TnT/CodeNet4Repair huggingface dataset:
Problem:
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.  Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Incorrect submission (verdict: Wrong Answer):
#bA
N = int(input())
d = [int(input()) for i in range(N)]
print(set(d))

Fixed, accepted submission:
#bA
N = int(input())
d = [int(input()) for i in range(N)]
print(len(set(d)))
Problem:
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key.  To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:    * The `0` key: a letter `0` will be inserted to the right of the string.   * The `1` key: a letter `1` will be inserted to the right of the string.   * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.  Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?

Incorrect submission (verdict: Wrong Answer):
s = ''
while True:
    try:
        x = input()
        if x == 'B' and len(s)!=0:
            s.rstrip()
        elif x == 'B' and len(s)==0:
            pass
        else:
            s = s+x
    except EOFError:
        break
print(s)

Fixed, accepted submission:
s=''
x=input()
for i in range(len(x)):
    if x[i]=='B' and len(s)!=0:s = s[:-1]
    elif x[i]=='B' and len(s)==0:pass
    else:s += x[i]
print(s)
Problem:
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?  Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.

Incorrect submission (verdict: Wrong Answer):
A=[]
for _ in range(3):
    A.append(int(input()))
a=A[0]
b=A[1]
c=A[2]
x=int(input())
count=0
for p in range(a):
    for q in range(b):
        for r in range(c):
            if 500*p+100*q+50*r==x:
                count=count+1
else:
    count=count
print(count)

Fixed, accepted submission:
A=[]
for _ in range(3):
    A.append(int(input()))
a=A[0]
b=A[1]
c=A[2]
x=int(input())
count=0
for p in range(a+1):
    for q in range(b+1):
        for r in range(c+1):
            if 500*p+100*q+50*r==x:
                count=count+1
            else:
              count=count
print(count)
Problem:
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i.  Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen.  Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.)  At most how much can he earn on balance?  Note that the same string may be written on multiple cards.

Incorrect submission (verdict: Wrong Answer):
from collections import Counter

n = int(input())
s = [input() for _ in range(n)]
m = int(input())
t = [input() for _ in range(m)]
output = 0

s_count, t_count = Counter(s), Counter(t)

for i in s_count:
    tmp_s = s_count[i]
    tmp_t = 0
    if i in t_count:
        tmp_t = t_count[i]

    if output > tmp_s - tmp_t:
        output = tmp_s - tmp_t

print(output)


Fixed, accepted submission:
from collections import Counter

n = int(input())
s = [input() for _ in range(n)]
m = int(input())
t = [input() for _ in range(m)]
output = 0

s_count, t_count = Counter(s), Counter(t)

for i in s_count:
    tmp_s = s_count[i]
    tmp_t = 0
    if i in t_count:
        tmp_t = t_count[i]

    if output < tmp_s - tmp_t:
        output = tmp_s - tmp_t

print(output)
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):
def main():
    grid = [list(map(int, input().split())) for i in range(3)]
    all = sum(sum(grid, []))
    a = [0] * 3
    b = [0] * 3
    for i1 in range(0, grid[0][0] + 1):
        a[0] = i1
        b[0] = grid[0][0] - i1
        for j2 in range(0, grid[1][1] + 1):
            a[1] = j2
            b[1] = grid[1][1] - j2
            for k3 in range(0, grid[2][2] + 1):
                a[2] = k3
                b[2] = grid[2][2] - k3
                if (sum(a) * 3) + (sum(b) * 3) == all:
                    print("Yes")
                    exit()
    print('No')

Fixed, accepted submission:
import itertools
def main():
    grid = [list(map(int, input().split())) for i in range(3)]
    all = sum(sum(grid, []))
    a = [0] * 3
    b = [0] * 3
    for i, j, k in itertools.product(range(grid[0][0] + 1), range(grid[1][1] + 1), range(grid[2][2] + 1)):
        a = [i, j, k]
        b = [grid[idx][idx] - v for idx, v in enumerate(a)]
        if (sum(a) * 3) + (sum(b) * 3) == all:
            print("Yes")
            exit()
    print('No')


if __name__ == '__main__':
    main()
Problem:
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?

Incorrect submission (verdict: Wrong Answer):
a = int(input())
print(a/2-1)

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

if a%2==0:
  print(int((a/2)-1))
elif a%2==1:
  print(int((a-1)/2))
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=list(map(int,input().split()))

if A<=X<=A+B:
    print('Yes')
else:
    print('No')

Fixed, accepted submission:
A,B,X=list(map(int,input().split()))

if A<=X<=A+B:
    print('YES')
else:
    print('NO')
Problem:
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j).   Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`.   However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.   Determine if square1001 can achieve his objective.

Incorrect submission (verdict: Wrong Answer):

H, W = map(int, input().split())
field = [input() for i in range(H)] 
count = 0

print(field)

for i in range(H):
    for j in range(W):
        
        
        if (i == 0 and j == 0) or (i == H-1 and j == 0) or (i == 0 and j == W-1) or (i == H-1 and j == W-1):
            pass
        elif i == 0:
            if field[i][j] == '#' and field[i][j-1] != '#' and field[i][j+1] != '#' and field[i+1][j] != '#':
                count += 1
        elif i == H-1:
            if field[i][j] == '#' and field[i][j-1] != '#' and field[i][j+1] != '#' and field[i-1][j] != '#':
                count += 1
        elif j == 0:
            if field[i][j] == '#' and field[i][j-1] != '#' and field[i][j+1] != '#' and field[i+1][j] != '#':
                count += 1
        elif j == W-1:
            if field[i][j] == '#' and field[i][j-1] != '#' and field[i+1][j] != '#' and field[i-1][j] != '#':
                count += 1
        elif field[i][j] == '#' and field[i][j-1] != '#' and field[i][j+1] != '#' and field[i+1][j] != '#' and field[i-1][j] != '#':
            count += 1
print('No' if count>=1 else 'Yes')

Fixed, accepted submission:

H, W = map(int, input().split())
field = [list(input()) for i in range(H)] 
count = 0

for i in range(H):
    for j in range(W):
        
        
        if (i == 0 and j == 0) or (i == H-1 and j == 0) or (i == 0 and j == W-1) or (i == H-1 and j == W-1):
            pass
        elif i == 0:
            if field[i][j] == '#' and field[i][j-1] != '#' and field[i][j+1] != '#' and field[i+1][j] != '#':
                count += 1
        elif i == H-1:
            if field[i][j] == '#' and field[i][j-1] != '#' and field[i][j+1] != '#' and field[i-1][j] != '#':
                count += 1
        elif j == 0:
            if field[i][j] == '#' and field[i][j-1] != '#' and field[i][j+1] != '#' and field[i+1][j] != '#':
                count += 1
        elif j == W-1:
            if field[i][j] == '#' and field[i][j-1] != '#' and field[i+1][j] != '#' and field[i-1][j] != '#':
                count += 1
        elif field[i][j] == '#' and field[i][j-1] != '#' and field[i][j+1] != '#' and field[i+1][j] != '#' and field[i-1][j] != '#':
            count += 1
print('No' if count>=1 else 'Yes')
Problem:
We have a grid with H rows and W columns, where all the squares are initially white.  You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions:    * Choose one row, then paint all the squares in that row black.   * Choose one column, then paint all the squares in that column black.  At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.

Incorrect submission (verdict: Wrong Answer):
H=int(input())
W=int(input())
N=int(input())
if H<W:
    H,W=W,H
ans = N // H
if N % H > 0: N += 1
print(ans)


Fixed, accepted submission:
H=int(input())
W=int(input())
N=int(input())
if H<W:
    H,W=W,H
ans = N // H
if N % H > 0: ans += 1
print(ans)
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())
ans = n % 1000
print(ans)


Fixed, accepted submission:
n = int(input())
mod = n % 1000
ans = 1000 - mod if mod != 0 else 0
print(ans)
Problem:
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.  Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Incorrect submission (verdict: Wrong Answer):
n_d = list(map(int, input().split()))
d = n_d[1:]
max_d = len(set(d))
print(max_d)

Fixed, accepted submission:
x = int(input())
y = [int(input()) for i in range(x)]
max_d = len(set(y))
print(max_d)
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())
l=[]
for i in range(n):
  s=input()
  l.append(s)
  
ac=l.count('AC')
wa=l.count('WA')
tle=l.count('TLE')
re=l.count('RE')

print('AC ' +'x' + str(ac))
print('WA ' +'x' + str(wa))
print('TLE ' +'x' + str(tle))
print('RE ' +'x' + str(re))


Fixed, accepted submission:
n=int(input())
l=[]
for i in range(n):
  s=input()
  l.append(s)
  
ac=l.count('AC')
wa=l.count('WA')
tle=l.count('TLE')
re=l.count('RE')

print('AC ' +'x ' + str(ac))
print('WA ' +'x ' + str(wa))
print('TLE ' +'x ' + str(tle))
print('RE ' +'x ' + str(re))
Problem:
There is a hotel with the following accommodation fee:    * X yen (the currency of Japan) per night, for the first K nights   * Y yen per night, for the (K+1)-th and subsequent nights  Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
k=int(input())
x=int(input())
y=int(input())
ans=0
for i in range(n):
    if n>k:
        ans+=y
    else:
        ans+=x
print(ans)

Fixed, accepted submission:
n=int(input())
k=int(input())
x=int(input())
y=int(input())
ans=0
for i in range(1,n+1):
    if i>k:
        ans+=y
    else:
        ans+=x
print(ans)
Problem:
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.  He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.

Incorrect submission (verdict: Wrong Answer):
A,V=map(int,input().split())
B,W=map(int,input().split())
T=int(input())
if W>=V:
  print('No')
else:
  key=abs(A-B)
  if key+W*T>V*T:
    print('No')
  else:
    print('Yes')

Fixed, accepted submission:
A,V=map(int,input().split())
B,W=map(int,input().split())
T=int(input())
if W>=V:
  print('NO')
else:
  key=abs(A-B)
  if key+W*T>V*T:
    print('NO')
  else:
    print('YES')
Problem:
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters.  Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s.  Given the name of the contest, print the abbreviation of the name.

Incorrect submission (verdict: Wrong Answer):
s=input()
print("A"+s[0]+"C")

Fixed, accepted submission:
a,s,c=input().split()
print("A"+s[0]+"C")
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):
from collections import defaultdict
def main():
    N = int(input())
    c = defaultdict(int)

    for i in range(N):
        c[input()] += 1

    print("AC × {}".format(c["AC"]))
    print("WA × {}".format(c["WA"]))
    print("TLE × {}".format(c["TLE"]))
    print("RE × {}".format(c["RE"]))

if __name__ == "__main__":
    main()

Fixed, accepted submission:
from collections import defaultdict
def main():
    N = int(input())
    c = defaultdict(int)

    for i in range(N):
        c[input()] += 1

    print("AC x {}".format(c["AC"]))
    print("WA x {}".format(c["WA"]))
    print("TLE x {}".format(c["TLE"]))
    print("RE x {}".format(c["RE"]))

if __name__ == "__main__":
    main()
Problem:
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements?  Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds:    * r_1 + c_1 = r_2 + c_2   * r_1 - c_1 = r_2 - c_2  For example, in the following figure, the bishop can move to any of the red squares in one move:  

Incorrect submission (verdict: Wrong Answer):
a,b=map(int,input().split())
print(-(int(-(a+b)/2)))

Fixed, accepted submission:
a,b=map(int,input().split())
if(a==1 or b==1):
  print(1)
else:
  print(int((a*b)/2)+(a*b)%2)
Problem:
Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten. He remembers that the remainder of a divided by b was greater than or equal to K. Find the number of possible pairs that he may have had.

Incorrect submission (verdict: Wrong Answer):
K = input().split()
N = int(K[0])
K = int(K[1])

count = 0
for b in range(K+1, N+1):
    for l in range(K, b):
        count += int((N-l)/b)

print(count)
if K != 0:
    for a in range(K, N+1):
        count += N-a
if K == 0:
        for a in range(1, N+1):
            count += N-a

print(count)

Fixed, accepted submission:
K = input().split()
N = int(K[0])
K = int(K[1])

count = 0
for b in range(K+1, N+1):
    siki1 = int((N-K)/b)
    siki2 = int((N-b+1)/b)
    if siki1 <= siki2:
        count += (b-K)*siki1
    elif siki1 > siki2:
        count += (b-K)*siki1
        count -= (b-N-1)*(siki1 - siki2)
        count -= b*(int((siki1*(siki1+1))/2) - int((siki2*(siki2+1))/2))

if K != 0:
    for a in range(K, N+1):
        count += N-a
if K == 0:
        for a in range(1, N+1):
            count += N-a

print(count)
Problem:
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}.  Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.

Incorrect submission (verdict: Wrong Answer):
C=list(open(0).read())
print('YES')if C[:3] == C[-1:-4:-1] else print('NO')

Fixed, accepted submission:
print('YES')if input() == input()[::-1] else print('NO')
Problem:
For an integer n not less than 0, let us define f(n) as follows:    * f(n) = 1 (if n < 2)   * f(n) = n f(n-2) (if n \geq 2)  Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

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

half = n // 2

ans = half / 5

print(ans)

Fixed, accepted submission:
n = int(input())
if (n % 2 == 1):
    print(0)
else:
    ans = 0
    d = 10
    while d <= n:
        ans += n // d
        d *= 5
    print(ans)
Problem:
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.  Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.

Incorrect submission (verdict: Wrong Answer):

def prime(n):
    s=2
    while(s*s <= n):
        if(n%s==0):
            return False
        s+=2
    return True

n=int(input())
for i in range(n):
    x = input()
    if prime(int(x)):
        print(x)

Fixed, accepted submission:

def prime(n):
    s=3
    while(s*s <= n):
        if(n%s==0):
            return False
        s+=2
    return True

n=int(input())
num=0
for i in range(n):
    x = int(input())
    if prime(x) and x%2!=0:
        num+=1
    if x==2:
        num+=1
print(num)
Problem:
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.  To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Incorrect submission (verdict: Wrong Answer):
abc = list(map(int, input().split()))
abc.sort()
print(abc)
if [5, 5, 7] == abc:
    ans = "YES"
else:
    ans = "NO"
print(ans)

Fixed, accepted submission:
abc = list(map(int, input().split()))
abc.sort()
if [5, 5, 7] == abc:
    ans = "YES"
else:
    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):
import re
import sys
import math
import itertools
import bisect
from copy import copy
from collections import deque,Counter
from decimal import Decimal
import functools
def v(): return input()
def k(): return int(input())
def S(): return input().split()
def I(): return map(int,input().split())
def X(): return list(input())
def L(): return list(input().split())
def l(): return list(map(int,input().split()))
def lcm(a,b): return a*b//math.gcd(a,b)
sys.setrecursionlimit(10 ** 9)
mod = 10**9+7
cnt = 0
ans = 0
inf = float("inf")

a,b,c = I()
if c > a+b:
    print("YES")
else:
    print("NO")


Fixed, accepted submission:
import re
import sys
import math
import itertools
import bisect
from copy import copy
from collections import deque,Counter
from decimal import Decimal
import functools
def v(): return input()
def k(): return int(input())
def S(): return input().split()
def I(): return map(int,input().split())
def X(): return list(input())
def L(): return list(input().split())
def l(): return list(map(int,input().split()))
def lcm(a,b): return a*b//math.gcd(a,b)
sys.setrecursionlimit(10 ** 9)
mod = 10**9+7
cnt = 0
ans = 0
inf = float("inf")

A,B,X=I()
print("YES" if A<=X<=A+B else "NO")
Problem:
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.

Incorrect submission (verdict: Wrong Answer):
s = input()
for i in range(len(s)):
    if s[i] in s[i+1:]:
        print('No')
        break
else:
    print('Yes')

Fixed, accepted submission:
N=input()
if len(N)==len(set(N)):
    print("yes")
else:
    print("no")
Problem:
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`.  You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.

Incorrect submission (verdict: Wrong Answer):
s=input()
print(s[0],len(s)-2,s[len(s)-1])

Fixed, accepted submission:
s=input()
print(s[0]+str(len(s)-2)+s[len(s)-1])