TnT/CodeNet4Repair huggingface dataset:
Problem:
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.  The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.  One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.  If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.    * X < Z \leq Y   * x_1, x_2, ..., x_N < Z   * y_1, y_2, ..., y_M \geq Z  Determine if war will break out.

Incorrect submission (verdict: Wrong Answer):
n,m,x,y=map(int,input().split())
X = list(map(int,input().split()))
Y = list(map(int,input().split()))
if max(X)+1 < min(Y):
  print("No War")
else:
  print("War")

Fixed, accepted submission:
n,m,x,y=map(int,input().split())
X = list(map(int,input().split()))
Y = list(map(int,input().split()))
if max(X) < min(Y) and max(X)+1 > x and max(X)+1 <= y:
  print("No War")
else:
  print("War")
Problem:
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N).  For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it.  We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied:    * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2.  Does there exist a good set of choices? If the answer is yes, find one good set of choices.

Incorrect submission (verdict: Wrong Answer):
def main():
    N = int(input())
    A = list(map(int, input().split()))
    A.insert(0, 1e10)
    ans = []
    for i in range(len(A) - 1, 0, -1):
        idx = i
        s = 0
        while idx <= len(A) - 1:
            s += A[idx]
            idx += i

        if s % 2 != 0:
            ans.append(i)
    
    if len(ans) > 0:
        print(*ans)
    else:
        print(0)


if __name__ == '__main__':
    main()


Fixed, accepted submission:
def main():
    _ = int(input())
    A = list(map(int, input().split()))
    A.insert(0, 1e10)
    ans = []
    B = [0] * len(A)
    for i in range(len(A)-1, 0, -1):
        idx = i
        s = 0
        while idx <= len(A) - 1:
            s += B[idx]
            idx += i

        if s % 2 != A[i]:
            ans.append(i)
            B[i] = 1

    if len(ans) > 0:
        print(len(ans))
        print(*ans)
    else:
        print(0)


if __name__ == '__main__':
    main()
Problem:
Write a program which reads a sequence and prints it in the reverse order.

Incorrect submission (verdict: Wrong Answer):
import sys

lines = [line for line in sys.stdin]

n = int(lines[0])
s = lines[1]
print (s[::-1])

Fixed, accepted submission:
import sys

lines = [line for line in sys.stdin]

n = int(lines[0])
s = list(map(int,lines[1].split()))
s.reverse()
s=map(str,s)
print(" ".join(s))
Problem:
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`.  You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Incorrect submission (verdict: Wrong Answer):
a,b = list(map(int,input().split()))
s = input()
num = s[:a] + s[a+1]
print('Yes' if s[a+1] == '-' and len(s) == a+b+1 and ('-' not in num) else 'No')


Fixed, accepted submission:
a,b = list(map(int,input().split()))
s = input()
num = s[:a] + s[a+1]
print('Yes' if s[a] == '-' and len(s) == a+b+1 and ('-' not in num) else 'No')
Problem:
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.  If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.

Incorrect submission (verdict: Wrong Answer):
n = int(input().split()[0])

txy_list = []

for i in range(0, n):
    work_list = list(map(int, input().split()))
    txy_list.append(work_list)

pre_txy = [0, 0, 0]
answer = False

for txy in txy_list:
    t, x, y = txy[0], txy[1], txy[2]
    pre_t, pre_x, pre_y = pre_txy[0], pre_txy[1], pre_txy[2]

    t_w, x_w, y_w = t - pre_t, abs(x - pre_x), abs(y - pre_y)
    work = t_w - (x_w + y_w)

    if (work % 2 == 0) and (work >= 0):
        pre_txy = txy
        continue
    else:
        break
else:
    answer = True

a = 'YES' if answer else 'NO'
print(a)

Fixed, accepted submission:


n = int(input().split()[0])

txy_list = []

for i in range(0, n):
    work_list = list(map(int, input().split()))
    txy_list.append(work_list)

pre_txy = [0, 0, 0]
answer = False

for txy in txy_list:
    t, x, y = txy[0], txy[1], txy[2]
    pre_t, pre_x, pre_y = pre_txy[0], pre_txy[1], pre_txy[2]

    t_w, x_w, y_w = t - pre_t, abs(x - pre_x), abs(y - pre_y)
    work = t_w - (x_w + y_w)

    if (work % 2 == 0) and (work >= 0):
        pre_txy = txy
        continue
    else:
        break
else:
    answer = True

a = 'Yes' if answer else 'No'
print(a)
Problem:
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament.  The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day.  Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.

Incorrect submission (verdict: Wrong Answer):
s = input()
if s.count("x")>7:
  print("No")
else:
  print("Yes")

Fixed, accepted submission:
s = input()
if s.count("x")>7:
  print("NO")
else:
  print("YES")
Problem:
An elementary school student Takahashi has come to a variety store.  He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it?  Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

Incorrect submission (verdict: Wrong Answer):
a, b, c = map(int, input().split()) 
print("YES" if a+b>c else "NO")


Fixed, accepted submission:
a, b, c = map(int, input().split()) 
print("Yes" if a+b>=c else "No")
Problem:
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

Incorrect submission (verdict: Wrong Answer):
n = int(input())

print(n*(n+1)/2)

Fixed, accepted submission:
n = int(input())
 
print(int(n*(n+1)/2))
Problem:
There is a cave.  The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside.  It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage.  Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1.    * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible.  Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.

Incorrect submission (verdict: Wrong Answer):
# D
from collections import deque

N,M=map(int,input().split())
AL=[[] for _ in range(N)]
for _ in range(M):
    A,B=map(int,input().split())
    AL[A-1].append(B-1)
    AL[B-1].append(A-1)
Q=deque([[0,-1]]) 
ANS=[-1 for _ in range(N)] 
ANS[0]=1
TF=[0 for _ in range(N)]
while Q:
    print(Q)
    i,c=Q.popleft()
    if TF[i]:
        continue
    TF[i]=1
    ANS[i]=c
    for j in AL[i]:
        if not TF[j]:
            Q.append([j,i])

print('Yes')
for i in range(1,N):
    print(ANS[i])

Fixed, accepted submission:
import sys
input = sys.stdin.readline

from collections import deque

N,M=map(int,input().split())
AL=[[] for _ in range(N)]
for _ in range(M):
    A,B=map(int,input().split())
    AL[A-1].append(B-1)
    AL[B-1].append(A-1)
Q=deque([[0,-1]]) 
ANS=[-1 for _ in range(N)] 
ANS[0]=1
TF=[0 for _ in range(N)]
while Q:
    i,c=Q.popleft()
    if TF[i]:
        continue
    TF[i]=1
    ANS[i]=c
    for j in AL[i]:
        if not TF[j]:
            Q.append([j,i])

print('Yes')
for i in range(1,N):
    print(ANS[i]+1)
Problem:
We ask you to select some number of positive integers, and calculate the sum of them.  It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.  Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.  If the objective is achievable, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
def inputs(num_of_input):
    ins = [input() for i in range(num_of_input)]
    return ins


def solve(inputs):
    [A, B, X] = list(map(lambda x: int(x), inputs[0].split()))
    num = (B // X) - ((A - 1) // X)
    return num


if __name__ == "__main__":
    ret = solve(inputs(1))
    print(ret)


Fixed, accepted submission:
def inputs(num_of_input):
    ins = [input() for i in range(num_of_input)]
    return ins


def solve(inputs):
    [A, B, C] = list(map(lambda x: int(x), inputs[0].split()))
    first_state = A % B
    state = None
    if first_state == C:
        return "YES"

    sum = A
    while state != first_state:
        sum += A
        sur = sum % B
        if sur == C:
            return "YES"
        state = sur

    return "NO"


if __name__ == "__main__":
    ret = solve(inputs(1))
    print(ret)
Problem:
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters.  Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way.  You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.

Incorrect submission (verdict: Wrong Answer):
text=input()

if not "c" in text or not "f" in text:
    print("no")

if text.find("c")<text.find("f"):
   print("yes")


Fixed, accepted submission:
s=str(input())
p=[chr(i) for i in range(65, 65+26)]
p.remove("C")
p.remove("F")
for i in p:
    s=s.replace(i,'')
 
if 'CF' in s:
    print('Yes')
else:
    print('No')
Problem:
We ask you to select some number of positive integers, and calculate the sum of them.  It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.  Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.  If the objective is achievable, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
a,b,c=map(int,input().split())
d=a%b
e=(d*2)%b
i=3
f=[]
f.append(d)
f.append(e)
while e!=d:
  e=d*i
  e%=b
  i+=1
  f.append(e)

if c in f:
  print('Yes')
else:
  print('No')






Fixed, accepted submission:
a,b,c=map(int,input().split())
d=a%b
e=(d*2)%b
i=3
f=[]
f.append(d)
f.append(e)
while e!=d:
  x=d*i
  e=x%b
  i+=1
  f.append(e)

if c in f:
  print('YES')
else:
  print('NO')
Problem:
We will buy a product for N yen (the currency of Japan) at a shop.  If we use only 1000-yen bills to pay the price, how much change will we receive?  Assume we use the minimum number of bills required.

Incorrect submission (verdict: Wrong Answer):
N = int(input("N:"))
sa = 10000 - N
ans = sa % 1000
print(ans)

Fixed, accepted submission:
N = int(input())

sa = 10000 - N
ans = sa % 1000

print(ans)
Problem:
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.  Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut:    * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).   * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.  Find the minimum amount of money needed before cutting the bar into two parts with the same length.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
a=list(map(int,input().split()))
temp=sum(a)
ans=0
t=999999999
tsum=0
for i in range(n):
  tsum=tsum+a[i]
  if abs(tsum-temp/2)<t:
    t=abs(tsum-temp/2)
print(int(t)*2-temp)

Fixed, accepted submission:
n=int(input())
a=list(map(int,input().split()))
temp=sum(a)
ans=0
t=9999999999999999
tsum=0
ta=0
for i in range(n):
  tsum=tsum+a[i]
  if abs(tsum-temp/2)<t:
    t=abs(tsum-temp/2)
    ta=tsum
print(abs(int(ta)*2-temp))
Problem:
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`.  You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Incorrect submission (verdict: Wrong Answer):
a, b = [int(i) for i in input().split()]
content = input()

if len(content) == a + b + 1:
    A = content[:a]
    B = content[-b:]
    hyphn = content[a]
    print(A, B)
    if A.isdecimal() and B.isdecimal() and hyphn == '-':
        print('Yes')
    else :
        print('No')


Fixed, accepted submission:
a, b = [int(i) for i in input().split()]
content = input()

if len(content) == a + b + 1:
    A = content[:a]
    B = content[-b:]
    hyphn = content[a]
    if A.isdecimal() and B.isdecimal() and hyphn == '-':
        print('Yes')
    else :
        print('No')
Problem:
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.

Incorrect submission (verdict: Wrong Answer):
n = int(input().replace(' ', ''))
for i in range(100100):
  if i*i == n:
    print('Yes')
else:
  print('No')

Fixed, accepted submission:
n = int(input().replace(' ', ''))
for i in range(100100):
  if i*i == n:
    print('Yes')
    break
else:
  print('No')
Problem:
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A.  The problem has N test cases.  For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively.  See the Output section for the output format.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
lst = []
for i in range(n):
	str1 = input()
	lst.append(str1)
print('AC *' , lst.count('AC'))
print('WA *' , lst.count('WA'))
print('TLE *' ,lst.count('TLE'))
print('RE *' , lst.count('RE'))

Fixed, accepted submission:
n = int(input())
lst = []
for i in range(n):
	str1 = input()
	lst.append(str1)
print('AC x' , lst.count('AC'))
print('WA x' , lst.count('WA'))
print('TLE x' ,lst.count('TLE'))
print('RE x' , lst.count('RE'))
Problem:
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.

Incorrect submission (verdict: Wrong Answer):
# # Make IO faster
# import sys
# input = sys.stdin.readline


# X = input()


# N = int(input())

# X, Y = map(int, input().split())
 for N lines
# XY = [list(map(int, input().split())) for _ in range(N)]

# from IPython import embed; embed(); exit();


import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import accumulate, permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from fractions import gcd
from heapq import heappush, heappop
from functools import reduce
import numpy as np
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7

S = input()
ans = 0
for s in ['a', 'b', 'c']:
    ans += S.count(s)

if ans == 3:
    print("Yes")
else:
    print("No")


Fixed, accepted submission:
s=input()
if s=="abc" or s=="acb" or s=="bac" or s=="bca" or s=="cab" or s=="cba":
    print("Yes")
else:
    print("No")
Problem:
太郎君はよくJOI雑貨店で買い物をする． JOI雑貨店には硬貨は500円，100円，50円，10円，5円，1円が十分な数だけあり，いつも最も枚数が少なくなるようなおつりの支払い方をする．太郎君がJOI雑貨店で買い物をしてレジで1000円札を1枚出した時，もらうおつりに含まれる硬貨の枚数を求めるプログラムを作成せよ．  例えば入力例1の場合は下の図に示すように，4を出力しなければならない．  

Incorrect submission (verdict: Wrong Answer):
taro_spend = []
spend_num = 0

for i in range(5):
    taro_spend.append(int(input()))
    if taro_spend[i] == 0:
        taro_spend.pop()
        break
    spend_num += 1

changes = [1000 - taro_spend[i] for i in range(spend_num)]
print(changes)
for change in changes:
    change_coin = 0

    if change >= 500:
        change -= 500
        change_coin += 1

    if change >= 100:
        while change >= 100:
            change -= 100
            change_coin += 1
            
    if change >= 50:
        change -= 50
        change_coin += 1
        
    if change >= 10:
        while change >= 10:
            change -= 10
            change_coin += 1

    if change >= 5:
        change -= 5
        change_coin += 1

    if change >= 1:
        while change >= 1:
            change -= 1
            change_coin += 1

    print(change_coin)

Fixed, accepted submission:
taro_spend = []
spend_num = 0

for i in range(5):
    taro_spend.append(int(input()))
    if taro_spend[i] == 0:
        taro_spend.pop()
        break
    spend_num += 1

changes = [1000 - taro_spend[i] for i in range(spend_num)]

for change in changes:
    change_coin = 0

    if change >= 500:
        change -= 500
        change_coin += 1

    if change >= 100:
        while change >= 100:
            change -= 100
            change_coin += 1
            
    if change >= 50:
        change -= 50
        change_coin += 1
        
    if change >= 10:
        while change >= 10:
            change -= 10
            change_coin += 1

    if change >= 5:
        change -= 5
        change_coin += 1

    if change >= 1:
        while change >= 1:
            change -= 1
            change_coin += 1

    print(change_coin)
Problem:
Snuke has a favorite restaurant.  The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer.  So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.

Incorrect submission (verdict: Wrong Answer):
a = int(input())
x = a*800
y = a/15*200
print(x-y)

Fixed, accepted submission:
a = int(input())
x = a*800
n = int(a/15)
y = n*200
print(int(x-y))
Problem:
How many multiples of d are there among the integers between L and R (inclusive)?

Incorrect submission (verdict: Wrong Answer):
L, R, D = map(int, input().split())
cnt = 0
for i in range(L, R+1):
    i % D == 0
    cnt += 1
print(cnt)


Fixed, accepted submission:
L, R, D = map(int, input().split())
cnt = 0
for i in range(L, R+1):
    if i % D == 0:
        cnt += 1
print(cnt)
Problem:
E869120 found a chest which is likely to contain treasure.   However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.   He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`.  One more thing he found is a sheet of paper with the following facts written on it:    * Condition 1: The string S contains a string T as a contiguous substring.   * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.  Print the string S.   If such a string does not exist, print `UNRESTORABLE`.

Incorrect submission (verdict: Wrong Answer):
from copy import deepcopy
s = list(input())
t = list(input())
lens = len(s)
lent = len(t)
ans = deepcopy(s)
for i in range(len(s)):
  if s[i] == "?":
    ans[i] = "a"
def match(moji):
  for i in range(lent):
    if moji[i] != t[i] and moji[i] != "?":
      return False
  return True
for i in range(lent, lens):
  ns = s[lens-i:lens-i+lent]
  if match(ns):
    ans[lens-i:lens-i+lent] = t
    print(ans)
    exit()
print("UNRESTORABLE")

Fixed, accepted submission:
from copy import deepcopy
s = list(input())
t = list(input())
lens = len(s)
lent = len(t)
ans = deepcopy(s)
for i in range(len(s)):
  if s[i] == "?":
    ans[i] = "a"
def match(moji):
  for i in range(lent):
    if moji[i] != t[i] and moji[i] != "?":
      return False
  return True
for i in range(lent, lens+1):
  ns = s[lens-i:lens-i+lent]
  if match(ns):
    ans[lens-i:lens-i+lent] = t
    print("".join(ans))
    exit()
print("UNRESTORABLE")