????NN",
"output": "NO"
},
{
"input": "100 8\nNYNNY?YNNNNNN?NNNNNYNY?YYNYNN?NNNY??NNYNYNNNYNNNYNNNNNNNNY?NNNYNYN?NNNY?YY?NNYNN?NNNYNNYNNYN?NNYNYNN",
"output": "YES"
},
{
"input": "10 1\nNY???NY?Y?",
"output": "Y... | 1,492,502,454 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 342 | 12,288,000 | from sys import stdin
from functools import lru_cache
from collections import defaultdict
infile = stdin
N, K = map(int, infile.readline().split())
watched = infile.readline().strip()
mem = set()
mem.add((0,0,0))
for i, letter in enumerate(watched):
for d in range(K+1):
for c in range(K+1):
if (i,d,c) in mem:
if letter in ('N', '?'):
mem.add((i+1, max(d, c+1), c+1))
if letter in ('Y', '?'):
mem.add((i+1, d, 0))
def good():
for i,d,c in mem:
if i == N and d == K:
return 'YES'
return 'NO'
print(good()) | Title: Stepan's Series
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Well, the series which Stepan watched for a very long time, ended. In total, the series had *n* episodes. For each of them, Stepan remembers either that he definitely has watched it, or that he definitely hasn't watched it, or he is unsure, has he watched this episode or not.
Stepan's dissatisfaction is the maximum number of consecutive series that Stepan did not watch.
Your task is to determine according to Stepan's memories if his dissatisfaction could be exactly equal to *k*.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=*n*) — the number of episodes in the series and the dissatisfaction which should be checked.
The second line contains the sequence which consists of *n* symbols "Y", "N" and "?". If the *i*-th symbol equals "Y", Stepan remembers that he has watched the episode number *i*. If the *i*-th symbol equals "N", Stepan remembers that he hasn't watched the epizode number *i*. If the *i*-th symbol equals "?", Stepan doesn't exactly remember if he has watched the episode number *i* or not.
Output Specification:
If Stepan's dissatisfaction can be exactly equal to *k*, then print "YES" (without qoutes). Otherwise print "NO" (without qoutes).
Demo Input:
['5 2\nNYNNY\n', '6 1\n????NN\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test Stepan remembers about all the episodes whether he has watched them or not. His dissatisfaction is 2, because he hasn't watch two episodes in a row — the episode number 3 and the episode number 4. The answer is "YES", because *k* = 2.
In the second test *k* = 1, Stepan's dissatisfaction is greater than or equal to 2 (because he remembers that he hasn't watch at least two episodes in a row — number 5 and number 6), even if he has watched the episodes from the first to the fourth, inclusive. | ```python
from sys import stdin
from functools import lru_cache
from collections import defaultdict
infile = stdin
N, K = map(int, infile.readline().split())
watched = infile.readline().strip()
mem = set()
mem.add((0,0,0))
for i, letter in enumerate(watched):
for d in range(K+1):
for c in range(K+1):
if (i,d,c) in mem:
if letter in ('N', '?'):
mem.add((i+1, max(d, c+1), c+1))
if letter in ('Y', '?'):
mem.add((i+1, d, 0))
def good():
for i,d,c in mem:
if i == N and d == K:
return 'YES'
return 'NO'
print(good())
``` | 3 | |
22 | A | Second Order Statistics | PROGRAMMING | 800 | [
"brute force"
] | A. Second Order Statistics | 2 | 256 | Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. | If the given sequence has the second order statistics, output this order statistics, otherwise output NO. | [
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] | [
"1\n",
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 ... | 1,686,491,836 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | t = int(input())
a = list(map(int, input().split()[:t]))
new_a = sorted(set(a))
if len(new_a) < 2:
print("NO")
else:
print(new_a[1]) | Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
t = int(input())
a = list(map(int, input().split()[:t]))
new_a = sorted(set(a))
if len(new_a) < 2:
print("NO")
else:
print(new_a[1])
``` | 3.977 |
712 | C | Memory and De-Evolution | PROGRAMMING | 1,600 | [
"greedy",
"math"
] | null | null | Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length *x*, and he wishes to perform operations to obtain an equilateral triangle of side length *y*.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y*? | The first and only line contains two integers *x* and *y* (3<=≤<=*y*<=<<=*x*<=≤<=100<=000) — the starting and ending equilateral triangle side lengths respectively. | Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y* if he starts with the equilateral triangle of side length *x*. | [
"6 3\n",
"8 5\n",
"22 4\n"
] | [
"4\n",
"3\n",
"6\n"
] | In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides *a*, *b*, and *c* as (*a*, *b*, *c*). Then, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18af21f738bad490df83097a90e1f2879a4b21c6.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcfd51d1b2d764a1cf5fbc255cc02e6f5aaed3b1.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test, Memory can do: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0969b7d413854c1e7528991d926bef1f7ffba008.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/63e9e66b882c03e4c73e93ad92204dc255329309.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,500 | [
{
"input": "6 3",
"output": "4"
},
{
"input": "8 5",
"output": "3"
},
{
"input": "22 4",
"output": "6"
},
{
"input": "4 3",
"output": "3"
},
{
"input": "57 27",
"output": "4"
},
{
"input": "61 3",
"output": "9"
},
{
"input": "5 4",
"out... | 1,595,600,263 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 2,000 | 21,401,600 | x,y = map(int,input().split())
turn = cnt = 0
a = b = c = y
while True:
if a >= x and b >= x and c >= x:
print(cnt)
break
cnt += 1
if turn % 3 == 0:
a = b + c - 1
if turn % 3 == 1:
b = a + c - 1
if turn % 3 == 2:
c = a + b - 1
| Title: Memory and De-Evolution
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length *x*, and he wishes to perform operations to obtain an equilateral triangle of side length *y*.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y*?
Input Specification:
The first and only line contains two integers *x* and *y* (3<=≤<=*y*<=<<=*x*<=≤<=100<=000) — the starting and ending equilateral triangle side lengths respectively.
Output Specification:
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y* if he starts with the equilateral triangle of side length *x*.
Demo Input:
['6 3\n', '8 5\n', '22 4\n']
Demo Output:
['4\n', '3\n', '6\n']
Note:
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides *a*, *b*, and *c* as (*a*, *b*, *c*). Then, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18af21f738bad490df83097a90e1f2879a4b21c6.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcfd51d1b2d764a1cf5fbc255cc02e6f5aaed3b1.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test, Memory can do: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0969b7d413854c1e7528991d926bef1f7ffba008.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/63e9e66b882c03e4c73e93ad92204dc255329309.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
x,y = map(int,input().split())
turn = cnt = 0
a = b = c = y
while True:
if a >= x and b >= x and c >= x:
print(cnt)
break
cnt += 1
if turn % 3 == 0:
a = b + c - 1
if turn % 3 == 1:
b = a + c - 1
if turn % 3 == 2:
c = a + b - 1
``` | 0 | |
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"... | 1,660,557,157 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 21 | 77 | 204,800 | a,b,n=map(int, input().split())
num = a
flag=0
ans=str(a)
for i in range(n):
if num%b==0:
ans+="0"
else:
ost = (num*10)%b
if (b-ost)>9:
print(-1)
flag=1
break
else:
#num = (num*10)+b-ost
ans+=str(b-ost)
num = 0
if flag==0:
print(ans) | 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())
num = a
flag=0
ans=str(a)
for i in range(n):
if num%b==0:
ans+="0"
else:
ost = (num*10)%b
if (b-ost)>9:
print(-1)
flag=1
break
else:
#num = (num*10)+b-ost
ans+=str(b-ost)
num = 0
if flag==0:
print(ans)
``` | 0 | |
583 | A | Asphalting Roads | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct. | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | [
"2\n1 1\n1 2\n2 1\n2 2\n",
"1\n1 1\n"
] | [
"1 4 \n",
"1 \n"
] | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | 500 | [
{
"input": "2\n1 1\n1 2\n2 1\n2 2",
"output": "1 4 "
},
{
"input": "1\n1 1",
"output": "1 "
},
{
"input": "2\n1 1\n2 2\n1 2\n2 1",
"output": "1 2 "
},
{
"input": "2\n1 2\n2 2\n2 1\n1 1",
"output": "1 3 "
},
{
"input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 ... | 1,579,110,279 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 109 | 307,200 | a = int(input())
x = []
y = []
res = []
for i in range(a**2):
a = [i for i in input().split()]
if a[0] not in x and a[1] not in y:
res.append(int(i)+1)
x.append(a[0])
y.append(a[1])
print(*res) | Title: Asphalting Roads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
Output Specification:
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
Demo Input:
['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n']
Demo Output:
['1 4 \n', '1 \n']
Note:
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | ```python
a = int(input())
x = []
y = []
res = []
for i in range(a**2):
a = [i for i in input().split()]
if a[0] not in x and a[1] not in y:
res.append(int(i)+1)
x.append(a[0])
y.append(a[1])
print(*res)
``` | 3 | |
602 | B | Approximating a Constant Range | PROGRAMMING | 1,400 | [
"dp",
"implementation",
"two pointers"
] | null | null | When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range. | The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000). | Print a single number — the maximum length of an almost constant range of the given sequence. | [
"5\n1 2 3 3 2\n",
"11\n5 4 5 5 6 7 8 8 8 7 6\n"
] | [
"4\n",
"5\n"
] | In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10]. | 1,000 | [
{
"input": "5\n1 2 3 3 2",
"output": "4"
},
{
"input": "11\n5 4 5 5 6 7 8 8 8 7 6",
"output": "5"
},
{
"input": "2\n3 2",
"output": "2"
},
{
"input": "4\n1001 1000 1000 1001",
"output": "4"
},
{
"input": "4\n1 1 2 3",
"output": "3"
},
{
"input": "3\n1 ... | 1,621,024,660 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 77 | 0 | n = int(input())
arr = list(map(int, input().split()))
maxi = (-1)*1e10
for i in range(len(arr)):
cnt = 0
for j in range(i, len(arr)):
if abs(arr[j] - arr[i]) > 1:
maxi = max(maxi, cnt)
break
else:
cnt += 1
print(maxi)
| Title: Approximating a Constant Range
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
Output Specification:
Print a single number — the maximum length of an almost constant range of the given sequence.
Demo Input:
['5\n1 2 3 3 2\n', '11\n5 4 5 5 6 7 8 8 8 7 6\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10]. | ```python
n = int(input())
arr = list(map(int, input().split()))
maxi = (-1)*1e10
for i in range(len(arr)):
cnt = 0
for j in range(i, len(arr)):
if abs(arr[j] - arr[i]) > 1:
maxi = max(maxi, cnt)
break
else:
cnt += 1
print(maxi)
``` | 0 | |
350 | A | TL | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"implementation"
] | null | null | Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist. | The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds. | If there is a valid TL value, print it. Otherwise, print -1. | [
"3 6\n4 5 2\n8 9 6 10 7 11\n",
"3 1\n3 4 5\n6\n"
] | [
"5",
"-1\n"
] | none | 500 | [
{
"input": "3 6\n4 5 2\n8 9 6 10 7 11",
"output": "5"
},
{
"input": "3 1\n3 4 5\n6",
"output": "-1"
},
{
"input": "2 5\n45 99\n49 41 77 83 45",
"output": "-1"
},
{
"input": "50 50\n18 13 5 34 10 36 36 12 15 11 16 17 14 36 23 45 32 24 31 18 24 32 7 1 31 3 49 8 16 23 3 39 47 43... | 1,575,912,581 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 248 | 0 | n, k = [int(i) for i in input().split()] #int(input())
corr = [int(i) for i in input().split()]
wrong = [int(i) for i in input().split()]
max_corr = max(corr)
min_corr = min(corr)
min_wrong = min(wrong)
best_tl = max(max_corr, 2*min_corr)
if best_tl<min_wrong:
print(best_tl)
else:
print(-1) | Title: TL
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
Output Specification:
If there is a valid TL value, print it. Otherwise, print -1.
Demo Input:
['3 6\n4 5 2\n8 9 6 10 7 11\n', '3 1\n3 4 5\n6\n']
Demo Output:
['5', '-1\n']
Note:
none | ```python
n, k = [int(i) for i in input().split()] #int(input())
corr = [int(i) for i in input().split()]
wrong = [int(i) for i in input().split()]
max_corr = max(corr)
min_corr = min(corr)
min_wrong = min(wrong)
best_tl = max(max_corr, 2*min_corr)
if best_tl<min_wrong:
print(best_tl)
else:
print(-1)
``` | 3 | |
740 | B | Alyona and flowers | PROGRAMMING | 1,200 | [
"constructive algorithms"
] | null | null | Little Alyona is celebrating Happy Birthday! Her mother has an array of *n* flowers. Each flower has some mood, the mood of *i*-th flower is *a**i*. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1,<=<=-<=2,<=1,<=3,<=<=-<=4. Suppose the mother suggested subarrays (1,<=<=-<=2), (3,<=<=-<=4), (1,<=3), (1,<=<=-<=2,<=1,<=3). Then if the girl chooses the third and the fourth subarrays then:
- the first flower adds 1·1<==<=1 to the girl's happiness, because he is in one of chosen subarrays, - the second flower adds (<=-<=2)·1<==<=<=-<=2, because he is in one of chosen subarrays, - the third flower adds 1·2<==<=2, because he is in two of chosen subarrays, - the fourth flower adds 3·2<==<=6, because he is in two of chosen subarrays, - the fifth flower adds (<=-<=4)·0<==<=0, because he is in no chosen subarrays.
Thus, in total 1<=+<=(<=-<=2)<=+<=2<=+<=6<=+<=0<==<=7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=100<=≤<=*a**i*<=≤<=100).
The next *m* lines contain the description of the subarrays suggested by the mother. The *i*-th of these lines contain two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) denoting the subarray *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*].
Each subarray can encounter more than once. | Print single integer — the maximum possible value added to the Alyona's happiness. | [
"5 4\n1 -2 1 3 -4\n1 2\n4 5\n3 4\n1 4\n",
"4 3\n1 2 3 4\n1 3\n2 4\n1 1\n",
"2 2\n-1 -2\n1 1\n1 2\n"
] | [
"7\n",
"16\n",
"0\n"
] | The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | 1,000 | [
{
"input": "5 4\n1 -2 1 3 -4\n1 2\n4 5\n3 4\n1 4",
"output": "7"
},
{
"input": "4 3\n1 2 3 4\n1 3\n2 4\n1 1",
"output": "16"
},
{
"input": "2 2\n-1 -2\n1 1\n1 2",
"output": "0"
},
{
"input": "5 6\n1 1 1 -1 0\n2 4\n1 3\n4 5\n1 5\n1 4\n4 5",
"output": "8"
},
{
"inpu... | 1,479,988,553 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 78 | 0 | inp = [int(x) for x in input().split()]
l = [int(x) for x in input().split()]
total = 0
ans = 0
for x in range(inp[1]):
val = [int(y) for y in input().split()]
total = sum(l[val[0]-1:val[1]])
if(total > 0):
ans = ans + total
print(ans) | Title: Alyona and flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Alyona is celebrating Happy Birthday! Her mother has an array of *n* flowers. Each flower has some mood, the mood of *i*-th flower is *a**i*. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1,<=<=-<=2,<=1,<=3,<=<=-<=4. Suppose the mother suggested subarrays (1,<=<=-<=2), (3,<=<=-<=4), (1,<=3), (1,<=<=-<=2,<=1,<=3). Then if the girl chooses the third and the fourth subarrays then:
- the first flower adds 1·1<==<=1 to the girl's happiness, because he is in one of chosen subarrays, - the second flower adds (<=-<=2)·1<==<=<=-<=2, because he is in one of chosen subarrays, - the third flower adds 1·2<==<=2, because he is in two of chosen subarrays, - the fourth flower adds 3·2<==<=6, because he is in two of chosen subarrays, - the fifth flower adds (<=-<=4)·0<==<=0, because he is in no chosen subarrays.
Thus, in total 1<=+<=(<=-<=2)<=+<=2<=+<=6<=+<=0<==<=7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=100<=≤<=*a**i*<=≤<=100).
The next *m* lines contain the description of the subarrays suggested by the mother. The *i*-th of these lines contain two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) denoting the subarray *a*[*l**i*],<=*a*[*l**i*<=+<=1],<=...,<=*a*[*r**i*].
Each subarray can encounter more than once.
Output Specification:
Print single integer — the maximum possible value added to the Alyona's happiness.
Demo Input:
['5 4\n1 -2 1 3 -4\n1 2\n4 5\n3 4\n1 4\n', '4 3\n1 2 3 4\n1 3\n2 4\n1 1\n', '2 2\n-1 -2\n1 1\n1 2\n']
Demo Output:
['7\n', '16\n', '0\n']
Note:
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | ```python
inp = [int(x) for x in input().split()]
l = [int(x) for x in input().split()]
total = 0
ans = 0
for x in range(inp[1]):
val = [int(y) for y in input().split()]
total = sum(l[val[0]-1:val[1]])
if(total > 0):
ans = ans + total
print(ans)
``` | 3 | |
746 | A | Compote | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Nikolay has *a* lemons, *b* apples and *c* pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio 1:<=2:<=4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can't crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.
Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can't use any fruits, in this case print 0. | The first line contains the positive integer *a* (1<=≤<=*a*<=≤<=1000) — the number of lemons Nikolay has.
The second line contains the positive integer *b* (1<=≤<=*b*<=≤<=1000) — the number of apples Nikolay has.
The third line contains the positive integer *c* (1<=≤<=*c*<=≤<=1000) — the number of pears Nikolay has. | Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. | [
"2\n5\n7\n",
"4\n7\n13\n",
"2\n3\n2\n"
] | [
"7\n",
"21\n",
"0\n"
] | In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1 + 2 + 4 = 7.
In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3 + 6 + 12 = 21.
In the third example Nikolay don't have enough pears to cook any compote, so the answer is 0. | 500 | [
{
"input": "2\n5\n7",
"output": "7"
},
{
"input": "4\n7\n13",
"output": "21"
},
{
"input": "2\n3\n2",
"output": "0"
},
{
"input": "1\n1\n1",
"output": "0"
},
{
"input": "1\n2\n4",
"output": "7"
},
{
"input": "1000\n1000\n1000",
"output": "1750"
}... | 1,609,357,406 | 2,147,483,647 | Python 3 | OK | TESTS | 84 | 109 | 0 | a=int(input())
b=int(input())
c=int(input())
for i in range (a,-1,-1):
if(b>=2*i and c>=4*i):
print(7*i)
break
elif(i==0):
print(0)
| Title: Compote
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nikolay has *a* lemons, *b* apples and *c* pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio 1:<=2:<=4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can't crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.
Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can't use any fruits, in this case print 0.
Input Specification:
The first line contains the positive integer *a* (1<=≤<=*a*<=≤<=1000) — the number of lemons Nikolay has.
The second line contains the positive integer *b* (1<=≤<=*b*<=≤<=1000) — the number of apples Nikolay has.
The third line contains the positive integer *c* (1<=≤<=*c*<=≤<=1000) — the number of pears Nikolay has.
Output Specification:
Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote.
Demo Input:
['2\n5\n7\n', '4\n7\n13\n', '2\n3\n2\n']
Demo Output:
['7\n', '21\n', '0\n']
Note:
In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1 + 2 + 4 = 7.
In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3 + 6 + 12 = 21.
In the third example Nikolay don't have enough pears to cook any compote, so the answer is 0. | ```python
a=int(input())
b=int(input())
c=int(input())
for i in range (a,-1,-1):
if(b>=2*i and c>=4*i):
print(7*i)
break
elif(i==0):
print(0)
``` | 3 | |
322 | B | Ciel and Flowers | PROGRAMMING | 1,600 | [
"combinatorics",
"math"
] | null | null | Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make. | The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers. | Print the maximal number of bouquets Fox Ciel can make. | [
"3 6 9\n",
"4 4 4\n",
"0 0 0\n"
] | [
"6\n",
"4\n",
"0\n"
] | In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet. | 1,000 | [
{
"input": "3 6 9",
"output": "6"
},
{
"input": "4 4 4",
"output": "4"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "0 3 6",
"output": "3"
},
{
"input": "7 8 9",
"output": "7"
},
{
"input": "8 8 9",
"output": "8"
},
{
"input": "15 3 9... | 1,634,912,512 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 122 | 28,364,800 | r,g,b=map(int,input().split())
ans=0
for i in range(3):
t=0
rt=r
gt=g
bt=b
if rt>=i and gt>=i and bt>=i:
rt-=i
gt-=i
bt-=i
t+=i
t+=rt//3+gt//3+bt//3
ans=max(ans,t)
print(ans)
| Title: Ciel and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
Input Specification:
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
Output Specification:
Print the maximal number of bouquets Fox Ciel can make.
Demo Input:
['3 6 9\n', '4 4 4\n', '0 0 0\n']
Demo Output:
['6\n', '4\n', '0\n']
Note:
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet. | ```python
r,g,b=map(int,input().split())
ans=0
for i in range(3):
t=0
rt=r
gt=g
bt=b
if rt>=i and gt>=i and bt>=i:
rt-=i
gt-=i
bt-=i
t+=i
t+=rt//3+gt//3+bt//3
ans=max(ans,t)
print(ans)
``` | 3 | |
208 | A | Dubstep | PROGRAMMING | 900 | [
"strings"
] | null | null | Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. | The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. | Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. | [
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] | [
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] | In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | 500 | [
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "... | 1,689,314,936 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 92 | 0 | s = input()
check = s.replace("WUB","")
print(check) | Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | ```python
s = input()
check = s.replace("WUB","")
print(check)
``` | 0 | |
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": ... | 1,642,654,100 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | l=-2e9
r=2e9
n=int(input())
for i in range(0,n):
listi=input().split()
f=int(listi[1])
if listi[-1]=='Y':
if listi[0]=='>=':
l=f
if listi[0]=='>':
l=f+1
if listi[0]=='<=':
r=f
if listi[0]=='<':
r=f-1
else:
if listi[0]=='>=':
r=f-1
if listi[0]=='>':
r=f
if listi[0]=='<=':
l=f+1
if listi[0]=='<':
l=f
if l<=r:
print(l)
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
l=-2e9
r=2e9
n=int(input())
for i in range(0,n):
listi=input().split()
f=int(listi[1])
if listi[-1]=='Y':
if listi[0]=='>=':
l=f
if listi[0]=='>':
l=f+1
if listi[0]=='<=':
r=f
if listi[0]=='<':
r=f-1
else:
if listi[0]=='>=':
r=f-1
if listi[0]=='>':
r=f
if listi[0]=='<=':
l=f+1
if listi[0]=='<':
l=f
if l<=r:
print(l)
else:
print("Impossible")
``` | 0 | |
373 | B | Making Sequences is Fun | PROGRAMMING | 1,600 | [
"binary search",
"implementation",
"math"
] | null | null | We'll define *S*(*n*) for positive integer *n* as follows: the number of the *n*'s digits in the decimal base. For example, *S*(893)<==<=3, *S*(114514)<==<=6.
You want to make a consecutive integer sequence starting from number *m* (*m*,<=*m*<=+<=1,<=...). But you need to pay *S*(*n*)·*k* to add the number *n* to the sequence.
You can spend a cost up to *w*, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length. | The first line contains three integers *w* (1<=≤<=*w*<=≤<=1016), *m* (1<=≤<=*m*<=≤<=1016), *k* (1<=≤<=*k*<=≤<=109).
Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. | The first line should contain a single integer — the answer to the problem. | [
"9 1 1\n",
"77 7 7\n",
"114 5 14\n",
"1 1 2\n"
] | [
"9\n",
"7\n",
"6\n",
"0\n"
] | none | 1,000 | [
{
"input": "9 1 1",
"output": "9"
},
{
"input": "77 7 7",
"output": "7"
},
{
"input": "114 5 14",
"output": "6"
},
{
"input": "1 1 2",
"output": "0"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "462 183 8",
"output": "19"
},
{
"i... | 1,563,651,338 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 2,000 | 409,600 | a=input().split()
l=len(a[1])
p=int(a[1])
d=10**l
m=int(a[2])
w=int(a[0])
c=0
while l<w:
r=min(int(w/(l*m)),d-p)
c+=r
w-=r*l*m
p=d
d*=10
l+=1
print(c)
| Title: Making Sequences is Fun
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We'll define *S*(*n*) for positive integer *n* as follows: the number of the *n*'s digits in the decimal base. For example, *S*(893)<==<=3, *S*(114514)<==<=6.
You want to make a consecutive integer sequence starting from number *m* (*m*,<=*m*<=+<=1,<=...). But you need to pay *S*(*n*)·*k* to add the number *n* to the sequence.
You can spend a cost up to *w*, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.
Input Specification:
The first line contains three integers *w* (1<=≤<=*w*<=≤<=1016), *m* (1<=≤<=*m*<=≤<=1016), *k* (1<=≤<=*k*<=≤<=109).
Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
The first line should contain a single integer — the answer to the problem.
Demo Input:
['9 1 1\n', '77 7 7\n', '114 5 14\n', '1 1 2\n']
Demo Output:
['9\n', '7\n', '6\n', '0\n']
Note:
none | ```python
a=input().split()
l=len(a[1])
p=int(a[1])
d=10**l
m=int(a[2])
w=int(a[0])
c=0
while l<w:
r=min(int(w/(l*m)),d-p)
c+=r
w-=r*l*m
p=d
d*=10
l+=1
print(c)
``` | 0 | |
11 | A | Increasing Sequence | PROGRAMMING | 900 | [
"constructive algorithms",
"implementation",
"math"
] | A. Increasing Sequence | 1 | 64 | A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing? | The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106). | Output the minimal number of moves needed to make the sequence increasing. | [
"4 2\n1 3 3 2\n"
] | [
"3\n"
] | none | 0 | [
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
... | 1,669,999,168 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 60 | 0 | n, k = map(int,input().split())
a = list(map(int,input().split()))
print(a[k]) | Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none | ```python
n, k = map(int,input().split())
a = list(map(int,input().split()))
print(a[k])
``` | 0 |
928 | B | Chat | PROGRAMMING | 1,400 | [
"*special",
"dp"
] | null | null | There are times you recall a good old friend and everything you've come through together. Luckily there are social networks — they store all your message history making it easy to know what you argued over 10 years ago.
More formal, your message history is a sequence of messages ordered by time sent numbered from 1 to *n* where *n* is the total number of messages in the chat.
Each message might contain a link to an earlier message which it is a reply to. When opening a message *x* or getting a link to it, the dialogue is shown in such a way that *k* previous messages, message *x* and *k* next messages are visible (with respect to message *x*). In case there are less than *k* messages somewhere, they are yet all shown.
Digging deep into your message history, you always read all visible messages and then go by the link in the current message *x* (if there is one) and continue reading in the same manner.
Determine the number of messages you'll read if your start from message number *t* for all *t* from 1 to *n*. Calculate these numbers independently. If you start with message *x*, the initial configuration is *x* itself, *k* previous and *k* next messages. Messages read multiple times are considered as one. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*) — the total amount of messages and the number of previous and next messages visible.
The second line features a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=<<=*i*), where *a**i* denotes the *i*-th message link destination or zero, if there's no link from *i*. All messages are listed in chronological order. It's guaranteed that the link from message *x* goes to message with number strictly less than *x*. | Print *n* integers with *i*-th denoting the number of distinct messages you can read starting from message *i* and traversing the links while possible. | [
"6 0\n0 1 1 2 3 2\n",
"10 1\n0 1 0 3 4 5 2 3 7 0\n",
"2 2\n0 1\n"
] | [
"1 2 2 3 3 3 \n",
"2 3 3 4 5 6 6 6 8 2 \n",
"2 2 \n"
] | Consider *i* = 6 in sample case one. You will read message 6, then 2, then 1 and then there will be no link to go.
In the second sample case *i* = 6 gives you messages 5, 6, 7 since *k* = 1, then 4, 5, 6, then 2, 3, 4 and then the link sequence breaks. The number of distinct messages here is equal to 6. | 1,250 | [
{
"input": "6 0\n0 1 1 2 3 2",
"output": "1 2 2 3 3 3 "
},
{
"input": "10 1\n0 1 0 3 4 5 2 3 7 0",
"output": "2 3 3 4 5 6 6 6 8 2 "
},
{
"input": "2 2\n0 1",
"output": "2 2 "
},
{
"input": "1 1\n0",
"output": "1 "
},
{
"input": "5 2\n0 1 2 3 1",
"output": "3 4... | 1,665,137,409 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 118 | 140 | 16,281,600 | n,k=map(int,input().split())
arr=list(map(int,input().split()))
dp=[0]*(n)
arr=[el-1 for el in arr]
for i in range(n):
if arr[i]==-1:
dp[i]=1+min(n-1-i,k)+min(k,i)
else :
dp[i]=dp[arr[i]]+min(1+min(n-1-i,k)+min(i,k),min(n-1,i+k)-min(n-1,arr[i]+k))
print(*dp) | Title: Chat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are times you recall a good old friend and everything you've come through together. Luckily there are social networks — they store all your message history making it easy to know what you argued over 10 years ago.
More formal, your message history is a sequence of messages ordered by time sent numbered from 1 to *n* where *n* is the total number of messages in the chat.
Each message might contain a link to an earlier message which it is a reply to. When opening a message *x* or getting a link to it, the dialogue is shown in such a way that *k* previous messages, message *x* and *k* next messages are visible (with respect to message *x*). In case there are less than *k* messages somewhere, they are yet all shown.
Digging deep into your message history, you always read all visible messages and then go by the link in the current message *x* (if there is one) and continue reading in the same manner.
Determine the number of messages you'll read if your start from message number *t* for all *t* from 1 to *n*. Calculate these numbers independently. If you start with message *x*, the initial configuration is *x* itself, *k* previous and *k* next messages. Messages read multiple times are considered as one.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*) — the total amount of messages and the number of previous and next messages visible.
The second line features a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=<<=*i*), where *a**i* denotes the *i*-th message link destination or zero, if there's no link from *i*. All messages are listed in chronological order. It's guaranteed that the link from message *x* goes to message with number strictly less than *x*.
Output Specification:
Print *n* integers with *i*-th denoting the number of distinct messages you can read starting from message *i* and traversing the links while possible.
Demo Input:
['6 0\n0 1 1 2 3 2\n', '10 1\n0 1 0 3 4 5 2 3 7 0\n', '2 2\n0 1\n']
Demo Output:
['1 2 2 3 3 3 \n', '2 3 3 4 5 6 6 6 8 2 \n', '2 2 \n']
Note:
Consider *i* = 6 in sample case one. You will read message 6, then 2, then 1 and then there will be no link to go.
In the second sample case *i* = 6 gives you messages 5, 6, 7 since *k* = 1, then 4, 5, 6, then 2, 3, 4 and then the link sequence breaks. The number of distinct messages here is equal to 6. | ```python
n,k=map(int,input().split())
arr=list(map(int,input().split()))
dp=[0]*(n)
arr=[el-1 for el in arr]
for i in range(n):
if arr[i]==-1:
dp[i]=1+min(n-1-i,k)+min(k,i)
else :
dp[i]=dp[arr[i]]+min(1+min(n-1-i,k)+min(i,k),min(n-1,i+k)-min(n-1,arr[i]+k))
print(*dp)
``` | 3 | |
991 | A | If at first you don't succeed... | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time? | The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$). | If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$. | [
"10 10 5 20\n",
"2 2 0 4\n",
"2 2 2 1\n"
] | [
"5",
"-1",
"-1"
] | The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible. | 500 | [
{
"input": "10 10 5 20",
"output": "5"
},
{
"input": "2 2 0 4",
"output": "-1"
},
{
"input": "2 2 2 1",
"output": "-1"
},
{
"input": "98 98 97 100",
"output": "1"
},
{
"input": "1 5 2 10",
"output": "-1"
},
{
"input": "5 1 2 10",
"output": "-1"
}... | 1,548,725,565 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 108 | 0 | a, b, c, n = map(int, input().split())
total = a + b - c
if total >= n:
print('-1')
else:
print(n - total) | Title: If at first you don't succeed...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?
Input Specification:
The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$).
Output Specification:
If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$.
Demo Input:
['10 10 5 20\n', '2 2 0 4\n', '2 2 2 1\n']
Demo Output:
['5', '-1', '-1']
Note:
The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible. | ```python
a, b, c, n = map(int, input().split())
total = a + b - c
if total >= n:
print('-1')
else:
print(n - total)
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are *n* (*n* is always even) possible starting positions for characters marked along the *Ox* axis. The positions are given by their distinct coordinates *x*1,<=*x*2,<=...,<=*x**n*, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be *n*<=-<=2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally. | The first line on the input contains a single integer *n* (2<=≤<=*n*<=≤<=200<=000, *n* is even) — the number of positions available initially. The second line contains *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109), giving the coordinates of the corresponding positions. | Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally. | [
"6\n0 1 3 7 15 31\n",
"2\n73 37\n"
] | [
"7\n",
"36\n"
] | In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15; 1. Lesha bans the position at coordinate 3; 1. Vova bans the position at coordinate 31; 1. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans. | 0 | [
{
"input": "6\n0 1 3 7 15 31",
"output": "7"
},
{
"input": "2\n73 37",
"output": "36"
},
{
"input": "2\n0 1000000000",
"output": "1000000000"
},
{
"input": "8\n729541013 135019377 88372488 319157478 682081360 558614617 258129110 790518782",
"output": "470242129"
},
{
... | 1,447,001,835 | 1,635 | Python 3 | SKIPPED | TESTS | 0 | 46 | 204,800 | __author__ = 'cmashinho'
n = int(input())
l = list(map(int, input().split()))
def findMax():
maxx = 0
coord = 0
for i in range(1, len(l)):
if max(l[i], l[i - 1]) - min(l[i], l[i - 1]) > maxx:
maxx = max(l[i], l[i - 1]) - min(l[i], l[i - 1])
coord = i - 1
return coord
def findMin():
minn = 10 ** 18
coord = 0
for i in range(1, len(l)):
if max(l[i], l[i - 1]) - min(l[i], l[i - 1]) < minn:
minn = max(l[i], l[i - 1]) - min(l[i], l[i - 1])
coord = i - 1
return coord
if n == 2:
print(max(l) - min(l))
elif n == 3:
minn = 10 ** 18
if max(l[0], l[1]) - min(l[0], l[1]) < minn:
minn = max(l[0], l[1]) - min(l[0], l[1])
if max(l[1], l[2]) - min(l[1], l[2]) < minn:
minn = max(l[1], l[2]) - min(l[1], l[2])
if max(l[0], l[2]) - min(l[0], l[2]) < minn:
minn = max(l[0], l[2]) - min(l[0], l[2])
print(minn)
else:
current = 0
while(len(l) != 2):
if current == 0:
l.pop(findMax() + 1)
current = 1
else:
l.pop(findMin() + 1)
current = 0
print(max(l) - min(l))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are *n* (*n* is always even) possible starting positions for characters marked along the *Ox* axis. The positions are given by their distinct coordinates *x*1,<=*x*2,<=...,<=*x**n*, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be *n*<=-<=2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
Input Specification:
The first line on the input contains a single integer *n* (2<=≤<=*n*<=≤<=200<=000, *n* is even) — the number of positions available initially. The second line contains *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109), giving the coordinates of the corresponding positions.
Output Specification:
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Demo Input:
['6\n0 1 3 7 15 31\n', '2\n73 37\n']
Demo Output:
['7\n', '36\n']
Note:
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15; 1. Lesha bans the position at coordinate 3; 1. Vova bans the position at coordinate 31; 1. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans. | ```python
__author__ = 'cmashinho'
n = int(input())
l = list(map(int, input().split()))
def findMax():
maxx = 0
coord = 0
for i in range(1, len(l)):
if max(l[i], l[i - 1]) - min(l[i], l[i - 1]) > maxx:
maxx = max(l[i], l[i - 1]) - min(l[i], l[i - 1])
coord = i - 1
return coord
def findMin():
minn = 10 ** 18
coord = 0
for i in range(1, len(l)):
if max(l[i], l[i - 1]) - min(l[i], l[i - 1]) < minn:
minn = max(l[i], l[i - 1]) - min(l[i], l[i - 1])
coord = i - 1
return coord
if n == 2:
print(max(l) - min(l))
elif n == 3:
minn = 10 ** 18
if max(l[0], l[1]) - min(l[0], l[1]) < minn:
minn = max(l[0], l[1]) - min(l[0], l[1])
if max(l[1], l[2]) - min(l[1], l[2]) < minn:
minn = max(l[1], l[2]) - min(l[1], l[2])
if max(l[0], l[2]) - min(l[0], l[2]) < minn:
minn = max(l[0], l[2]) - min(l[0], l[2])
print(minn)
else:
current = 0
while(len(l) != 2):
if current == 0:
l.pop(findMax() + 1)
current = 1
else:
l.pop(findMin() + 1)
current = 0
print(max(l) - min(l))
``` | -1 | |
2 | A | Winner | PROGRAMMING | 1,500 | [
"hashing",
"implementation"
] | A. Winner | 1 | 64 | The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. | The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. | Print the name of the winner. | [
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] | [
"andrew\n",
"andrew\n"
] | none | 0 | [
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},... | 1,660,576,309 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | Tn=[]
Ts=[]
for _ in range(int(input())):
S=input()
Tn.append(S[:S.find(' ')])
Ts.append(int(S[S.find(' ')+1:]))
if max(Ts)==Ts[-1] and Ts.count(max(Ts))==1:
print(Tn[Ts.index(max(Ts[:-1]))])
else:
print(Tn[Ts.index(max(Ts))])
| Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none | ```python
Tn=[]
Ts=[]
for _ in range(int(input())):
S=input()
Tn.append(S[:S.find(' ')])
Ts.append(int(S[S.find(' ')+1:]))
if max(Ts)==Ts[-1] and Ts.count(max(Ts))==1:
print(Tn[Ts.index(max(Ts[:-1]))])
else:
print(Tn[Ts.index(max(Ts))])
``` | 0 |
235 | A | LCM Challenge | PROGRAMMING | 1,600 | [
"number theory"
] | null | null | Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement. | Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*. | [
"9\n",
"7\n"
] | [
"504\n",
"210\n"
] | The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | 500 | [
{
"input": "9",
"output": "504"
},
{
"input": "7",
"output": "210"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "60"
},
{
"input": "33",
"output": "32736"
},
{
"input": "21",
"output": ... | 1,559,187,059 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 31 | 310 | 0 | import math
n = int(input())
if(n<=2):
print(n)
exit()
if(n%2!=0):
print(n*(n-1)*(n-2))
else:
print(max(((n*(n-2))//math.gcd(n,n-2))*(n-1),((n*(n-2))//math.gcd(n,n-3))*(n-1)))
| Title: LCM Challenge
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement.
Output Specification:
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*.
Demo Input:
['9\n', '7\n']
Demo Output:
['504\n', '210\n']
Note:
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | ```python
import math
n = int(input())
if(n<=2):
print(n)
exit()
if(n%2!=0):
print(n*(n-1)*(n-2))
else:
print(max(((n*(n-2))//math.gcd(n,n-2))*(n-1),((n*(n-2))//math.gcd(n,n-3))*(n-1)))
``` | 0 | |
839 | A | Arya and Bran | PROGRAMMING | 900 | [
"implementation"
] | null | null | Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are *n* days, at the *i*-th day, Arya finds *a**i* candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. Formally, you need to output the minimum day index to the end of which *k* candies will be given out (the days are indexed from 1 to *n*).
Print -1 if she can't give him *k* candies during *n* given days. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10000).
The second line contains *n* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). | If it is impossible for Arya to give Bran *k* candies within *n* days, print -1.
Otherwise print a single integer — the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. | [
"2 3\n1 2\n",
"3 17\n10 10 10\n",
"1 9\n10\n"
] | [
"2",
"3",
"-1"
] | In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day. | 500 | [
{
"input": "2 3\n1 2",
"output": "2"
},
{
"input": "3 17\n10 10 10",
"output": "3"
},
{
"input": "1 9\n10",
"output": "-1"
},
{
"input": "10 70\n6 5 2 3 3 2 1 4 3 2",
"output": "-1"
},
{
"input": "20 140\n40 4 81 40 10 54 34 50 84 60 16 1 90 78 38 93 99 60 81 99",... | 1,647,282,632 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 31 | 0 | n,k=[int(i) for i in input().split()]
konfeta=[int(i) for i in input().split()]
sum_konfet=0
for i in range(0,n):
sum_konfet+=konfeta[i]
if k%8==0:
min_kolvo_days=k//8
else:
min_kolvo_days=k//8+1
if min_kolvo_days>n or sum_konfet<k:
print(-1)
else:
print(n)
| Title: Arya and Bran
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.
At first, Arya and Bran have 0 Candies. There are *n* days, at the *i*-th day, Arya finds *a**i* candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.
Your task is to find the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day. Formally, you need to output the minimum day index to the end of which *k* candies will be given out (the days are indexed from 1 to *n*).
Print -1 if she can't give him *k* candies during *n* given days.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10000).
The second line contains *n* integers *a*1,<=*a*2,<=*a*3,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
If it is impossible for Arya to give Bran *k* candies within *n* days, print -1.
Otherwise print a single integer — the minimum number of days Arya needs to give Bran *k* candies before the end of the *n*-th day.
Demo Input:
['2 3\n1 2\n', '3 17\n10 10 10\n', '1 9\n10\n']
Demo Output:
['2', '3', '-1']
Note:
In the first sample, Arya can give Bran 3 candies in 2 days.
In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.
In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day. | ```python
n,k=[int(i) for i in input().split()]
konfeta=[int(i) for i in input().split()]
sum_konfet=0
for i in range(0,n):
sum_konfet+=konfeta[i]
if k%8==0:
min_kolvo_days=k//8
else:
min_kolvo_days=k//8+1
if min_kolvo_days>n or sum_konfet<k:
print(-1)
else:
print(n)
``` | 0 | |
239 | A | Two Bags of Potatoes | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math"
] | null | null | Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order. | The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105). | Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1. | [
"10 1 10\n",
"10 6 40\n"
] | [
"-1\n",
"2 8 14 20 26 \n"
] | none | 500 | [
{
"input": "10 1 10",
"output": "-1"
},
{
"input": "10 6 40",
"output": "2 8 14 20 26 "
},
{
"input": "10 1 20",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "1 10000 1000000000",
"output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 12999... | 1,614,580,714 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 92 | 0 | def potatoes(y, k ,n):
arr = list(range(k, n, k))
arr = [i - y for i in arr if i-y >= 1]
if len(arr) > 0:
print(*arr)
else:
print(-1)
y, k, n = map(int, input().split())
potatoes(y, k ,n) | Title: Two Bags of Potatoes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
Input Specification:
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Output Specification:
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1.
Demo Input:
['10 1 10\n', '10 6 40\n']
Demo Output:
['-1\n', '2 8 14 20 26 \n']
Note:
none | ```python
def potatoes(y, k ,n):
arr = list(range(k, n, k))
arr = [i - y for i in arr if i-y >= 1]
if len(arr) > 0:
print(*arr)
else:
print(-1)
y, k, n = map(int, input().split())
potatoes(y, k ,n)
``` | 0 | |
208 | A | Dubstep | PROGRAMMING | 900 | [
"strings"
] | null | null | Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. | The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. | Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. | [
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] | [
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] | In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | 500 | [
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "... | 1,699,382,194 | 2,147,483,647 | PyPy 3 | OK | TESTS | 71 | 186 | 0 | s = input()
og = ''
i = 0
while i < len(s):
if s[i:i+3] != "WUB":
og = og + s[i]
i = i + 1
else:
if (len(og) != 0) and (og[-1] != " "):
og = og + " "
i = i + 3
print(og) | Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | ```python
s = input()
og = ''
i = 0
while i < len(s):
if s[i:i+3] != "WUB":
og = og + s[i]
i = i + 1
else:
if (len(og) != 0) and (og[-1] != " "):
og = og + " "
i = i + 3
print(og)
``` | 3 | |
644 | A | Parliament of Berland | PROGRAMMING | 1,000 | [
"*special",
"constructive algorithms"
] | null | null | There are *n* parliamentarians in Berland. They are numbered with integers from 1 to *n*. It happened that all parliamentarians with odd indices are Democrats and all parliamentarians with even indices are Republicans.
New parliament assembly hall is a rectangle consisting of *a*<=×<=*b* chairs — *a* rows of *b* chairs each. Two chairs are considered neighbouring if they share as side. For example, chair number 5 in row number 2 is neighbouring to chairs number 4 and 6 in this row and chairs with number 5 in rows 1 and 3. Thus, chairs have four neighbours in general, except for the chairs on the border of the hall
We know that if two parliamentarians from one political party (that is two Democrats or two Republicans) seat nearby they spent all time discussing internal party issues.
Write the program that given the number of parliamentarians and the sizes of the hall determine if there is a way to find a seat for any parliamentarian, such that no two members of the same party share neighbouring seats. | The first line of the input contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=10<=000, 1<=≤<=*a*,<=*b*<=≤<=100) — the number of parliamentarians, the number of rows in the assembly hall and the number of seats in each row, respectively. | If there is no way to assigns seats to parliamentarians in a proper way print -1.
Otherwise print the solution in *a* lines, each containing *b* integers. The *j*-th integer of the *i*-th line should be equal to the index of parliamentarian occupying this seat, or 0 if this seat should remain empty. If there are multiple possible solution, you may print any of them. | [
"3 2 2\n",
"8 4 3\n",
"10 2 2\n"
] | [
"0 3\n1 2\n",
"7 8 3\n0 1 4\n6 0 5\n0 2 0\n",
"-1\n"
] | In the first sample there are many other possible solutions. For example,
and
The following assignment
is incorrect, because parliamentarians 1 and 3 are both from Democrats party but will occupy neighbouring seats. | 500 | [
{
"input": "3 2 2",
"output": "1 2 \n0 3 "
},
{
"input": "8 4 3",
"output": "1 2 3 \n4 5 6 \n7 8 0 \n0 0 0 "
},
{
"input": "10 2 2",
"output": "-1"
},
{
"input": "1 1 1",
"output": "1 "
},
{
"input": "8 3 3",
"output": "1 2 3 \n4 5 6 \n7 8 0 "
},
{
"in... | 1,458,235,636 | 116,836 | Python 3 | COMPILATION_ERROR | PRETESTS | 0 | 0 | 0 | # :)
l = raw_input().split(" ")
n = int(l[0])
a = int(l[1])
b = int(l[2])
Matrix = [[0 for x in range(b)] for x in range(a)]
alter = -1
odd_even = [1,2]
if a*b < n:
print -1,
else:
for i in range(a):
alter += 1
row = i
col = 0
while(row>=0 and row<a and col>=0 and col<b):
if odd_even[alter%2] <= n:
Matrix[row][col] = odd_even[alter%2]
odd_even[alter%2] += 2
else:
Matrix[row][col] = 0
row -= 1
col += 1
for j in range(1,b):
alter += 1
row = a-1
col = j
while(row>=0 and row<a and col>=0 and col<b):
if odd_even[alter%2] <= n:
Matrix[row][col] = odd_even[alter%2]
odd_even[alter%2] += 2
else:
Matrix[row][col] = 0
row -= 1
col += 1
for i in range(a):
for j in range(b):
print Matrix[i][j],
if i != (a-1):
print | Title: Parliament of Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* parliamentarians in Berland. They are numbered with integers from 1 to *n*. It happened that all parliamentarians with odd indices are Democrats and all parliamentarians with even indices are Republicans.
New parliament assembly hall is a rectangle consisting of *a*<=×<=*b* chairs — *a* rows of *b* chairs each. Two chairs are considered neighbouring if they share as side. For example, chair number 5 in row number 2 is neighbouring to chairs number 4 and 6 in this row and chairs with number 5 in rows 1 and 3. Thus, chairs have four neighbours in general, except for the chairs on the border of the hall
We know that if two parliamentarians from one political party (that is two Democrats or two Republicans) seat nearby they spent all time discussing internal party issues.
Write the program that given the number of parliamentarians and the sizes of the hall determine if there is a way to find a seat for any parliamentarian, such that no two members of the same party share neighbouring seats.
Input Specification:
The first line of the input contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=10<=000, 1<=≤<=*a*,<=*b*<=≤<=100) — the number of parliamentarians, the number of rows in the assembly hall and the number of seats in each row, respectively.
Output Specification:
If there is no way to assigns seats to parliamentarians in a proper way print -1.
Otherwise print the solution in *a* lines, each containing *b* integers. The *j*-th integer of the *i*-th line should be equal to the index of parliamentarian occupying this seat, or 0 if this seat should remain empty. If there are multiple possible solution, you may print any of them.
Demo Input:
['3 2 2\n', '8 4 3\n', '10 2 2\n']
Demo Output:
['0 3\n1 2\n', '7 8 3\n0 1 4\n6 0 5\n0 2 0\n', '-1\n']
Note:
In the first sample there are many other possible solutions. For example,
and
The following assignment
is incorrect, because parliamentarians 1 and 3 are both from Democrats party but will occupy neighbouring seats. | ```python
# :)
l = raw_input().split(" ")
n = int(l[0])
a = int(l[1])
b = int(l[2])
Matrix = [[0 for x in range(b)] for x in range(a)]
alter = -1
odd_even = [1,2]
if a*b < n:
print -1,
else:
for i in range(a):
alter += 1
row = i
col = 0
while(row>=0 and row<a and col>=0 and col<b):
if odd_even[alter%2] <= n:
Matrix[row][col] = odd_even[alter%2]
odd_even[alter%2] += 2
else:
Matrix[row][col] = 0
row -= 1
col += 1
for j in range(1,b):
alter += 1
row = a-1
col = j
while(row>=0 and row<a and col>=0 and col<b):
if odd_even[alter%2] <= n:
Matrix[row][col] = odd_even[alter%2]
odd_even[alter%2] += 2
else:
Matrix[row][col] = 0
row -= 1
col += 1
for i in range(a):
for j in range(b):
print Matrix[i][j],
if i != (a-1):
print
``` | -1 | |
53 | C | Little Frog | PROGRAMMING | 1,200 | [
"constructive algorithms"
] | C. Little Frog | 2 | 256 | Once upon a time a little frog whose name was Vasya decided to travel around his home swamp. Overall there are *n* mounds on the swamp, located on one line. The distance between the neighboring mounds is one meter. Vasya wants to visit all the mounds in one day; besides, he wants to visit each one exactly once. For that he makes a route plan, to decide the order in which to jump on the mounds. Vasya can pick any mound as the first one. He thinks it boring to jump two times at the same distance. That's why he wants any two jumps on his route to have different lengths. Help Vasya the Frog and make the plan for him. | The single line contains a number *n* (1<=≤<=*n*<=≤<=104) which is the number of mounds. | Print *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) which are the frog's route plan.
- All the *p**i*'s should be mutually different. - All the |*p**i*–*p**i*<=+<=1|'s should be mutually different (1<=≤<=*i*<=≤<=*n*<=-<=1).
If there are several solutions, output any. | [
"2\n",
"3\n"
] | [
"1 2 ",
"1 3 2 "
] | none | 1,500 | [
{
"input": "2",
"output": "1 2 "
},
{
"input": "3",
"output": "1 3 2 "
},
{
"input": "4",
"output": "1 4 2 3 "
},
{
"input": "5",
"output": "1 5 2 4 3 "
},
{
"input": "6",
"output": "1 6 2 5 3 4 "
},
{
"input": "1",
"output": "1 "
},
{
"inp... | 1,643,366,404 | 2,147,483,647 | Python 3 | OK | TESTS | 22 | 122 | 0 | import math
def main_function():
n = int(input())
current_ = 1
direction_plus = True
dif = n - 1
for i in range(n):
if i == 0:
print(1, end=" ")
elif i != n - 1:
if direction_plus:
print(dif + current_, end=" ")
current_ += dif
else:
print(current_ - dif, end=" ")
current_ -= dif
dif -= 1
direction_plus = not direction_plus
else:
if direction_plus:
print(dif + current_)
current_ += dif
else:
print(current_ - dif,)
current_ -= dif
dif -= 1
direction_plus = not direction_plus
main_function() | Title: Little Frog
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once upon a time a little frog whose name was Vasya decided to travel around his home swamp. Overall there are *n* mounds on the swamp, located on one line. The distance between the neighboring mounds is one meter. Vasya wants to visit all the mounds in one day; besides, he wants to visit each one exactly once. For that he makes a route plan, to decide the order in which to jump on the mounds. Vasya can pick any mound as the first one. He thinks it boring to jump two times at the same distance. That's why he wants any two jumps on his route to have different lengths. Help Vasya the Frog and make the plan for him.
Input Specification:
The single line contains a number *n* (1<=≤<=*n*<=≤<=104) which is the number of mounds.
Output Specification:
Print *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) which are the frog's route plan.
- All the *p**i*'s should be mutually different. - All the |*p**i*–*p**i*<=+<=1|'s should be mutually different (1<=≤<=*i*<=≤<=*n*<=-<=1).
If there are several solutions, output any.
Demo Input:
['2\n', '3\n']
Demo Output:
['1 2 ', '1 3 2 ']
Note:
none | ```python
import math
def main_function():
n = int(input())
current_ = 1
direction_plus = True
dif = n - 1
for i in range(n):
if i == 0:
print(1, end=" ")
elif i != n - 1:
if direction_plus:
print(dif + current_, end=" ")
current_ += dif
else:
print(current_ - dif, end=" ")
current_ -= dif
dif -= 1
direction_plus = not direction_plus
else:
if direction_plus:
print(dif + current_)
current_ += dif
else:
print(current_ - dif,)
current_ -= dif
dif -= 1
direction_plus = not direction_plus
main_function()
``` | 3.9695 |
691 | B | s-palindrome | PROGRAMMING | 1,600 | [
"implementation",
"strings"
] | null | null | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
You are given a string *s*. Check if the string is "s-palindrome". | The only line contains the string *s* (1<=≤<=|*s*|<=≤<=1000) which consists of only English letters. | Print "TAK" if the string *s* is "s-palindrome" and "NIE" otherwise. | [
"oXoxoXo\n",
"bod\n",
"ER\n"
] | [
"TAK\n",
"TAK\n",
"NIE\n"
] | none | 0 | [
{
"input": "oXoxoXo",
"output": "TAK"
},
{
"input": "bod",
"output": "TAK"
},
{
"input": "ER",
"output": "NIE"
},
{
"input": "o",
"output": "TAK"
},
{
"input": "a",
"output": "NIE"
},
{
"input": "opo",
"output": "NIE"
},
{
"input": "HCMoxkg... | 1,655,008,376 | 2,147,483,647 | PyPy 3 | OK | TESTS | 168 | 93 | 0 | s = str(input())
t = 'AHIMOoTUVvWwXxY'
d = {'p' : 'q', 'q' : 'p', 'b' : 'd', 'd' : 'b'}
for c in t:
d[c] = c
for i, c in enumerate(s):
if c in d and s[len(s)-1-i] == d[c]:
continue
else:
print('NIE')
exit()
print('TAK')
| Title: s-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
You are given a string *s*. Check if the string is "s-palindrome".
Input Specification:
The only line contains the string *s* (1<=≤<=|*s*|<=≤<=1000) which consists of only English letters.
Output Specification:
Print "TAK" if the string *s* is "s-palindrome" and "NIE" otherwise.
Demo Input:
['oXoxoXo\n', 'bod\n', 'ER\n']
Demo Output:
['TAK\n', 'TAK\n', 'NIE\n']
Note:
none | ```python
s = str(input())
t = 'AHIMOoTUVvWwXxY'
d = {'p' : 'q', 'q' : 'p', 'b' : 'd', 'd' : 'b'}
for c in t:
d[c] = c
for i, c in enumerate(s):
if c in d and s[len(s)-1-i] == d[c]:
continue
else:
print('NIE')
exit()
print('TAK')
``` | 3 | |
134 | A | Average Numbers | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | You are given a sequence of positive integers *a*1,<=*a*2,<=...,<=*a**n*. Find all such indices *i*, that the *i*-th element equals the arithmetic mean of all other elements (that is all elements except for this one). | The first line contains the integer *n* (2<=≤<=*n*<=≤<=2·105). The second line contains elements of the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). All the elements are positive integers. | Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to *n*.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line. | [
"5\n1 2 3 4 5\n",
"4\n50 50 50 50\n"
] | [
"1\n3 ",
"4\n1 2 3 4 "
] | none | 500 | [
{
"input": "5\n1 2 3 4 5",
"output": "1\n3 "
},
{
"input": "4\n50 50 50 50",
"output": "4\n1 2 3 4 "
},
{
"input": "3\n2 3 1",
"output": "1\n1 "
},
{
"input": "2\n4 2",
"output": "0"
},
{
"input": "2\n1 1",
"output": "2\n1 2 "
},
{
"input": "10\n3 3 3 ... | 1,683,470,483 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 1,000 | 12,800,000 | n = int(input())
a = list(map(int, input().split()))
p = []
for i in range(n):
if float(a[i]) == (sum(a)-a[i])/(n-1):
p.append(i+1)
print(len(p))
print(*p)
# Sun May 07 2023 17:41:22 GMT+0300 (Moscow Standard Time)
| Title: Average Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence of positive integers *a*1,<=*a*2,<=...,<=*a**n*. Find all such indices *i*, that the *i*-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input Specification:
The first line contains the integer *n* (2<=≤<=*n*<=≤<=2·105). The second line contains elements of the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). All the elements are positive integers.
Output Specification:
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to *n*.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Demo Input:
['5\n1 2 3 4 5\n', '4\n50 50 50 50\n']
Demo Output:
['1\n3 ', '4\n1 2 3 4 ']
Note:
none | ```python
n = int(input())
a = list(map(int, input().split()))
p = []
for i in range(n):
if float(a[i]) == (sum(a)-a[i])/(n-1):
p.append(i+1)
print(len(p))
print(*p)
# Sun May 07 2023 17:41:22 GMT+0300 (Moscow Standard Time)
``` | 0 | |
522 | A | Reposts | PROGRAMMING | 1,200 | [
"*special",
"dfs and similar",
"dp",
"graphs",
"trees"
] | null | null | One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user. | Print a single integer — the maximum length of a repost chain. | [
"5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n",
"6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n",
"1\nSoMeStRaNgEgUe repos... | [
"6\n",
"2\n",
"2\n"
] | none | 500 | [
{
"input": "5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya",
"output": "6"
},
{
"input": "6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforc... | 1,679,136,475 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 102,400 | from collections import defaultdict
def max_depth(tree, root):
if len(tree[root]) == 0:
return 1
for child in tree[root]:
return max_depth(tree, child) + 1
n = int(input())
tree = defaultdict(list)
for _ in range(n):
reposter, _ , poster = input().split()
parent = poster.lower()
child = reposter.lower()
#populate tree
tree[parent].append(child)
height = max_depth(tree, 'polycarp')
print(height)
| Title: Reposts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user.
Output Specification:
Print a single integer — the maximum length of a repost chain.
Demo Input:
['5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n', '6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n', '1\nSoMeStRaNgEgUe reposted PoLyCaRp\n']
Demo Output:
['6\n', '2\n', '2\n']
Note:
none | ```python
from collections import defaultdict
def max_depth(tree, root):
if len(tree[root]) == 0:
return 1
for child in tree[root]:
return max_depth(tree, child) + 1
n = int(input())
tree = defaultdict(list)
for _ in range(n):
reposter, _ , poster = input().split()
parent = poster.lower()
child = reposter.lower()
#populate tree
tree[parent].append(child)
height = max_depth(tree, 'polycarp')
print(height)
``` | 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"... | 1,622,183,942 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 122 | 0 | n=int(input())
l=list(map(int,input().split(' ')))
for i in range(n-1):
if((l[i+1]-l[i])%2!=0):
print(i+2)
break | 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())
l=list(map(int,input().split(' ')))
for i in range(n-1):
if((l[i+1]-l[i])%2!=0):
print(i+2)
break
``` | 0 |
412 | B | Network Configuration | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | null | null | The R1 company wants to hold a web search championship. There were *n* computers given for the competition, each of them is connected to the Internet. The organizers believe that the data transfer speed directly affects the result. The higher the speed of the Internet is, the faster the participant will find the necessary information. Therefore, before the competition started, each computer had its maximum possible data transfer speed measured. On the *i*-th computer it was *a**i* kilobits per second.
There will be *k* participants competing in the championship, each should get a separate computer. The organizing company does not want any of the participants to have an advantage over the others, so they want to provide the same data transfer speed to each participant's computer. Also, the organizers want to create the most comfortable conditions for the participants, so the data transfer speed on the participants' computers should be as large as possible.
The network settings of the R1 company has a special option that lets you to cut the initial maximum data transfer speed of any computer to any lower speed. How should the R1 company configure the network using the described option so that at least *k* of *n* computers had the same data transfer speed and the data transfer speed on these computers was as large as possible? | The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100) — the number of computers and the number of participants, respectively. In the second line you have a space-separated sequence consisting of *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (16<=≤<=*a**i*<=≤<=32768); number *a**i* denotes the maximum data transfer speed on the *i*-th computer. | Print a single integer — the maximum Internet speed value. It is guaranteed that the answer to the problem is always an integer. | [
"3 2\n40 20 30\n",
"6 4\n100 20 40 20 50 50\n"
] | [
"30\n",
"40\n"
] | In the first test case the organizers can cut the first computer's speed to 30 kilobits. Then two computers (the first and the third one) will have the same speed of 30 kilobits. They should be used as the participants' computers. This answer is optimal. | 1,000 | [
{
"input": "3 2\n40 20 30",
"output": "30"
},
{
"input": "6 4\n100 20 40 20 50 50",
"output": "40"
},
{
"input": "1 1\n16",
"output": "16"
},
{
"input": "2 1\n10000 17",
"output": "10000"
},
{
"input": "2 2\n200 300",
"output": "200"
},
{
"input": "3 1... | 1,568,996,118 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 140 | 0 |
n, k = map(int, input().split())
a = list(reversed(sorted(map(int, input().split()))))
print(a[k - 1])
# ♥
| Title: Network Configuration
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The R1 company wants to hold a web search championship. There were *n* computers given for the competition, each of them is connected to the Internet. The organizers believe that the data transfer speed directly affects the result. The higher the speed of the Internet is, the faster the participant will find the necessary information. Therefore, before the competition started, each computer had its maximum possible data transfer speed measured. On the *i*-th computer it was *a**i* kilobits per second.
There will be *k* participants competing in the championship, each should get a separate computer. The organizing company does not want any of the participants to have an advantage over the others, so they want to provide the same data transfer speed to each participant's computer. Also, the organizers want to create the most comfortable conditions for the participants, so the data transfer speed on the participants' computers should be as large as possible.
The network settings of the R1 company has a special option that lets you to cut the initial maximum data transfer speed of any computer to any lower speed. How should the R1 company configure the network using the described option so that at least *k* of *n* computers had the same data transfer speed and the data transfer speed on these computers was as large as possible?
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100) — the number of computers and the number of participants, respectively. In the second line you have a space-separated sequence consisting of *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (16<=≤<=*a**i*<=≤<=32768); number *a**i* denotes the maximum data transfer speed on the *i*-th computer.
Output Specification:
Print a single integer — the maximum Internet speed value. It is guaranteed that the answer to the problem is always an integer.
Demo Input:
['3 2\n40 20 30\n', '6 4\n100 20 40 20 50 50\n']
Demo Output:
['30\n', '40\n']
Note:
In the first test case the organizers can cut the first computer's speed to 30 kilobits. Then two computers (the first and the third one) will have the same speed of 30 kilobits. They should be used as the participants' computers. This answer is optimal. | ```python
n, k = map(int, input().split())
a = list(reversed(sorted(map(int, input().split()))))
print(a[k - 1])
# ♥
``` | 3 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,614,968,352 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | t=input()
x=0
y=0
z=0
for i in range(t):
array=map(int,raw_input().split(" "))
x+=array[0]
y+=array[1]
z+=array[2]
print "YES" if (x==0 and y==0 and z==0) else "NO" | Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
t=input()
x=0
y=0
z=0
for i in range(t):
array=map(int,raw_input().split(" "))
x+=array[0]
y+=array[1]
z+=array[2]
print "YES" if (x==0 and y==0 and z==0) else "NO"
``` | -1 |
687 | B | Remainders Game | PROGRAMMING | 1,800 | [
"chinese remainder theorem",
"math",
"number theory"
] | null | null | Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*?
Note, that means the remainder of *x* after dividing it by *y*. | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000). | Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise. | [
"4 5\n2 3 5 12\n",
"2 7\n2 3\n"
] | [
"Yes\n",
"No\n"
] | In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7. | 1,000 | [
{
"input": "4 5\n2 3 5 12",
"output": "Yes"
},
{
"input": "2 7\n2 3",
"output": "No"
},
{
"input": "1 6\n8",
"output": "No"
},
{
"input": "2 3\n9 4",
"output": "Yes"
},
{
"input": "4 16\n19 16 13 9",
"output": "Yes"
},
{
"input": "5 10\n5 16 19 9 17",
... | 1,573,821,357 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 10 | 124 | 4,710,400 | def gcd(a, b):
if a % b == 0:
return b
return gcd(b, a % b)
def lcm(c):
n = len(c)
if n == 1: return c[0]
a = lcm(c[:n//2])
b = lcm(c[n//2:])
return a * b // gcd(a, b)
def main():
n, k = map(lambda x : int(x), input().split())
c = list(map(lambda x : int(x), input().split()))
if gcd(k, lcm(c))==k: print("Yes")
else: print("No")
main() | Title: Remainders Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*?
Note, that means the remainder of *x* after dividing it by *y*.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000).
Output Specification:
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise.
Demo Input:
['4 5\n2 3 5 12\n', '2 7\n2 3\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7. | ```python
def gcd(a, b):
if a % b == 0:
return b
return gcd(b, a % b)
def lcm(c):
n = len(c)
if n == 1: return c[0]
a = lcm(c[:n//2])
b = lcm(c[n//2:])
return a * b // gcd(a, b)
def main():
n, k = map(lambda x : int(x), input().split())
c = list(map(lambda x : int(x), input().split()))
if gcd(k, lcm(c))==k: print("Yes")
else: print("No")
main()
``` | -1 | |
39 | B | Company Income Growth | PROGRAMMING | 1,300 | [
"greedy"
] | B. Company Income Growth | 2 | 64 | Petya works as a PR manager for a successful Berland company BerSoft. He needs to prepare a presentation on the company income growth since 2001 (the year of its founding) till now. Petya knows that in 2001 the company income amounted to *a*1 billion bourles, in 2002 — to *a*2 billion, ..., and in the current (2000<=+<=*n*)-th year — *a**n* billion bourles. On the base of the information Petya decided to show in his presentation the linear progress history which is in his opinion perfect. According to a graph Petya has already made, in the first year BerSoft company income must amount to 1 billion bourles, in the second year — 2 billion bourles etc., each following year the income increases by 1 billion bourles. Unfortunately, the real numbers are different from the perfect ones. Among the numbers *a**i* can even occur negative ones that are a sign of the company’s losses in some years. That is why Petya wants to ignore some data, in other words, cross some numbers *a**i* from the sequence and leave only some subsequence that has perfect growth.
Thus Petya has to choose a sequence of years *y*1, *y*2, ..., *y**k*,so that in the year *y*1 the company income amounted to 1 billion bourles, in the year *y*2 — 2 billion bourles etc., in accordance with the perfect growth dynamics. Help him to choose the longest such sequence. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). The next line contains *n* integers *a**i* (<=-<=100<=≤<=*a**i*<=≤<=100). The number *a**i* determines the income of BerSoft company in the (2000<=+<=*i*)-th year. The numbers in the line are separated by spaces. | Output *k* — the maximum possible length of a perfect sequence. In the next line output the sequence of years *y*1, *y*2, ..., *y**k*. Separate the numbers by spaces. If the answer is not unique, output any. If no solution exist, output one number 0. | [
"10\n-2 1 1 3 2 3 4 -10 -2 5\n",
"3\n-1 -2 -3\n"
] | [
"5\n2002 2005 2006 2007 2010\n",
"0\n"
] | none | 0 | [
{
"input": "10\n-2 1 1 3 2 3 4 -10 -2 5",
"output": "5\n2002 2005 2006 2007 2010 "
},
{
"input": "3\n-1 -2 -3",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1\n2002 "
},
{
"in... | 1,565,389,762 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | import sys
n = sys.stdin.readline()
arr = sys.stdin.readline().split()
arr = [ int(x) for x in arr]
res = []
contador = 1
for i in range(len(arr)):
print arr[i]
if arr[i] == contador:
contador += 1
res.append(2000 + i +1)
if res:
print(' '.join([str(y) for y in res]))
else:
print(0)
| Title: Company Income Growth
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Petya works as a PR manager for a successful Berland company BerSoft. He needs to prepare a presentation on the company income growth since 2001 (the year of its founding) till now. Petya knows that in 2001 the company income amounted to *a*1 billion bourles, in 2002 — to *a*2 billion, ..., and in the current (2000<=+<=*n*)-th year — *a**n* billion bourles. On the base of the information Petya decided to show in his presentation the linear progress history which is in his opinion perfect. According to a graph Petya has already made, in the first year BerSoft company income must amount to 1 billion bourles, in the second year — 2 billion bourles etc., each following year the income increases by 1 billion bourles. Unfortunately, the real numbers are different from the perfect ones. Among the numbers *a**i* can even occur negative ones that are a sign of the company’s losses in some years. That is why Petya wants to ignore some data, in other words, cross some numbers *a**i* from the sequence and leave only some subsequence that has perfect growth.
Thus Petya has to choose a sequence of years *y*1, *y*2, ..., *y**k*,so that in the year *y*1 the company income amounted to 1 billion bourles, in the year *y*2 — 2 billion bourles etc., in accordance with the perfect growth dynamics. Help him to choose the longest such sequence.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). The next line contains *n* integers *a**i* (<=-<=100<=≤<=*a**i*<=≤<=100). The number *a**i* determines the income of BerSoft company in the (2000<=+<=*i*)-th year. The numbers in the line are separated by spaces.
Output Specification:
Output *k* — the maximum possible length of a perfect sequence. In the next line output the sequence of years *y*1, *y*2, ..., *y**k*. Separate the numbers by spaces. If the answer is not unique, output any. If no solution exist, output one number 0.
Demo Input:
['10\n-2 1 1 3 2 3 4 -10 -2 5\n', '3\n-1 -2 -3\n']
Demo Output:
['5\n2002 2005 2006 2007 2010\n', '0\n']
Note:
none | ```python
import sys
n = sys.stdin.readline()
arr = sys.stdin.readline().split()
arr = [ int(x) for x in arr]
res = []
contador = 1
for i in range(len(arr)):
print arr[i]
if arr[i] == contador:
contador += 1
res.append(2000 + i +1)
if res:
print(' '.join([str(y) for y in res]))
else:
print(0)
``` | -1 |
608 | B | Hamming Distance Sum | PROGRAMMING | 1,500 | [
"combinatorics",
"strings"
] | null | null | Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|. | The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only. | Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|. | [
"01\n00111\n",
"0011\n0110\n"
] | [
"3\n",
"2\n"
] | For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement. | 1,000 | [
{
"input": "01\n00111",
"output": "3"
},
{
"input": "0011\n0110",
"output": "2"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "0\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1... | 1,660,802,172 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 8,806,400 | a = list(map(int, input()))
b = list(map(int, input()))
total = 0
for i in range(len(a)):
for j in range(i, i+len(b)-len(a)+1):
total += (a[i] + b[j]) % 2
print(total)
| Title: Hamming Distance Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Input Specification:
The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only.
Output Specification:
Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Demo Input:
['01\n00111\n', '0011\n0110\n']
Demo Output:
['3\n', '2\n']
Note:
For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement. | ```python
a = list(map(int, input()))
b = list(map(int, input()))
total = 0
for i in range(len(a)):
for j in range(i, i+len(b)-len(a)+1):
total += (a[i] + b[j]) % 2
print(total)
``` | 0 | |
629 | B | Far Relative’s Problem | PROGRAMMING | 1,100 | [
"brute force"
] | null | null | Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible.
Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.
Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends.
Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive. | Print the maximum number of people that may come to Famil Door's party. | [
"4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n",
"6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n"
] | [
"2\n",
"4\n"
] | In the first sample, friends 3 and 4 can come on any day in range [117, 128].
In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140. | 1,000 | [
{
"input": "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128",
"output": "2"
},
{
"input": "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200",
"output": "4"
},
{
"input": "1\nF 68 307",
"output": "0"
},
{
"input": "40\nM 55 363\nF 117 252\nM 157 282\nF 322 345\... | 1,666,718,529 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 76 | 93 | 1,843,200 | from sys import stdin,stdout
input=stdin.readline
def print(*args, end='\n', sep=' ') -> None:
stdout.write(sep.join(map(str, args)) + end)
n=int(input()) ; female=[0]*368 ; male=[0]*368 ; ans=0
for i in range(n):
gen,st,end=input().split()
if gen=="F":
female[int(st)-1]+=1
female[int(end)]-=1
else:
male[int(st)-1]+=1
male[int(end)]-=1
for i in range(367):
female[i+1]+=female[i]
male[i+1]+=male[i]
for i in range(368):
ans=max(min(female[i],male[i]),ans)
print(ans*2) | Title: Far Relative’s Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible.
Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.
Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends.
Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive.
Output Specification:
Print the maximum number of people that may come to Famil Door's party.
Demo Input:
['4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n', '6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample, friends 3 and 4 can come on any day in range [117, 128].
In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140. | ```python
from sys import stdin,stdout
input=stdin.readline
def print(*args, end='\n', sep=' ') -> None:
stdout.write(sep.join(map(str, args)) + end)
n=int(input()) ; female=[0]*368 ; male=[0]*368 ; ans=0
for i in range(n):
gen,st,end=input().split()
if gen=="F":
female[int(st)-1]+=1
female[int(end)]-=1
else:
male[int(st)-1]+=1
male[int(end)]-=1
for i in range(367):
female[i+1]+=female[i]
male[i+1]+=male[i]
for i in range(368):
ans=max(min(female[i],male[i]),ans)
print(ans*2)
``` | 3 | |
466 | A | Cheap Travel | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides? | The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket. | Print a single integer — the minimum sum in rubles that Ann will need to spend. | [
"6 2 1 2\n",
"5 2 2 3\n"
] | [
"6\n",
"8\n"
] | In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets. | 500 | [
{
"input": "6 2 1 2",
"output": "6"
},
{
"input": "5 2 2 3",
"output": "8"
},
{
"input": "10 3 5 1",
"output": "4"
},
{
"input": "1000 1 1000 1000",
"output": "1000000"
},
{
"input": "1000 3 1000 1000",
"output": "334000"
},
{
"input": "1 1 1 1",
"... | 1,690,718,007 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 30 | 0 | n,m,a,b = map(int,input().split())
z=b/m
if a<=z:
print(n*a)
elif b>a or n%m==0:
print(int(b*(n//m)+a*(n%m)))
else:
print(b*(n//m)+1) | Title: Cheap Travel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
Input Specification:
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
Output Specification:
Print a single integer — the minimum sum in rubles that Ann will need to spend.
Demo Input:
['6 2 1 2\n', '5 2 2 3\n']
Demo Output:
['6\n', '8\n']
Note:
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets. | ```python
n,m,a,b = map(int,input().split())
z=b/m
if a<=z:
print(n*a)
elif b>a or n%m==0:
print(int(b*(n//m)+a*(n%m)))
else:
print(b*(n//m)+1)
``` | 0 | |
11 | A | Increasing Sequence | PROGRAMMING | 900 | [
"constructive algorithms",
"implementation",
"math"
] | A. Increasing Sequence | 1 | 64 | A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing? | The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106). | Output the minimal number of moves needed to make the sequence increasing. | [
"4 2\n1 3 3 2\n"
] | [
"3\n"
] | none | 0 | [
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
... | 1,591,733,210 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 280 | 1,331,200 | from math import ceil
n,d=map(int,input().split())
p=-1
ans=0
for i in map(int,input().split()):
k=ceil((max(p-i+1,0))/d)
ans+=k
p=i+k*d
print(ans) | Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none | ```python
from math import ceil
n,d=map(int,input().split())
p=-1
ans=0
for i in map(int,input().split()):
k=ceil((max(p-i+1,0))/d)
ans+=k
p=i+k*d
print(ans)
``` | 3.850082 |
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... | 1,687,956,015 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 92 | 0 | n=int(input())
vm=maxi=0
for i in range(n):
a,b=input().split(" ")
vm-=int(a)
vm+=int(b)
maxi=max(maxi,vm)
print(maxi) | 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
n=int(input())
vm=maxi=0
for i in range(n):
a,b=input().split(" ")
vm-=int(a)
vm+=int(b)
maxi=max(maxi,vm)
print(maxi)
``` | 3 | |
126 | B | Password | PROGRAMMING | 1,700 | [
"binary search",
"dp",
"hashing",
"string suffix structures",
"strings"
] | null | null | Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend. | You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters. | Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes. | [
"fixprefixsuffix\n",
"abcdabc\n"
] | [
"fix",
"Just a legend"
] | none | 1,000 | [
{
"input": "fixprefixsuffix",
"output": "fix"
},
{
"input": "abcdabc",
"output": "Just a legend"
},
{
"input": "qwertyqwertyqwerty",
"output": "qwerty"
},
{
"input": "papapapap",
"output": "papap"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaa"
},
{
"... | 1,543,733,508 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 248 | 0 | s = input()
n = len(s)
if n==1 or n==2:
print("Just a legend")
else:
lista = []
for i in range(n):
lista.append(0)
l = 0
i = 1
while i < n:
if s[l] == s[i]:
lista[i] = l
l += 1
i += 1
else:
if l != 0:
l = lista[l-1]
else:
lista[i] = 0
i += 1
if (lista[-1] in lista[:-1]) and (lista[-1] != 0):
print(s[:value])
elif lista[lista[-1] - 1]==0:
print("Just a legend")
elif lista[lista[-1] - 1]:
print(s[:lista[lista[-1] - 1]])
else:
print("Just a legend")
| Title: Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
Input Specification:
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
Output Specification:
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
Demo Input:
['fixprefixsuffix\n', 'abcdabc\n']
Demo Output:
['fix', 'Just a legend']
Note:
none | ```python
s = input()
n = len(s)
if n==1 or n==2:
print("Just a legend")
else:
lista = []
for i in range(n):
lista.append(0)
l = 0
i = 1
while i < n:
if s[l] == s[i]:
lista[i] = l
l += 1
i += 1
else:
if l != 0:
l = lista[l-1]
else:
lista[i] = 0
i += 1
if (lista[-1] in lista[:-1]) and (lista[-1] != 0):
print(s[:value])
elif lista[lista[-1] - 1]==0:
print("Just a legend")
elif lista[lista[-1] - 1]:
print(s[:lista[lista[-1] - 1]])
else:
print("Just a legend")
``` | -1 | |
350 | A | TL | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"implementation"
] | null | null | Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist. | The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds. | If there is a valid TL value, print it. Otherwise, print -1. | [
"3 6\n4 5 2\n8 9 6 10 7 11\n",
"3 1\n3 4 5\n6\n"
] | [
"5",
"-1\n"
] | none | 500 | [
{
"input": "3 6\n4 5 2\n8 9 6 10 7 11",
"output": "5"
},
{
"input": "3 1\n3 4 5\n6",
"output": "-1"
},
{
"input": "2 5\n45 99\n49 41 77 83 45",
"output": "-1"
},
{
"input": "50 50\n18 13 5 34 10 36 36 12 15 11 16 17 14 36 23 45 32 24 31 18 24 32 7 1 31 3 49 8 16 23 3 39 47 43... | 1,526,725,639 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 154 | 0 | a,b=input().split()
a,b,list1,list2=int(a),int(b),[],[]
list1=[int(x) for x in input().split()]
list2=[int(x) for x in input().split()]
c,d,e=max(list1),min(list2),min(list1)
if c>d:print(-1)
elif c<d:
if 2*e<=c:print(c)
else:print(-1)
| Title: TL
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
Output Specification:
If there is a valid TL value, print it. Otherwise, print -1.
Demo Input:
['3 6\n4 5 2\n8 9 6 10 7 11\n', '3 1\n3 4 5\n6\n']
Demo Output:
['5', '-1\n']
Note:
none | ```python
a,b=input().split()
a,b,list1,list2=int(a),int(b),[],[]
list1=[int(x) for x in input().split()]
list2=[int(x) for x in input().split()]
c,d,e=max(list1),min(list2),min(list1)
if c>d:print(-1)
elif c<d:
if 2*e<=c:print(c)
else:print(-1)
``` | 0 | |
611 | B | New Year and Old Property | PROGRAMMING | 1,300 | [
"bitmasks",
"brute force",
"implementation"
] | null | null | The year 2015 is almost over.
Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510<==<=111110111112. Note that he doesn't care about the number of zeros in the decimal representation.
Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?
Assume that all positive integers are always written without leading zeros. | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=1018) — the first year and the last year in Limak's interval respectively. | Print one integer – the number of years Limak will count in his chosen interval. | [
"5 10\n",
"2015 2015\n",
"100 105\n",
"72057594000000000 72057595000000000\n"
] | [
"2\n",
"1\n",
"0\n",
"26\n"
] | In the first sample Limak's interval contains numbers 5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>, 6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>, 7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>, 8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>, 9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub> and 10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>. Two of them (101<sub class="lower-index">2</sub> and 110<sub class="lower-index">2</sub>) have the described property. | 750 | [
{
"input": "5 10",
"output": "2"
},
{
"input": "2015 2015",
"output": "1"
},
{
"input": "100 105",
"output": "0"
},
{
"input": "72057594000000000 72057595000000000",
"output": "26"
},
{
"input": "1 100",
"output": "16"
},
{
"input": "100000000000000000... | 1,571,426,708 | 2,147,483,647 | Python 3 | OK | TESTS | 103 | 109 | 0 | countYears = 0
def newYearCount(startYear, endYear, baseYear, zeros):
global countYears
if (baseYear > endYear):
return None
if (startYear <= baseYear and baseYear<= endYear and zeros == 1):
countYears += 1
if (zeros == 0):
newYearCount(startYear, endYear, baseYear << 1, zeros + 1)
newYearCount(startYear, endYear, (baseYear << 1) + 1, zeros)
startYear, endYear = map(int, input().split())
newYearCount(startYear, endYear, 1, 0)
print(countYears) | Title: New Year and Old Property
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The year 2015 is almost over.
Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510<==<=111110111112. Note that he doesn't care about the number of zeros in the decimal representation.
Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?
Assume that all positive integers are always written without leading zeros.
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=1018) — the first year and the last year in Limak's interval respectively.
Output Specification:
Print one integer – the number of years Limak will count in his chosen interval.
Demo Input:
['5 10\n', '2015 2015\n', '100 105\n', '72057594000000000 72057595000000000\n']
Demo Output:
['2\n', '1\n', '0\n', '26\n']
Note:
In the first sample Limak's interval contains numbers 5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>, 6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>, 7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>, 8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>, 9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub> and 10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>. Two of them (101<sub class="lower-index">2</sub> and 110<sub class="lower-index">2</sub>) have the described property. | ```python
countYears = 0
def newYearCount(startYear, endYear, baseYear, zeros):
global countYears
if (baseYear > endYear):
return None
if (startYear <= baseYear and baseYear<= endYear and zeros == 1):
countYears += 1
if (zeros == 0):
newYearCount(startYear, endYear, baseYear << 1, zeros + 1)
newYearCount(startYear, endYear, (baseYear << 1) + 1, zeros)
startYear, endYear = map(int, input().split())
newYearCount(startYear, endYear, 1, 0)
print(countYears)
``` | 3 | |
1,003 | A | Polycarp's Pockets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that. | The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins. | Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket. | [
"6\n1 2 4 3 3 2\n",
"1\n100\n"
] | [
"2\n",
"1\n"
] | none | 0 | [
{
"input": "6\n1 2 4 3 3 2",
"output": "2"
},
{
"input": "1\n100",
"output": "1"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100... | 1,609,938,341 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 77 | 0 | # import os
t = int(input())
a = list(map(int, input().split()))
r = 0
for item in list(set(a)):
new = a.count(item)
if new > r:
r = new
print(new)
# 03/01 - 1
# 04/01 - 21
# 05/01 - 27
# 06/01 - 4
| Title: Polycarp's Pockets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.
Output Specification:
Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.
Demo Input:
['6\n1 2 4 3 3 2\n', '1\n100\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
# import os
t = int(input())
a = list(map(int, input().split()))
r = 0
for item in list(set(a)):
new = a.count(item)
if new > r:
r = new
print(new)
# 03/01 - 1
# 04/01 - 21
# 05/01 - 27
# 06/01 - 4
``` | 0 | |
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",
... | 1,642,948,510 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 218 | 3,788,800 | from fractions import Fraction
x,y = list(map(int, input().split()))
print(Fraction(y/x)) | 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
from fractions import Fraction
x,y = list(map(int, input().split()))
print(Fraction(y/x))
``` | 0 |
598 | A | Tricky Sum | PROGRAMMING | 900 | [
"math"
] | null | null | In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum.
For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for *t* values of *n*. | The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed.
Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109). | Print the requested sum for each of *t* integers *n* given in the input. | [
"2\n4\n1000000000\n"
] | [
"-4\n499999998352516354\n"
] | The answer for the first sample is explained in the statement. | 0 | [
{
"input": "2\n4\n1000000000",
"output": "-4\n499999998352516354"
},
{
"input": "10\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "-1\n-3\n0\n-4\n1\n7\n14\n6\n15\n25"
},
{
"input": "10\n10\n9\n47\n33\n99\n83\n62\n1\n100\n53",
"output": "25\n15\n1002\n435\n4696\n3232\n1827\n-1\n4796\n130... | 1,636,427,808 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 4,505,600 | from math import log2, floor
from time import time
def solve(n):
maxPower = floor(log2(n))
negSum = 0
for i in range(maxPower + 1):
negSum -= 2 ** i
Sum = int(0.5 * n * (1 + n))
Sum += 2 * negSum
return Sum
if __name__ == "__main__":
numInputs = int(input('t?'))
for i in range(numInputs):
num = int(input('n?'))
start = time()
print(solve(num))
end = time()
print('%.3es' % (end - start)) | Title: Tricky Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum.
For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for *t* values of *n*.
Input Specification:
The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed.
Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the requested sum for each of *t* integers *n* given in the input.
Demo Input:
['2\n4\n1000000000\n']
Demo Output:
['-4\n499999998352516354\n']
Note:
The answer for the first sample is explained in the statement. | ```python
from math import log2, floor
from time import time
def solve(n):
maxPower = floor(log2(n))
negSum = 0
for i in range(maxPower + 1):
negSum -= 2 ** i
Sum = int(0.5 * n * (1 + n))
Sum += 2 * negSum
return Sum
if __name__ == "__main__":
numInputs = int(input('t?'))
for i in range(numInputs):
num = int(input('n?'))
start = time()
print(solve(num))
end = time()
print('%.3es' % (end - start))
``` | 0 | |
34 | A | Reconnaissance 2 | PROGRAMMING | 800 | [
"implementation"
] | A. Reconnaissance 2 | 2 | 256 | *n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit. | The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction. | Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle. | [
"5\n10 12 13 15 10\n",
"4\n10 20 30 40\n"
] | [
"5 1\n",
"1 2\n"
] | none | 500 | [
{
"input": "5\n10 12 13 15 10",
"output": "5 1"
},
{
"input": "4\n10 20 30 40",
"output": "1 2"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "2 3"
},
{
"input": "5\n826 747 849 687 437",
"output": "1 2"
},
{
"input": "5\n999 999 993 969 999",
"output"... | 1,676,608,911 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 92 | 0 | x=int(input())
a=list(map(int,input().split()))
c=abs(a[1]-a[0])
l=1
y=2
for i in range(1,x):
if c>abs(a[i]-a[i-1]):
c=abs(a[i]-a[i-1])
l=i+1
y=i
if c>abs(a[x-1]-a[0]):
c=abs(a[x-1]-a[0])
l=x
y=1
print(l,y)
| Title: Reconnaissance 2
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output Specification:
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Demo Input:
['5\n10 12 13 15 10\n', '4\n10 20 30 40\n']
Demo Output:
['5 1\n', '1 2\n']
Note:
none | ```python
x=int(input())
a=list(map(int,input().split()))
c=abs(a[1]-a[0])
l=1
y=2
for i in range(1,x):
if c>abs(a[i]-a[i-1]):
c=abs(a[i]-a[i-1])
l=i+1
y=i
if c>abs(a[x-1]-a[0]):
c=abs(a[x-1]-a[0])
l=x
y=1
print(l,y)
``` | 3.977 |
858 | A | k-rounding | PROGRAMMING | 1,100 | [
"brute force",
"math",
"number theory"
] | null | null | For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*. | The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8). | Print the *k*-rounding of *n*. | [
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] | [
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] | none | 750 | [
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output":... | 1,505,659,530 | 6,030 | Python 3 | WRONG_ANSWER | TESTS | 30 | 170 | 5,529,600 | n,k=map(int,input().split())
for i in range(1,100000):
m=str(n*i)
m=m[len(m)-k:len(m)]
if(m=='0'*k):
print(n*i)
break
i+=1
else:
print(str(n)+('0'*k))
| Title: k-rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none | ```python
n,k=map(int,input().split())
for i in range(1,100000):
m=str(n*i)
m=m[len(m)-k:len(m)]
if(m=='0'*k):
print(n*i)
break
i+=1
else:
print(str(n)+('0'*k))
``` | 0 | |
652 | A | Gabriel and Caterpillar | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | The 9-th grade student Gabriel noticed a caterpillar on a tree when walking around in a forest after the classes. The caterpillar was on the height *h*1 cm from the ground. On the height *h*2 cm (*h*2<=><=*h*1) on the same tree hung an apple and the caterpillar was crawling to the apple.
Gabriel is interested when the caterpillar gets the apple. He noted that the caterpillar goes up by *a* cm per hour by day and slips down by *b* cm per hour by night.
In how many days Gabriel should return to the forest to see the caterpillar get the apple. You can consider that the day starts at 10 am and finishes at 10 pm. Gabriel's classes finish at 2 pm. You can consider that Gabriel noticed the caterpillar just after the classes at 2 pm.
Note that the forest is magic so the caterpillar can slip down under the ground and then lift to the apple. | The first line contains two integers *h*1,<=*h*2 (1<=≤<=*h*1<=<<=*h*2<=≤<=105) — the heights of the position of the caterpillar and the apple in centimeters.
The second line contains two integers *a*,<=*b* (1<=≤<=*a*,<=*b*<=≤<=105) — the distance the caterpillar goes up by day and slips down by night, in centimeters per hour. | Print the only integer *k* — the number of days Gabriel should wait to return to the forest and see the caterpillar getting the apple.
If the caterpillar can't get the apple print the only integer <=-<=1. | [
"10 30\n2 1\n",
"10 13\n1 1\n",
"10 19\n1 2\n",
"1 50\n5 4\n"
] | [
"1\n",
"0\n",
"-1\n",
"1\n"
] | In the first example at 10 pm of the first day the caterpillar gets the height 26. At 10 am of the next day it slips down to the height 14. And finally at 6 pm of the same day the caterpillar gets the apple.
Note that in the last example the caterpillar was slipping down under the ground and getting the apple on the next day. | 0 | [
{
"input": "10 30\n2 1",
"output": "1"
},
{
"input": "10 13\n1 1",
"output": "0"
},
{
"input": "10 19\n1 2",
"output": "-1"
},
{
"input": "1 50\n5 4",
"output": "1"
},
{
"input": "1 1000\n2 1",
"output": "82"
},
{
"input": "999 1000\n1 1",
"output"... | 1,460,552,891 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 62 | 0 | h1, h2 = input().split(" ")
a, b = input().split(" ")
h1 = int(h1)
h2 = int(h2)
a = int(a)
b = int(b)
if a < b:
print(-1)
exit()
hours_elapsed = 0
time = 14
current_hight = h1
while True:
if time >= 24:
time = 0
if time >= 10 and time <= 22:
current_hight += a
else:
current_hight -= b
hours_elapsed += 1
time += 1
if current_hight >= h2:
break
print(round(hours_elapsed / 24)) | Title: Gabriel and Caterpillar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The 9-th grade student Gabriel noticed a caterpillar on a tree when walking around in a forest after the classes. The caterpillar was on the height *h*1 cm from the ground. On the height *h*2 cm (*h*2<=><=*h*1) on the same tree hung an apple and the caterpillar was crawling to the apple.
Gabriel is interested when the caterpillar gets the apple. He noted that the caterpillar goes up by *a* cm per hour by day and slips down by *b* cm per hour by night.
In how many days Gabriel should return to the forest to see the caterpillar get the apple. You can consider that the day starts at 10 am and finishes at 10 pm. Gabriel's classes finish at 2 pm. You can consider that Gabriel noticed the caterpillar just after the classes at 2 pm.
Note that the forest is magic so the caterpillar can slip down under the ground and then lift to the apple.
Input Specification:
The first line contains two integers *h*1,<=*h*2 (1<=≤<=*h*1<=<<=*h*2<=≤<=105) — the heights of the position of the caterpillar and the apple in centimeters.
The second line contains two integers *a*,<=*b* (1<=≤<=*a*,<=*b*<=≤<=105) — the distance the caterpillar goes up by day and slips down by night, in centimeters per hour.
Output Specification:
Print the only integer *k* — the number of days Gabriel should wait to return to the forest and see the caterpillar getting the apple.
If the caterpillar can't get the apple print the only integer <=-<=1.
Demo Input:
['10 30\n2 1\n', '10 13\n1 1\n', '10 19\n1 2\n', '1 50\n5 4\n']
Demo Output:
['1\n', '0\n', '-1\n', '1\n']
Note:
In the first example at 10 pm of the first day the caterpillar gets the height 26. At 10 am of the next day it slips down to the height 14. And finally at 6 pm of the same day the caterpillar gets the apple.
Note that in the last example the caterpillar was slipping down under the ground and getting the apple on the next day. | ```python
h1, h2 = input().split(" ")
a, b = input().split(" ")
h1 = int(h1)
h2 = int(h2)
a = int(a)
b = int(b)
if a < b:
print(-1)
exit()
hours_elapsed = 0
time = 14
current_hight = h1
while True:
if time >= 24:
time = 0
if time >= 10 and time <= 22:
current_hight += a
else:
current_hight -= b
hours_elapsed += 1
time += 1
if current_hight >= h2:
break
print(round(hours_elapsed / 24))
``` | 0 | |
797 | C | Minimal string | PROGRAMMING | 1,700 | [
"data structures",
"greedy",
"strings"
] | null | null | Petya recieved a gift of a string *s* with length up to 105 characters for his birthday. He took two more empty strings *t* and *u* and decided to play a game. This game has two possible moves:
- Extract the first character of *s* and append *t* with this character. - Extract the last character of *t* and append *u* with this character.
Petya wants to get strings *s* and *t* empty and string *u* lexicographically minimal.
You should write a program that will help Petya win the game. | First line contains non-empty string *s* (1<=≤<=|*s*|<=≤<=105), consisting of lowercase English letters. | Print resulting string *u*. | [
"cab\n",
"acdb\n"
] | [
"abc\n",
"abdc\n"
] | none | 0 | [
{
"input": "cab",
"output": "abc"
},
{
"input": "acdb",
"output": "abdc"
},
{
"input": "a",
"output": "a"
},
{
"input": "ab",
"output": "ab"
},
{
"input": "ba",
"output": "ab"
},
{
"input": "dijee",
"output": "deeji"
},
{
"input": "bhrmc",
... | 1,619,513,343 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 16 | 171 | 6,451,200 | import sys
input=sys.stdin.readline
from collections import defaultdict as dc
from collections import Counter
from bisect import bisect_right, bisect_left
import math
from operator import itemgetter
from heapq import heapify, heappop, heappush
from queue import PriorityQueue as pq
s=input()[:-1]
n=len(s)
p=[0]
t=[-1]*n
for i in range(1,n):
if not p:
p.append(i)
else:
while(p and s[p[-1]]>s[i]):
x=p.pop(-1)
t[x]=i
p.append(i)
while p:
x=p.pop(-1)
t[x]=-1
p=""
f=dc(int)
x=-1
for i in range(n):
if t[i]==-1:
p+=s[i]
f[i]=1
x=i
j=x+1
i=x-1
#print(t,s)
while(i>=0 and j<n):
if f[i]==1 or f[j]==1:
if f[i]:
i-=1
else:
j+=1
else:
if s[j]<=s[i]:
p+=s[j]
j+=1
else:
p+=s[i]
i-=1
while(i>=0):
if f[i]==0:
p+=s[i]
i-=1
while(j<n):
if f[j]==0:
p+=s[j]
j+=1
print(p)
| Title: Minimal string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya recieved a gift of a string *s* with length up to 105 characters for his birthday. He took two more empty strings *t* and *u* and decided to play a game. This game has two possible moves:
- Extract the first character of *s* and append *t* with this character. - Extract the last character of *t* and append *u* with this character.
Petya wants to get strings *s* and *t* empty and string *u* lexicographically minimal.
You should write a program that will help Petya win the game.
Input Specification:
First line contains non-empty string *s* (1<=≤<=|*s*|<=≤<=105), consisting of lowercase English letters.
Output Specification:
Print resulting string *u*.
Demo Input:
['cab\n', 'acdb\n']
Demo Output:
['abc\n', 'abdc\n']
Note:
none | ```python
import sys
input=sys.stdin.readline
from collections import defaultdict as dc
from collections import Counter
from bisect import bisect_right, bisect_left
import math
from operator import itemgetter
from heapq import heapify, heappop, heappush
from queue import PriorityQueue as pq
s=input()[:-1]
n=len(s)
p=[0]
t=[-1]*n
for i in range(1,n):
if not p:
p.append(i)
else:
while(p and s[p[-1]]>s[i]):
x=p.pop(-1)
t[x]=i
p.append(i)
while p:
x=p.pop(-1)
t[x]=-1
p=""
f=dc(int)
x=-1
for i in range(n):
if t[i]==-1:
p+=s[i]
f[i]=1
x=i
j=x+1
i=x-1
#print(t,s)
while(i>=0 and j<n):
if f[i]==1 or f[j]==1:
if f[i]:
i-=1
else:
j+=1
else:
if s[j]<=s[i]:
p+=s[j]
j+=1
else:
p+=s[i]
i-=1
while(i>=0):
if f[i]==0:
p+=s[i]
i-=1
while(j<n):
if f[j]==0:
p+=s[j]
j+=1
print(p)
``` | 0 | |
731 | F | Video Cards | PROGRAMMING | 1,900 | [
"brute force",
"data structures",
"implementation",
"math",
"number theory"
] | null | null | Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, Vlad immediately bought only to find out his computer is too old for the new game and needs to be updated.
There are *n* video cards in the shop, the power of the *i*-th video card is equal to integer value *a**i*. As Vlad wants to be sure the new game will work he wants to buy not one, but several video cards and unite their powers using the cutting-edge technology. To use this technology one of the cards is chosen as the leading one and other video cards are attached to it as secondary. For this new technology to work it's required that the power of each of the secondary video cards is divisible by the power of the leading video card. In order to achieve that the power of any secondary video card can be reduced to any integer value less or equal than the current power. However, the power of the leading video card should remain unchanged, i.e. it can't be reduced.
Vlad has an infinite amount of money so he can buy any set of video cards. Help him determine which video cards he should buy such that after picking the leading video card and may be reducing some powers of others to make them work together he will get the maximum total value of video power. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of video cards in the shop.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=200<=000) — powers of video cards. | The only line of the output should contain one integer value — the maximum possible total power of video cards working together. | [
"4\n3 2 15 9\n",
"4\n8 2 2 7\n"
] | [
"27\n",
"18\n"
] | In the first sample, it would be optimal to buy video cards with powers 3, 15 and 9. The video card with power 3 should be chosen as the leading one and all other video cards will be compatible with it. Thus, the total power would be 3 + 15 + 9 = 27. If he buys all the video cards and pick the one with the power 2 as the leading, the powers of all other video cards should be reduced by 1, thus the total power would be 2 + 2 + 14 + 8 = 26, that is less than 27. Please note, that it's not allowed to reduce the power of the leading video card, i.e. one can't get the total power 3 + 1 + 15 + 9 = 28.
In the second sample, the optimal answer is to buy all video cards and pick the one with the power 2 as the leading. The video card with the power 7 needs it power to be reduced down to 6. The total power would be 8 + 2 + 2 + 6 = 18. | 3,000 | [
{
"input": "4\n3 2 15 9",
"output": "27"
},
{
"input": "4\n8 2 2 7",
"output": "18"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n123819",
"output": "123819"
},
{
"input": "10\n9 6 8 5 5 2 8 9 2 2",
"output": "52"
},
{
"input": "100\n17 23 71 2... | 1,689,345,926 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | print("_RANDOM_GUESS_1689345926.373158")# 1689345926.373172 | Title: Video Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, Vlad immediately bought only to find out his computer is too old for the new game and needs to be updated.
There are *n* video cards in the shop, the power of the *i*-th video card is equal to integer value *a**i*. As Vlad wants to be sure the new game will work he wants to buy not one, but several video cards and unite their powers using the cutting-edge technology. To use this technology one of the cards is chosen as the leading one and other video cards are attached to it as secondary. For this new technology to work it's required that the power of each of the secondary video cards is divisible by the power of the leading video card. In order to achieve that the power of any secondary video card can be reduced to any integer value less or equal than the current power. However, the power of the leading video card should remain unchanged, i.e. it can't be reduced.
Vlad has an infinite amount of money so he can buy any set of video cards. Help him determine which video cards he should buy such that after picking the leading video card and may be reducing some powers of others to make them work together he will get the maximum total value of video power.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of video cards in the shop.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=200<=000) — powers of video cards.
Output Specification:
The only line of the output should contain one integer value — the maximum possible total power of video cards working together.
Demo Input:
['4\n3 2 15 9\n', '4\n8 2 2 7\n']
Demo Output:
['27\n', '18\n']
Note:
In the first sample, it would be optimal to buy video cards with powers 3, 15 and 9. The video card with power 3 should be chosen as the leading one and all other video cards will be compatible with it. Thus, the total power would be 3 + 15 + 9 = 27. If he buys all the video cards and pick the one with the power 2 as the leading, the powers of all other video cards should be reduced by 1, thus the total power would be 2 + 2 + 14 + 8 = 26, that is less than 27. Please note, that it's not allowed to reduce the power of the leading video card, i.e. one can't get the total power 3 + 1 + 15 + 9 = 28.
In the second sample, the optimal answer is to buy all video cards and pick the one with the power 2 as the leading. The video card with the power 7 needs it power to be reduced down to 6. The total power would be 8 + 2 + 2 + 6 = 18. | ```python
print("_RANDOM_GUESS_1689345926.373158")# 1689345926.373172
``` | 0 | |
384 | A | Coder | PROGRAMMING | 800 | [
"implementation"
] | null | null | Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (*x*,<=*y*), he can move to (or attack) positions (*x*<=+<=1,<=*y*), (*x*–1,<=*y*), (*x*,<=*y*<=+<=1) and (*x*,<=*y*–1).
Iahub wants to know how many Coders can be placed on an *n*<=×<=*n* chessboard, so that no Coder attacks any other Coder. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000). | On the first line print an integer, the maximum number of Coders that can be placed on the chessboard.
On each of the next *n* lines print *n* characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'.
If there are multiple correct answers, you can print any. | [
"2\n"
] | [
"2\nC.\n.C\n"
] | none | 500 | [
{
"input": "2",
"output": "2\nC.\n.C"
},
{
"input": "3",
"output": "5\nC.C\n.C.\nC.C"
},
{
"input": "4",
"output": "8\nC.C.\n.C.C\nC.C.\n.C.C"
},
{
"input": "10",
"output": "50\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C\nC.C.C.C.C.\n.C.C.C.C.C... | 1,577,529,630 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 93 | 0 | n = int(input())
print(n)
board = [ ['.'] * n for i in range(n) ]
for i in range(n):
for j in range(n):
if (i + j) % 2 == 0:
board[i][j] = 'C'
for line in board:
print(''.join(line))
| Title: Coder
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (*x*,<=*y*), he can move to (or attack) positions (*x*<=+<=1,<=*y*), (*x*–1,<=*y*), (*x*,<=*y*<=+<=1) and (*x*,<=*y*–1).
Iahub wants to know how many Coders can be placed on an *n*<=×<=*n* chessboard, so that no Coder attacks any other Coder.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000).
Output Specification:
On the first line print an integer, the maximum number of Coders that can be placed on the chessboard.
On each of the next *n* lines print *n* characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'.
If there are multiple correct answers, you can print any.
Demo Input:
['2\n']
Demo Output:
['2\nC.\n.C\n']
Note:
none | ```python
n = int(input())
print(n)
board = [ ['.'] * n for i in range(n) ]
for i in range(n):
for j in range(n):
if (i + j) % 2 == 0:
board[i][j] = 'C'
for line in board:
print(''.join(line))
``` | 0 | |
405 | A | Gravity Flip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch! | The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column. | Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch. | [
"4\n3 2 1 2\n",
"3\n2 3 8\n"
] | [
"1 2 2 3 \n",
"2 3 8 \n"
] | The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | 500 | [
{
"input": "4\n3 2 1 2",
"output": "1 2 2 3 "
},
{
"input": "3\n2 3 8",
"output": "2 3 8 "
},
{
"input": "5\n2 1 2 1 2",
"output": "1 1 2 2 2 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n4 3",
"output": "3 4 "
},
{
"input": "6\n100 40 60 20... | 1,690,568,204 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 0 | n = int(input())
s = input().split()
lst = [int(w) for w in s]
lst2=[]
for i in range(1,n+1):
while(min(lst)!=lst[0] and len(lst)!=1):
lst[1]+=1
lst[0]-=1
lst2.append(str(lst[0]))
lst.pop(0)
print(" ".join(lst2))
| Title: Gravity Flip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch!
Input Specification:
The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column.
Output Specification:
Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch.
Demo Input:
['4\n3 2 1 2\n', '3\n2 3 8\n']
Demo Output:
['1 2 2 3 \n', '2 3 8 \n']
Note:
The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | ```python
n = int(input())
s = input().split()
lst = [int(w) for w in s]
lst2=[]
for i in range(1,n+1):
while(min(lst)!=lst[0] and len(lst)!=1):
lst[1]+=1
lst[0]-=1
lst2.append(str(lst[0]))
lst.pop(0)
print(" ".join(lst2))
``` | 0 | |
962 | A | Equator | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator. | The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day. | Print the index of the day when Polycarp will celebrate the equator. | [
"4\n1 3 2 1\n",
"6\n2 2 2 2 2 2\n"
] | [
"2\n",
"3\n"
] | In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | 0 | [
{
"input": "4\n1 3 2 1",
"output": "2"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "3"
},
{
"input": "1\n10000",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "2"
},
{
"input": "4\n2 1 1 3",
"output": "3"
}... | 1,523,467,456 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 61 | 6,963,200 | n = int(input())
s = [int(x) for x in input().split()]
summa = sum(s)
q = 0
for i in range(n):
q += s[i]
if a * 2 >= summa:
print(i + 1)
exit()
| Title: Equator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day.
Output Specification:
Print the index of the day when Polycarp will celebrate the equator.
Demo Input:
['4\n1 3 2 1\n', '6\n2 2 2 2 2 2\n']
Demo Output:
['2\n', '3\n']
Note:
In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | ```python
n = int(input())
s = [int(x) for x in input().split()]
summa = sum(s)
q = 0
for i in range(n):
q += s[i]
if a * 2 >= summa:
print(i + 1)
exit()
``` | -1 | |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
... | 1,675,017,588 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 11 | 46 | 0 | inp = list(map(int, input().split()))
inpu = []
res = []
for i in range(inp[0]):
inpu.append(input().split())
for i in inpu:
if 'C' in i or 'M' in i or 'Y' in i or 'G' in i:
res.append('#Color')
break
if len(res) == 0:
res.append('#Black&White')
[print(i) for i in res]
| Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
inp = list(map(int, input().split()))
inpu = []
res = []
for i in range(inp[0]):
inpu.append(input().split())
for i in inpu:
if 'C' in i or 'M' in i or 'Y' in i or 'G' in i:
res.append('#Color')
break
if len(res) == 0:
res.append('#Black&White')
[print(i) for i in res]
``` | 0 | |
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... | 1,473,527,422 | 1,522 | Python 3 | WRONG_ANSWER | PRETESTS | 3 | 62 | 0 | s = input()
l = [0]*4
for el in s:
if el == 'L':
l[0]+=1
if el == 'R':
l[1]+=1
if el == 'U':
l[2]+=1
if el == 'D':
l[3]+=1
if sum(l)%4!=0:
print(-1)
else:
k = sum(l)//4
ans = 0
for el in l:
ans+=abs(k-el)
print(ans//2)
| 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
s = input()
l = [0]*4
for el in s:
if el == 'L':
l[0]+=1
if el == 'R':
l[1]+=1
if el == 'U':
l[2]+=1
if el == 'D':
l[3]+=1
if sum(l)%4!=0:
print(-1)
else:
k = sum(l)//4
ans = 0
for el in l:
ans+=abs(k-el)
print(ans//2)
``` | 0 | |
373 | B | Making Sequences is Fun | PROGRAMMING | 1,600 | [
"binary search",
"implementation",
"math"
] | null | null | We'll define *S*(*n*) for positive integer *n* as follows: the number of the *n*'s digits in the decimal base. For example, *S*(893)<==<=3, *S*(114514)<==<=6.
You want to make a consecutive integer sequence starting from number *m* (*m*,<=*m*<=+<=1,<=...). But you need to pay *S*(*n*)·*k* to add the number *n* to the sequence.
You can spend a cost up to *w*, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length. | The first line contains three integers *w* (1<=≤<=*w*<=≤<=1016), *m* (1<=≤<=*m*<=≤<=1016), *k* (1<=≤<=*k*<=≤<=109).
Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. | The first line should contain a single integer — the answer to the problem. | [
"9 1 1\n",
"77 7 7\n",
"114 5 14\n",
"1 1 2\n"
] | [
"9\n",
"7\n",
"6\n",
"0\n"
] | none | 1,000 | [
{
"input": "9 1 1",
"output": "9"
},
{
"input": "77 7 7",
"output": "7"
},
{
"input": "114 5 14",
"output": "6"
},
{
"input": "1 1 2",
"output": "0"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "462 183 8",
"output": "19"
},
{
"i... | 1,668,030,397 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | def s(n):
s = str(n)
return len(s)
def solve(w, m , k):
f = 0
n = m
nb = 0
while f<w:
f += s(n)*k
n += 1
nb +=1
return nb
w, m, k = [int(x) for x in input().split()]
print(solve(w,m,k))
| Title: Making Sequences is Fun
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We'll define *S*(*n*) for positive integer *n* as follows: the number of the *n*'s digits in the decimal base. For example, *S*(893)<==<=3, *S*(114514)<==<=6.
You want to make a consecutive integer sequence starting from number *m* (*m*,<=*m*<=+<=1,<=...). But you need to pay *S*(*n*)·*k* to add the number *n* to the sequence.
You can spend a cost up to *w*, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.
Input Specification:
The first line contains three integers *w* (1<=≤<=*w*<=≤<=1016), *m* (1<=≤<=*m*<=≤<=1016), *k* (1<=≤<=*k*<=≤<=109).
Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
The first line should contain a single integer — the answer to the problem.
Demo Input:
['9 1 1\n', '77 7 7\n', '114 5 14\n', '1 1 2\n']
Demo Output:
['9\n', '7\n', '6\n', '0\n']
Note:
none | ```python
def s(n):
s = str(n)
return len(s)
def solve(w, m , k):
f = 0
n = m
nb = 0
while f<w:
f += s(n)*k
n += 1
nb +=1
return nb
w, m, k = [int(x) for x in input().split()]
print(solve(w,m,k))
``` | 0 | |
551 | A | GukiZ and Contest | PROGRAMMING | 800 | [
"brute force",
"implementation",
"sortings"
] | null | null | Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected. | The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*). | In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input. | [
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] | [
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] | In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position. | 500 | [
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4... | 1,597,865,259 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 858 | 409,600 | students = int(input())
rating = list(map(int, input().split(" ")))
standing = 1
ratingCopy = rating.copy()
res = [0] * students
ratingCopy.sort()
while len(ratingCopy) > 0:
bestRating = ratingCopy[len(ratingCopy) - 1]
ratingCopy = [i for i in ratingCopy if i != bestRating]
counter = 0
for i in range(students):
if rating[i] == bestRating:
res[i] = standing
counter = counter + 1
standing = counter + standing
for n in res:
print(n) | Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position. | ```python
students = int(input())
rating = list(map(int, input().split(" ")))
standing = 1
ratingCopy = rating.copy()
res = [0] * students
ratingCopy.sort()
while len(ratingCopy) > 0:
bestRating = ratingCopy[len(ratingCopy) - 1]
ratingCopy = [i for i in ratingCopy if i != bestRating]
counter = 0
for i in range(students):
if rating[i] == bestRating:
res[i] = standing
counter = counter + 1
standing = counter + standing
for n in res:
print(n)
``` | 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... | 1,440,614,493 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 124 | 0 | n = int(input())
array = list(map(int, input().split()))
a, b = map(int, input().split())
summ = 0
for i in range(a - 1, b - 1):
summ += array[i]
print(summ) | 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
n = int(input())
array = list(map(int, input().split()))
a, b = map(int, input().split())
summ = 0
for i in range(a - 1, b - 1):
summ += array[i]
print(summ)
``` | 3.969 |
476 | B | Dreamoon and WiFi | PROGRAMMING | 1,300 | [
"bitmasks",
"brute force",
"combinatorics",
"dp",
"math",
"probabilities"
] | null | null | Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.
Each command is one of the following two types:
1. Go 1 unit towards the positive direction, denoted as '+' 1. Go 1 unit towards the negative direction, denoted as '-'
But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5).
You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands? | The first line contains a string *s*1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}.
The second line contains a string *s*2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.
Lengths of two strings are equal and do not exceed 10. | Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=9. | [
"++-+-\n+-+-+\n",
"+-+-\n+-??\n",
"+++\n??-\n"
] | [
"1.000000000000\n",
"0.500000000000\n",
"0.000000000000\n"
] | For the first sample, both *s*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> will lead Dreamoon to finish at the same position + 1.
For the second sample, *s*<sub class="lower-index">1</sub> will lead Dreamoon to finish at position 0, while there are four possibilites for *s*<sub class="lower-index">2</sub>: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5.
For the third sample, *s*<sub class="lower-index">2</sub> could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position + 3 is 0. | 1,500 | [
{
"input": "++-+-\n+-+-+",
"output": "1.000000000000"
},
{
"input": "+-+-\n+-??",
"output": "0.500000000000"
},
{
"input": "+++\n??-",
"output": "0.000000000000"
},
{
"input": "++++++++++\n+++??++?++",
"output": "0.125000000000"
},
{
"input": "--+++---+-\n????????... | 1,663,059,325 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 31 | 62 | 0 | from math import comb
def main():
s1 = input()
s2 = input()
orginal = 0
for s in s1:
orginal += (1 if s == '+' else -1 )
sofar = 0
unk = 0
for s in s2:
if s == '?':
unk += 1
continue
sofar += (1 if s == '+' else -1)
offset = abs(orginal - sofar)
total_comb = 1 << unk
print(comb(unk,(unk + offset) // 2) / total_comb)
main() | Title: Dreamoon and WiFi
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.
Each command is one of the following two types:
1. Go 1 unit towards the positive direction, denoted as '+' 1. Go 1 unit towards the negative direction, denoted as '-'
But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5).
You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?
Input Specification:
The first line contains a string *s*1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}.
The second line contains a string *s*2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.
Lengths of two strings are equal and do not exceed 10.
Output Specification:
Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=9.
Demo Input:
['++-+-\n+-+-+\n', '+-+-\n+-??\n', '+++\n??-\n']
Demo Output:
['1.000000000000\n', '0.500000000000\n', '0.000000000000\n']
Note:
For the first sample, both *s*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> will lead Dreamoon to finish at the same position + 1.
For the second sample, *s*<sub class="lower-index">1</sub> will lead Dreamoon to finish at position 0, while there are four possibilites for *s*<sub class="lower-index">2</sub>: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5.
For the third sample, *s*<sub class="lower-index">2</sub> could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position + 3 is 0. | ```python
from math import comb
def main():
s1 = input()
s2 = input()
orginal = 0
for s in s1:
orginal += (1 if s == '+' else -1 )
sofar = 0
unk = 0
for s in s2:
if s == '?':
unk += 1
continue
sofar += (1 if s == '+' else -1)
offset = abs(orginal - sofar)
total_comb = 1 << unk
print(comb(unk,(unk + offset) // 2) / total_comb)
main()
``` | 3 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,683,978,061 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 92 | 0 | n = int(input())
# Initialize the vector sum to the zero vector
vector_sum = [0, 0, 0]
for i in range(n):
# Read the force vector
x, y, z = map(int, input().split())
# Add the force vector to the vector sum
vector_sum[0] += x
vector_sum[1] += y
vector_sum[2] += z
# Check if the vector sum is zero
if vector_sum == [0, 0, 0]:
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n = int(input())
# Initialize the vector sum to the zero vector
vector_sum = [0, 0, 0]
for i in range(n):
# Read the force vector
x, y, z = map(int, input().split())
# Add the force vector to the vector sum
vector_sum[0] += x
vector_sum[1] += y
vector_sum[2] += z
# Check if the vector sum is zero
if vector_sum == [0, 0, 0]:
print("YES")
else:
print("NO")
``` | 3.977 |
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... | 1,634,035,567 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 124 | 6,758,400 | M,N = map(int, input().split())
Area = M*N
print(Area//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())
Area = M*N
print(Area//2)
``` | 3.956411 |
328 | B | Sheldon and Ice Pieces | PROGRAMMING | 1,500 | [
"greedy"
] | null | null | Do you remember how Kai constructed the word "eternity" using pieces of ice as components?
Little Sheldon plays with pieces of ice, each piece has exactly one digit between 0 and 9. He wants to construct his favourite number *t*. He realized that digits 6 and 9 are very similar, so he can rotate piece of ice with 6 to use as 9 (and vice versa). Similary, 2 and 5 work the same. There is no other pair of digits with similar effect. He called this effect "Digital Mimicry".
Sheldon favourite number is *t*. He wants to have as many instances of *t* as possible. How many instances he can construct using the given sequence of ice pieces. He can use any piece at most once. | The first line contains integer *t* (1<=≤<=*t*<=≤<=10000). The second line contains the sequence of digits on the pieces. The length of line is equal to the number of pieces and between 1 and 200, inclusive. It contains digits between 0 and 9. | Print the required number of instances. | [
"42\n23454\n",
"169\n12118999\n"
] | [
"2\n",
"1\n"
] | This problem contains very weak pretests. | 500 | [
{
"input": "42\n23454",
"output": "2"
},
{
"input": "169\n12118999",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "7\n777",
"output": "3"
},
{
"input": "18\n8118",
"output": "2"
},
{
"input": "33\n33333333",
"output": "4"
},
... | 1,570,036,616 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 310 | 0 | # import numpy as np
def solution():
number = input()
book = {}
book_of_repeats = {}
for i in number:
intermidiate = i
if intermidiate == "5":
intermidiate = "2"
elif intermidiate == "9":
intermidiate = "6"
book_of_repeats[intermidiate] = 0
if intermidiate in book:
book[intermidiate] += 1
else:
book[intermidiate] = 1
cubs = input()
for i in cubs:
intermidiate = i
if intermidiate == "5":
intermidiate = "2"
elif intermidiate == "9":
intermidiate = "6"
if intermidiate in book_of_repeats:
book_of_repeats[intermidiate] += 1
result = len(cubs)//len(number)
for i in book:
intermidiate = book_of_repeats[i]//book[i]
if intermidiate < result:
result = intermidiate
print(result)
def to_mixed_frac(first, second):
while True:
for i in range(2, first + 1):
if first % i == 0 and second % i == 0:
first //= i
second //= i
break
else:
break
return str(first) + "/" + str(second)
def array_to_int(array):
for i in range(len(array)):
array[i] = int(array[i])
return array
def join0(array):
result = ""
for i in array:
result += str(i)
return result
solution()
# input-output by console
| Title: Sheldon and Ice Pieces
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Do you remember how Kai constructed the word "eternity" using pieces of ice as components?
Little Sheldon plays with pieces of ice, each piece has exactly one digit between 0 and 9. He wants to construct his favourite number *t*. He realized that digits 6 and 9 are very similar, so he can rotate piece of ice with 6 to use as 9 (and vice versa). Similary, 2 and 5 work the same. There is no other pair of digits with similar effect. He called this effect "Digital Mimicry".
Sheldon favourite number is *t*. He wants to have as many instances of *t* as possible. How many instances he can construct using the given sequence of ice pieces. He can use any piece at most once.
Input Specification:
The first line contains integer *t* (1<=≤<=*t*<=≤<=10000). The second line contains the sequence of digits on the pieces. The length of line is equal to the number of pieces and between 1 and 200, inclusive. It contains digits between 0 and 9.
Output Specification:
Print the required number of instances.
Demo Input:
['42\n23454\n', '169\n12118999\n']
Demo Output:
['2\n', '1\n']
Note:
This problem contains very weak pretests. | ```python
# import numpy as np
def solution():
number = input()
book = {}
book_of_repeats = {}
for i in number:
intermidiate = i
if intermidiate == "5":
intermidiate = "2"
elif intermidiate == "9":
intermidiate = "6"
book_of_repeats[intermidiate] = 0
if intermidiate in book:
book[intermidiate] += 1
else:
book[intermidiate] = 1
cubs = input()
for i in cubs:
intermidiate = i
if intermidiate == "5":
intermidiate = "2"
elif intermidiate == "9":
intermidiate = "6"
if intermidiate in book_of_repeats:
book_of_repeats[intermidiate] += 1
result = len(cubs)//len(number)
for i in book:
intermidiate = book_of_repeats[i]//book[i]
if intermidiate < result:
result = intermidiate
print(result)
def to_mixed_frac(first, second):
while True:
for i in range(2, first + 1):
if first % i == 0 and second % i == 0:
first //= i
second //= i
break
else:
break
return str(first) + "/" + str(second)
def array_to_int(array):
for i in range(len(array)):
array[i] = int(array[i])
return array
def join0(array):
result = ""
for i in array:
result += str(i)
return result
solution()
# input-output by console
``` | 3 | |
433 | A | Kitahara Haruki's Gift | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple. | In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes). | [
"3\n100 200 100\n",
"4\n100 100 100 200\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | 500 | [
{
"input": "3\n100 200 100",
"output": "YES"
},
{
"input": "4\n100 100 100 200",
"output": "NO"
},
{
"input": "1\n100",
"output": "NO"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "2\n100 100",
"output": "YES"
},
{
"input": "2\n200 200",
"o... | 1,667,102,218 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
using namespace std;
int main()
{
int n, a;
cin >> n;
int c1 = 0, c2 = 0;
for (int i = 0; i < n; i++)
{
cin >> a;
if (a == 100)
{
c1++;
}
}
c2 = n - c1;
if (c1 % 2 == 0 && c2 % 2 == 0)
{
cout << "YES";
}
else if (c1 % 2 != 0 && c2 % 2 == 0)
{
cout << "NO";
}
else if (c1 % 2 == 0 && c2 % 2 != 0)
{
if (c1 == 0)
cout << "NO";
else
cout << "YES";
}
else
{
cout << "NO";
}
} | Title: Kitahara Haruki's Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
Output Specification:
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
Demo Input:
['3\n100 200 100\n', '4\n100 100 100 200\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | ```python
#include <iostream>
using namespace std;
int main()
{
int n, a;
cin >> n;
int c1 = 0, c2 = 0;
for (int i = 0; i < n; i++)
{
cin >> a;
if (a == 100)
{
c1++;
}
}
c2 = n - c1;
if (c1 % 2 == 0 && c2 % 2 == 0)
{
cout << "YES";
}
else if (c1 % 2 != 0 && c2 % 2 == 0)
{
cout << "NO";
}
else if (c1 % 2 == 0 && c2 % 2 != 0)
{
if (c1 == 0)
cout << "NO";
else
cout << "YES";
}
else
{
cout << "NO";
}
}
``` | -1 | |
222 | A | Shooshuns and Sequence | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same. | The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found. | Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1. | [
"3 2\n3 1 1\n",
"3 1\n3 1 1\n"
] | [
"1\n",
"-1\n"
] | In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1. | 500 | [
{
"input": "3 2\n3 1 1",
"output": "1"
},
{
"input": "3 1\n3 1 1",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 1\n2 1",
"output": "-1"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
}... | 1,605,054,231 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 26,316,800 | n,k = map(int, input().split())
arr = list(map(int, input().split()))
def f():
arr.append(arr[k - 1])
arr.pop(0)
return False
def giveHash(arr):
hash = 0
temp = [str(i) for i in arr]
return " ".join(temp)
count = 0
seen = set()
seen.add(giveHash(arr))
while(True):
if(len(set(arr)) == 1):
print(count)
break
else:
f()
curr = giveHash(arr)
if(curr in seen):
print("-1")
break
else:
seen.add(curr)
count += 1 | Title: Shooshuns and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
Output Specification:
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Demo Input:
['3 2\n3 1 1\n', '3 1\n3 1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1. | ```python
n,k = map(int, input().split())
arr = list(map(int, input().split()))
def f():
arr.append(arr[k - 1])
arr.pop(0)
return False
def giveHash(arr):
hash = 0
temp = [str(i) for i in arr]
return " ".join(temp)
count = 0
seen = set()
seen.add(giveHash(arr))
while(True):
if(len(set(arr)) == 1):
print(count)
break
else:
f()
curr = giveHash(arr)
if(curr in seen):
print("-1")
break
else:
seen.add(curr)
count += 1
``` | 0 |
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