TnT/CodeNet4Repair huggingface dataset:
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):
N=int(input())
l=[]
m=[]
while N>0:
    a=input()
    l.append(a)
    N-=1
M=int(input())
while M>0:
    b=input()
    m.append(b)
    M-=1
print(sorted(l))
print(sorted(m))

Fixed, accepted submission:
n=int(input())
s=[input() for i in range(n)]
m=int(input())
t=[input() for i in range(m)]
li=list(set(s))
print(max(0,max(s.count(li[i])-t.count(li[i]) for i in range(len(li)))))
Problem:
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.  Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.  Find the inversion number of B, modulo 10^9 + 7.  Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Incorrect submission (verdict: Wrong Answer):
N,K = (int(i) for i in input().split()) 
A = [int(i) for i in input().split()] 

mod = 10**9+7
ans = 0

for i in range(N):
	bs = [1 if g < A[i] else 0 for g in A]
	print(bs)
	ans += 0.5*(sum(bs[i:]) + sum(bs[:i]))*K*(K+1) - sum(bs[:i])*K
print(int(ans)%mod)

Fixed, accepted submission:
N,K = (int(i) for i in input().split()) 
A = [int(i) for i in input().split()] 

mod = 10**9+7
ans = 0

for i in range(N):
	bs = [1 if g < A[i] else 0 for g in A]
	ans += (K+1)*K*sum(bs)//2 - sum(bs[:i])*K
print(int(ans)%mod)
Problem:
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**.  You are given a 4-digit integer N. Answer the question: Is N **good**?

Incorrect submission (verdict: Wrong Answer):
N=input();print('Yes')if set(N[:3])==1or set(N[1:])==1else print('No')

Fixed, accepted submission:
N=input();print('Yes')if len(set(N[:3]))==1or len(set(N[1:]))==1else print('No')
Problem:
AtCoDeer has three cards, one red, one green and one blue.   An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card.   We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer.   Is this integer a multiple of 4?

Incorrect submission (verdict: Wrong Answer):
r,g,b = map(int,input().split())
c = 100*r + 10*g + b
if c % 4 == 0:
  print('Yes')
else:
  print('No')

Fixed, accepted submission:
r,g,b = map(int,input().split())
c = 100*r + 10*g + b
if c % 4 == 0:
  print('YES')
else:
  print('NO')
Problem:
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value:    * Operation A: The displayed value is doubled.   * Operation B: The displayed value increases by K.  Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.

Incorrect submission (verdict: Wrong Answer):
num = int( input() )
k = int( input() )
ans = 1
for i in range( num ):
    if ( ans*2 > ans+k ):
        ans = ans * 2
    else:
        ans = ans + k
print( ans )


Fixed, accepted submission:
num = int( input() )
k = int( input() )
ans = 1
for i in range( num ):
    if ( ans*2 < ans+k ):
        ans = ans * 2
    else:
        ans = ans + k
print( ans )
Problem:
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.  You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.

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

for i in range(1,len(S)):
    if i%2 == 1:continue
    tmp = S[:i//2]
    print(tmp,S[i//2:i])
    if S[i//2:i] == tmp:
        ans = i
        
print(ans)

Fixed, accepted submission:
S = input()

for i in range(1,len(S)):
    if i%2 == 1:continue
    tmp = S[:i//2]
    if S[i//2:i] == tmp:
        ans = i
        
print(ans)
Problem:
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.  Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Incorrect submission (verdict: Wrong Answer):
a,b,c,x,y = map(int,input().split())
print( min(x,y)*min(a+b,2*c) + (max(x,y)-min(x,y))*(max(x,y)==y)*b + (max(x,y)-min(x,y))*(max(x,y)==a)*a)

Fixed, accepted submission:
a,b,c,x,y = map(int,input().split())
print(min((min(x,y)*min(a+b,2*c) + (max(x,y)-min(x,y))*(max(x,y)==y)*b + (max(x,y)-min(x,y))*(max(x,y)==x)*a), max(x,y)*2*c))
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 = input('N:')
k = input('K:')
x = input('X:')
y = input('Y:')

n = int(n)
k = int(k)
x = int(x)
y = int(y)

if n >= k:
    total = k * x + (n - k) * y
else:
    total = n * x
    
print(total)

Fixed, accepted submission:
n = input()
k = input()
x = input()
y = input()

n = int(n)
k = int(k)
x = int(x)
y = int(y)

if n >= k:
    total = k * x + (n - k) * y
else:
    total = n * x
    
print(total)
Problem:
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}.  Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.

Incorrect submission (verdict: Wrong Answer):
import sys
s = sys.stdin.readlines()

H, W = map(int, s[0].split())
a0 = '#' * (W + 2)
print(a0)
for i in range(H):
    print('#' + s[i + 1] + '#')
print(a0)

Fixed, accepted submission:
H, W = map(int, input().split())
a0 = '#' * (W + 2)
print(a0)
for i in range(H):
    print('#' + input() + '#')
print(a0)
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):
x = str(input().split()[1])
print("A{}C".format(x))

Fixed, accepted submission:
x = str(input().split()[1])
x = x[0]
print("A{}C".format(x))
Problem:
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank.  The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.)  Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

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

a=100

for i in range(3800):
  if a>X:
    print(i)
    break
  else:
    a=(a*1.01)//1
  if a>10**6:
    a/(10**6)
    X/(10**6)

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

a=100

for i in range(3800):
  if a>=X:
    print(i)
    break
  else:
    a=(a*1.01)//1
  if a>10**6:
    a/(10**6)
    X/(10**6)
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 = int(input())
nums = set(list(map(int, input().split())))
print(len(nums))

Fixed, accepted submission:
N = int(input())
nums = [input() for i in range(N)]
print(len(set(nums)))
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())
ans=0
cnt=5
if n%2==0:
    while True:
        tmp=n//cnt
        if tmp==0:
            break
        else:
            ans+=tmp
            cnt=cnt*5
            print(tmp)
print(ans//2)

Fixed, accepted submission:
n=int(input())
ans=0
if n%2==0:
    n=n//2
    while n>0:
        n=n//5
        ans+=n
print(ans)
Problem:
Sitting in a station waiting room, Joisino is gazing at her train ticket.  The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).  In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds.  The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Incorrect submission (verdict: Wrong Answer):
abcd = str(input())
abcd_list = list(abcd)
a = int(abcd_list[0])
b = int(abcd_list[1])
c = int(abcd_list[2])
d = int(abcd_list[3])

ans2='+'
ans3='+'
ans4='+'
for i in [a]:
    for j in [b,-b]:
        for k in [c,-c]:
            for l in [d,-d]:
                ans = i + j + k + l
                if ans == 7:
                    if j < 0:
                        ans2 = '-'
                    elif k < 0:
                        ans3 = '-'
                    elif l < 0:
                        ans4 = '-'

print(str(a)+ans2+str(b)+ans3+str(c)+ans4+str(d))      

Fixed, accepted submission:
s = str(input())
for bits in range(1 << 3):
    op1 = '+'
    op2 = '+'
    op3 = '+'
    for i in range(3):
        if (bits >> i & 1) == 1:
            if i == 0:
                op1 = '-'
            if i == 1:
                op2 = '-'
            if i == 2:
                op3 = '-'
    if eval(s[0]+op1+s[1]+op2+s[2]+op3+s[3]) == 7:
        print(s[0]+op1+s[1]+op2+s[2]+op3+s[3]+'=7')
        break
Problem:
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
an = list(map(int, input().split()))

maxi = max(an)
v = maxi
for a in an:
  if a == maxi:
    continue
  if abs(a-maxi) < abs(v-maxi):
    v = a
print(' '.join( map(str, [maxi, v]) ))

Fixed, accepted submission:
n = int(input())
an = list(map(int, input().split()))
if n == 2:
  print(' '.join( map(str, [max(an), min(an)]) ))
  exit()

maxi = max(an)
v = maxi
for a in an:
  if a == maxi:
    continue
  if abs(a-maxi/2) < abs(v-maxi/2):
    v = a
print(' '.join( map(str, [maxi, v]) ))
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 heapq
import sys

input = sys.stdin.readline

def Dijkstra_heap(s, edge):
    
    d = [10**6] * n
    used = [False] * n
    d[s] = 0
    used[s] = True
    edgelist = []
    for i in edge[s]:
        heapq.heappush(edgelist, i)
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        v = minedge % (10**6)
        
        if used[v]: continue
        d[v] = minedge // (10**6)
        used[v] = True
        for e in edge[v]:
            if not used[e]:
                heapq.heappush(edgelist, e+d[v]*(10**6))
    return d

n, m = map(int, input().split())
edge = [[] for _ in range(n)]

for _ in range(m):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    edge[a].append(b)
    edge[b].append(a)


dist = Dijkstra_heap(0, edge)
if 10**6 in dist:
  print("No")
  exit()
for i in dist[1:]:
  print(i)

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

f_inf = float("inf")

n, m = map(int, input().split())
to = [[] for _ in range(n)]

for _ in range(m):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    to[a].append(b)
    to[b].append(a)

que = deque([])
que.append(0)

dist = [f_inf for _ in range(n)]
dist[0] = 0

guide = [-1 for _ in range(n)]

while que:
    v = que.popleft()
    for u in to[v]:
        if dist[u] != f_inf: continue
        dist[u] = dist[v]+1
        guide[u] = v
        que.append(u)

print("Yes")
for i in range(1, n):
    ans = guide[i] + 1
    print(ans)
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):
def main():
    n, m = map(int, input().split())
    if n % 2 == 0 or n % 2 == 0:
        print('Even')
    else:
        print('Odd')

if __name__ == '__main__':
    main()

Fixed, accepted submission:
def main():
    n, m = map(int, input().split())
    if n % 2 == 0 or m % 2 == 0:
        print('Even')
    else:
        print('Odd')
if __name__ == '__main__':
    main()
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):
a, b = map(str,input().split())
c = int(a+b)
print(c)

Fixed, accepted submission:
a, b = map(str,input().split())
c = int(a+b)

d = c ** (1/2)
if d.is_integer():
  print("Yes")
else:
  print("No")
Problem:
A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination.  Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness.  For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions:    * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal.   * For every i, B_i \leq C_i holds.  If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
df = list([0 for i in range(n)])
ans = -1

if sum(a) < sum(b):
    ans = -1
else:
    for i in range(n):
        df[i] = a[i] - b[i]
    minus = list(filter(lambda x: x < 0, df))
    plus = list(filter(lambda x: x > 0, df))
    m_count = len(minus)
    m_sum = sum(minus)
    plus.sort(reverse=True)

    if m_count == 0:
        ans = 0
    else:
        print(minus)
        print(plus)
        for i in range(len(plus)):
            m_sum += plus[i]
            if m_sum >= 0:
                break

        if m_sum>=0:
            ans = m_count + i + 1
print(ans)


Fixed, accepted submission:
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
df = list([0 for i in range(n)])
ans = -1

if sum(a) < sum(b):
    ans = -1
else:
    for i in range(n):
        df[i] = a[i] - b[i]
    minus = list(filter(lambda x: x < 0, df))
    plus = list(filter(lambda x: x > 0, df))
    m_count = len(minus)
    m_sum = sum(minus)
    #print("m_count=%d"%(m_count))
    #print("m_sum=%d"%(m_sum))
    plus.sort(reverse=True)
    #print(df)
    

    if m_count == 0:
        ans = 0
    else:
        for i in range(len(plus)):
            m_sum += plus[i]
            if m_sum >= 0:
                break

        if m_sum>=0:
            ans = m_count + i + 1
print(ans)
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 = list(map(int, input().split()))

for i in range(a, a*(b+1), a):
  if i%b == c:
    print('Yes')
    exit()
print('No')

Fixed, accepted submission:
a, b, c = list(map(int, input().split()))

for i in range(a*b):
  if (a*i)%b == c:
    print('YES')
    exit()
print('NO')
Problem:
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R.  Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan.  Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.

Incorrect submission (verdict: Wrong Answer):
A, B, C, D = map(int, input().split())
L = A + B
R = C + D
if L < R:
	print('Left')
elif L==R:
	print('Balanced')
else:
	print('Right')

Fixed, accepted submission:
A, B, C, D = map(int, input().split())
L = A + B
R = C + D
if L > R:
	print('Left')
elif L==R:
	print('Balanced')
else:
	print('Right')
Problem:
Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Incorrect submission (verdict: Wrong Answer):
S = input()
ans = 0
for i in range(1,len(S)):
  if S[i-1] != S[i]:
    ans += 1

Fixed, accepted submission:
S = input()
ans = 0
for i in range(1,len(S)):
  if S[i] != S[i-1]:
    ans += 1
print(ans)
Problem:
Snuke loves puzzles.  Today, he is working on a puzzle using `S`\- and `c`-shaped pieces. In this puzzle, you can combine two `c`-shaped pieces into one `S`-shaped piece, as shown in the figure below:    Snuke decided to create as many `Scc` groups as possible by putting together one `S`-shaped piece and two `c`-shaped pieces.  Find the maximum number of `Scc` groups that can be created when Snuke has N `S`-shaped pieces and M `c`-shaped pieces.

Incorrect submission (verdict: Wrong Answer):
s,c = map(int,input().split())
ans = 0
if s*2 <= c:
    c = c - s*2
    ans += s
    s = 0
else:
    tmp = c//2
    s = s - tmp
    c = c%2
    ans += tmp
#print(s,c,ans)
ans += c//4
#print(s,c,ans)

Fixed, accepted submission:
n,m = map(int,input().split())

ans = 0
if m-n*2>0:
    ans += n
    tmp = m-n*2
    
    ans += tmp//4
else:
    ans += m//2
print(ans)
Problem:
Snuke is giving cookies to his three goats.  He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins).  Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.

Incorrect submission (verdict: Wrong Answer):
import sys

a, b = map(int, sys.stdin.readline().split())

if a % 3 == 0 or b % 3 == 0 or a + b % 3 == 0:
    print("Possible")
else:
    print("Impossible")

Fixed, accepted submission:
import sys

a, b = map(int, sys.stdin.readline().split())

if a % 3 == 0 or b % 3 == 0 or (a + b) % 3 == 0:
    print("Possible")
else:
    print("Impossible")
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] + str(len(s[1:-2])) + s[-1])


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