TnT/CodeNet4Repair huggingface dataset:
Problem:
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.  Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.  Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.  Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.

Incorrect submission (verdict: Wrong Answer):
N, C = map(int, input().split())
x = [None] * N
v = [None] * N
for i in range(N):
    x[i], v[i] = map(int, input().split())

a0 = [None] * N
a = v[0] - x[0]
a0[0] = max(0, a)
for i in range(1, N):
    a += v[i] - (x[i] - x[i - 1])
    a0[i] = max(a, a0[i - 1])

a1 = [None] * N
a = v[N - 1] - (C - x[N - 1])
a1[0] = max(0, a)
for i in range(1, N):
    print(a, (C - x[N - 1 - (i - 1)]), (C - x[N - 1 - i]))
    a += v[N - 1 - i] - ((C - x[N - 1 - (i - 1)]) - (C - x[N - 1 - i]))
    a1[i] = max(a, a1[i - 1])

result = max(a0[N - 1], a1[N - 1])
for i in range(N):
    result = max(result, a0[i] - x[i] + a1[N - 1 - (i + 1)])
for i in range(N):
    result = max(result, a1[i] - (C - x[N - 1 - i]) + a0[N - 1 - (i + 1)])
print(result)


Fixed, accepted submission:
N, C = map(int, input().split())
x = [None] * N
v = [None] * N
for i in range(N):
    x[i], v[i] = map(int, input().split())

a0 = [None] * N
a = v[0] - x[0]
a0[0] = max(0, a)
for i in range(1, N):
    a += v[i] - (x[i] - x[i - 1])
    a0[i] = max(a, a0[i - 1])

a1 = [None] * N
a = v[N - 1] - (C - x[N - 1])
a1[0] = max(0, a)
for i in range(1, N):
    a += v[N - 1 - i] - ((C - x[N - 1 - i]) - (C - x[N - 1 - (i - 1)]))
    a1[i] = max(a, a1[i - 1])

result = max(a0[N - 1], a1[N - 1])
for i in range(N - 1):
    result = max(result, a0[i] - x[i] + a1[N - 1 - (i + 1)])
for i in range(N - 1):
    result = max(result, a1[i] - (C - x[N - 1 - i]) + a0[N - 1 - (i + 1)])
print(result)
Problem:
Write a program which performs a sequence of commands to a given string $str$. The command is one of:    * print a b: print from the a-th character to the b-th character of $str$   * reverse a b: reverse from the a-th character to the b-th character of $str$   * replace a b p: replace from the a-th character to the b-th character of $str$ with p  Note that the indices of $str$ start with 0.

Incorrect submission (verdict: Wrong Answer):
S=input()
m=int(input())
for i in range(m):
    C=input().split()
    a=int(C[1])-1
    if a<0:
        a=0
    b=int(C[2])
    if C[0]=="print":
        print(S[a:b])
    elif C[0]=="reverse":
        S=S[:a]+S[b-len(S)-1:a-len(S)-1:-1]+S[b:]
        
        
    elif C[0]=="replace":
        S=S[:a]+C[3]+S[b:]
      

Fixed, accepted submission:
S=input()
m=int(input())
for i in range(m):
    C=input().split()
    a=int(C[1])
    b=int(C[2])+1
    if C[0]=="print":
        print(S[a:b])
    elif C[0]=="reverse":
        S=S[:a]+S[b-len(S)-1:a-len(S)-1:-1]+S[b:]
        
    elif C[0]=="replace":
        S=S[:a]+C[3]+S[b:]
Problem:
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets.  Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.

Incorrect submission (verdict: Wrong Answer):
count = 0
while count < 1000:
  count += 1
  print("Hello World")


Fixed, accepted submission:
count = 0
for i in range(10000):
  a = input()
  a = int(a)
  if a > 0:
    count += 1
    print('Case '+str(count)+': '+str(a))
  else:
    break
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):


import sys
from collections import defaultdict, deque

sys.setrecursionlimit(10 ** 7)
# ----------

INF = float("inf")
MOD = 10 ** 9 + 7
# ----------

N, M = map(int, input().split())
G = defaultdict(list)
for i in range(M):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    G[a].append(b)
    G[b].append(a)


q = deque()


ans = [-1 for _ in range(N)]
ans[0] = 0
q.append(0)

while q:
    v = q.popleft()
    for u in G[v]:
        if ans[u] != -1:
            continue
        ans[u] = v + 1
        q.append(u)

for i in range(1,N):
    print(ans[i])


Fixed, accepted submission:


import sys
from collections import defaultdict, deque

sys.setrecursionlimit(10 ** 7)
# ----------

INF = float("inf")
MOD = 10 ** 9 + 7
# ----------

N, M = map(int, input().split())
G = defaultdict(list)
for i in range(M):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    G[a].append(b)
    G[b].append(a)


q = deque()


ans = [-1 for _ in range(N)]
ans[0] = 0
q.append(0)

while q:
    v = q.popleft()
    for u in G[v]:
        if ans[u] != -1:
            continue
        ans[u] = v + 1
        q.append(u)

print('Yes')
for i in range(1,N):
    print(ans[i])
Problem:
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

Incorrect submission (verdict: Wrong Answer):
a,b=map(int,input().split())
product=a*b
if product%2==0:
  print("even")
else:
  print("odd")

Fixed, accepted submission:
a,b=map(int,input().split())
product=a*b
if product%2==0:
  print("Even")
else:
  print("Odd")
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):
if __name__ == '__main__':
    N = int(input()) + 1
    P = [list(map(int, input().split())) for x in range(N-1)]
    P.insert(0, [0,0,0])
    #print(N, P)
    for i in range(1, N):
        t = abs(P[i-1][0] - P[i][0])
        
        if t%2 == 0:
            if (abs( sum(P[i-1][1:]) - sum(P[i][1:]) ) <= t) and (abs( sum(P[i-1][1:]) - sum(P[i][1:]))) % 2 == 0:
                pass
            else:
                break
        
        else:
            if (abs( sum(P[i-1][1:]) - sum(P[i][1:]) ) <= t) and (abs( sum(P[i-1][1:]) - sum(P[i][1:]))) % 2 == 1:
                pass
            else:
                break
    else:
        print('OK')
        exit()
    print('NO')

Fixed, accepted submission:
if __name__ == '__main__':
    N = int(input()) + 1
    P = [list(map(int, input().split())) for x in range(N-1)]
    P.insert(0, [0,0,0])
    #print(N, P)
    for i in range(1, N):
        t = abs(P[i-1][0] - P[i][0])
        
        if t%2 == 0:
            if (abs( sum(P[i-1][1:]) - sum(P[i][1:]) ) <= t) and (abs( sum(P[i-1][1:]) - sum(P[i][1:]))) % 2 == 0:
                pass
            else:
                break
        
        else:
            if (abs( sum(P[i-1][1:]) - sum(P[i][1:]) ) <= t) and (abs( sum(P[i-1][1:]) - sum(P[i][1:]))) % 2 == 1:
                pass
            else:
                break
    else:
        print('Yes')
        exit()
    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):
import math
def py():
    print("Yes")
def pn():
    print("No")
def iin():
    x = int(input())
    return x

neko = 0
nya = 0
nuko = 0
a,b = input().split()
neko = int(a + b)
for i in range(100):
    if i*i == neko:
        print(i)
        nya = 1
        break

if(nya == 0):
    pn()

Fixed, accepted submission:
import math
def py():
    print("Yes")
def pn():
    print("No")
def iin():
    x = int(input())
    return x

neko = 0
nya = 0
nuko = 0
a,b = input().split()
neko = int(a + b)
for i in range(1000):
    if i*i == neko:
        py()
        nya = 1
        break

if(nya == 0):
    pn()
Problem:
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand.  How many grams of sand will the upper bulb contains after t seconds?

Incorrect submission (verdict: Wrong Answer):
X, t = map(int, input().split())
print(min(0, X-t))

Fixed, accepted submission:
X, t = map(int, input().split())
print(max(0, X-t))
Problem:
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.   According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j.   Determine if he is correct.

Incorrect submission (verdict: Wrong Answer):
c=[]
for _ in range(3):
  c.append( list(map(int,input().split())))

ans = True
a=[c[0][0],c[1][0],c[2][0]]
b=[c[0][0]-a[0],c[0][1]-a[0],c[0][2]-a[0]]
for i in range(3):
    for j in range(3):
        if c[i][j] != a[i]+b[j]:
            ans = False
print(ans)

Fixed, accepted submission:
c=[]
for _ in range(3):
  c.append( list(map(int,input().split())))
 
ans = "Yes"
a=[c[0][0],c[1][0],c[2][0]]
b=[c[0][0]-a[0],c[0][1]-a[0],c[0][2]-a[0]]
for i in range(3):
    for j in range(3):
        if c[i][j] != a[i]+b[j]:
            ans = "No"
print(ans)
Problem:
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs.  Determine if it is possible that there are exactly X cats among these A + B animals.

Incorrect submission (verdict: Wrong Answer):
A,B,X = map(int,input().split())
print("Yes" if X<=B else "No")

Fixed, accepted submission:
# Python3 (3.4.3)
import sys
input = sys.stdin.readline

# -------------------------------------------------------------
# function
# -------------------------------------------------------------


# -------------------------------------------------------------
# main
# -------------------------------------------------------------
A,B,X = map(int,input().split())

print("YES" if A <= X <= A+B else "NO")
Problem:
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`.  Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
s = input()
cnt = 0
if '
    for i in range(1,N):
        if s[i] == '.' and s[i-1] == '#':
            cnt += 1

print(cnt)

Fixed, accepted submission:
N = int(input())
s = input()
white = s.count('.')
black = 0
cnt = black + white
for i in s:
    if i == '#':
        black += 1
    else:
        white -= 1
    cnt = min(cnt,black + white)
    
print(cnt)
Problem:
There are N people numbered 1 to N. Each person wears a red hat or a blue hat.  You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`.  Determine if there are more people wearing a red hat than people wearing a blue hat.

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

color = input()

r_c = 0
b_c = 0

for i in range(cnt):
    if color[i] == 'R':
        r_c += 1
    elif color[i] == 'B':
        b_c += 1

if r_c > b_c:
    print("YES")
else:
    print("NO")


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

color = input()

r_c = 0
b_c = 0

for i in range(cnt):
    if color[i] == 'R':
        r_c += 1
    elif color[i] == 'B':
        b_c += 1

if r_c > b_c:
    print("Yes")
else:
    print("No")
Problem:
We have a board with a 2 \times N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1 \times 2 or 2 \times 1 square.  Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors.  Find the number of such ways to paint the dominoes, modulo 1000000007.  The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner:    * Each domino is represented by a different English letter (lowercase or uppercase).   * The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
S1 = input()
S2 = input()
def prepare(s1,s2,n):
    result = []
    i = 0
    while i < n:
        if s1[i] == s2[i]:
            result.append(1)
            i += 1
        else:
            result.append(0)
            i += 2

    return result



def calculate(arr):
    ans = 3
    for index in range(1, len(arr)):
        #xo
        if (arr[index - 1] == 1) and (arr[index] == 0):
            ans *= 2

        #xx
        if (arr[index - 1] == 1) and (arr[index] == 1):
            ans *= 2

        #oo
        if (arr[index - 1] == 0) and (arr[index] == 0):
            ans *= 3

        #ox
        if (arr[index - 1] == 0) and (arr[index] == 1):
            ans *= 1

    print(ans % 1000000007)



arr = prepare(S1,S2,N)
calculate(arr)

Fixed, accepted submission:
N = int(input())
S1 = input()
S2 = input()

def prepare(s1,s2,n):
    result = []
    i = 0
    while i < n:
        if s1[i] == s2[i]:
            result.append(1)
            i += 1
        else:
            result.append(0)
            i += 2

    return result



def calculate(arr):
    ans = 1
    for index in range(len(arr)):
        if index == 0:
            if arr[index] == 0:
                ans = 6
            else:
                ans = 3
            continue
        #xo
        if (arr[index - 1] == 1) and (arr[index] == 0):
            ans *= 2

        #xx
        if (arr[index - 1] == 1) and (arr[index] == 1):
            ans *= 2

        #oo
        if (arr[index - 1] == 0) and (arr[index] == 0):
            ans *= 3

        #ox
        if (arr[index - 1] == 0) and (arr[index] == 1):
            ans *= 1

    print(ans % 1000000007)



arr = prepare(S1,S2,N)
calculate(arr)
Problem:
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately.  He has very high awareness of safety, and decides to buy two bells, one for each hand.  The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.

Incorrect submission (verdict: Wrong Answer):
#s = list(input())
s = input()
n = len(s)
count = 0   

if n % 2 == 1:
    count += 1

while s[0 : int((n-count)/2)] != s[int((n-count)/2) : n-count]:
    count += 2

print(n-count)


Fixed, accepted submission:
#a, b, c = map(int, input().split())
abc = list(int(i) for i in input().split())

abc.sort()
z = abc[0] + abc[1]
print(z)
Problem:
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode:                 for i = 1 to A.length-1     key = A[i]     /* insert A[i] into the sorted sequence A[0,...,j-1] */     j = i - 1     while j >= 0 and A[j] > key         A[j+1] = A[j]         j--     A[j+1] = key   Note that, indices for array elements are based on 0-origin.  To illustrate the algorithms, your program should trace intermediate result for each step.

Incorrect submission (verdict: Wrong Answer):
# -*- coding: utf-8 -*-

input_num = int(input())
num_list = [int(i) for i in input().split()]

for i in range(1, len(num_list)):
    v = num_list[i]
    j = i - 1
    while j >= 0 and num_list[j] > v:
        num_list[j+1] = num_list[j]
        j -= 1
    num_list[j+1] = v
    for i in range(len(num_list)):
        print(str(num_list[i]), end="")
        if i < len(num_list)-1:
            print(" ", end="")
        else:
            print()

Fixed, accepted submission:
def insertion_sort(A):
    for i in range(1, len(A)):
        v = A[i]
        j = i - 1
        print(*A)
        while j >= 0 and A[j] > v:
            A[j+1] = A[j]
            A[j] = v
            j -= 1
    return A

n = input()
A = [int(x) for x in input().split()]
print(*insertion_sort(A))
Problem:
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i.  Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.

Incorrect submission (verdict: Wrong Answer):
n,l=int(input()),list(input())
f=any(l.count(l[i])!=2 and l.count(0)==1 for i in range(n))
print(2**(n//2)%1000000007 if f is True else 0)

Fixed, accepted submission:
def main():
    from collections import Counter
    n, *a = map(int, open(0).read().split())
    c = Counter(a)
    if n % 2 == 0:
        f = all(x == 2 for x in c.values())
    else:
        g = c.pop(0) == 1
        f = all(x == 2 for x in c.values()) and g
    ans = 2 ** (n // 2) % 1000000007 if f is True else 0
    print(ans)


if __name__ == '__main__':
    main()
Problem:
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.     

Incorrect submission (verdict: Wrong Answer):
nums = [int(e) for e in input().split()]

if (nums[2]+nums[4])<=nums[0] and (nums[3]+nums[4])<=nums[1] and (nums[3]-nums[4])>=nums[1] and (nums[3]-nums[4])>=nums[1]:
    print("Yes")
else:
    print("No")


Fixed, accepted submission:
nums = [int(e) for e in input().split()]

if (nums[2]+nums[4])<=nums[0] and (nums[3]+nums[4])<=nums[1] and (nums[2]-nums[4])>=0 and (nums[3]-nums[4])>=0:
    print("Yes")
else:
    print("No")
Problem:
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l.  You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i.  Find B_i for each i = 1, 2, ..., N.

Incorrect submission (verdict: Wrong Answer):
N, *X = map(int, open(0).read().split())
mid_l = sorted(X)[(N + 1) // 2 - 1]
mid_r = sorted(X)[(N + 1) // 2]
print([mid_l if v > mid_l else mid_r for v in X])

Fixed, accepted submission:
N, *X = map(int, open(0).read().split())
mid_l = sorted(X)[(N + 1) // 2 - 1]
mid_r = sorted(X)[(N + 1) // 2]
print(*[mid_l if v > mid_l else mid_r for v in X], sep='\n')
Problem:
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets.  Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.

Incorrect submission (verdict: Wrong Answer):
counter = 0

while True:
    x = input()
    if x == "0":
        break
    print("Case " + str(counter) + ":" + str(x))
    counter += 1

Fixed, accepted submission:
counter = 1

while True:
    x = input()
    if x == "0":
        break
    print("Case " + str(counter) + ": " + str(x))
    counter += 1
Problem:
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`.  Consider doing the following operation:    * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.  You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

Incorrect submission (verdict: Wrong Answer):
h,w,r = map(int,input().split())
l=[];an=0
cr=[];k=[0]*w
for i in range(h):
  x=list(input())
  l.append(x)
  cr.append(x.count('#'))
  for j in range(w):
    if x[j]=='#':
      k[j]+=1
ar=cr+k;tot=sum(cr)
tt=len(ar)
for i in range(1<<tt):
  re=tot;rr=[]
  cc=[]
  for j in range(tt):
    if i&(1<<j):
      re-=ar[j]
      if j>=h:
        cc.append(h-j)
      else:
        rr.append(j)
  #print(i,rr,cc,re)
  if re==r:
    an+=1
  #print(i,re)
print(an)

Fixed, accepted submission:
h,w,r = map(int,input().split())
l=[];an=0
cr=[];k=[0]*w
for i in range(h):
  x=list(input())
  l.append(x)
  cr.append(x.count('#'))
  for j in range(w):
    if x[j]=='#':
      k[j]+=1
ar=cr+k;tot=sum(cr)
tt=len(ar)
for i in range(1<<tt):
  re=tot;rr=[]
  cc=[]
  for j in range(tt):
    if i&(1<<j):
      re-=ar[j]
      if j>=h:
        cc.append(j-h)
      else:
        rr.append(j)
  for j in rr:
    for y in cc:
      if l[j][y]=='#':
        re+=1
  #print(i,rr,cc,re)
  if re==r:
    an+=1
  #print(i,re)
print(an)