TnT/CodeNet4Repair huggingface dataset:
Problem:
There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle.  Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.

Incorrect submission (verdict: Wrong Answer):
H , W = map(int , input().split())
h_ans = 100000
w_ans = 100000

for h in range(1,H):
    Sa = h * W
    Sb = (H - h) * (W // 2)
    Sc = (H - h) * (W - ( W //2 ))
    if max(Sa,Sb,Sc) - min(Sa,Sb,Sc) < h_ans:
        h_ans = max(Sa,Sb,Sc) - min(Sa,Sb,Sc)
    
for w in range(1,W):
    Sa = H * w
    Sb = (H // 2) * (W - w)
    Sc = (H - (H //2)) * (W - w)
    if max(Sa,Sb,Sc) - min(Sa,Sb,Sc) < w_ans:
        w_ans = max(Sa,Sb,Sc) - min(Sa,Sb,Sc)
        
print((min(h_ans,w_ans)))

Fixed, accepted submission:
H , W = map(int , input().split())
h_ans = 100000
w_ans = 100000

for h in range(1,H):
    Sa = h * W
    Sb = (H - h) * (W // 2)
    Sc = (H - h) * (W - ( W //2 ))
    if max(Sa,Sb,Sc) - min(Sa,Sb,Sc) < h_ans:
        h_ans = max(Sa,Sb,Sc) - min(Sa,Sb,Sc)
        
for h in range(1,H):
    Sa = h * W
    Sb = ((H - h)//2) * W
    Sc = (H - (h + (H-h)//2)) * W
    if max(Sa,Sb,Sc) - min(Sa,Sb,Sc) < h_ans:
        h_ans = max(Sa,Sb,Sc) - min(Sa,Sb,Sc)
    
for w in range(1,W):
    Sa = H * w
    Sb = (H // 2) * (W - w)
    Sc = (H - (H //2)) * (W - w)
    if max(Sa,Sb,Sc) - min(Sa,Sb,Sc) < w_ans:
        w_ans = max(Sa,Sb,Sc) - min(Sa,Sb,Sc)
    
for w in range(1,W):
    Sa = H * w
    Sb = H * ((W -w)//2)
    Sc = H * (W - (w + (W -w)//2) ) 
    if max(Sa,Sb,Sc) - min(Sa,Sb,Sc) < w_ans:
        w_ans = max(Sa,Sb,Sc) - min(Sa,Sb,Sc)
    
        
print((min(h_ans,w_ans)))
Problem:
For a given weighted graph $G = (V, E)$, find the minimum spanning tree (MST) of $G$ and print total weight of edges belong to the MST.

Incorrect submission (verdict: Wrong Answer):
MAX = 2001


def prim(tree: [int], adjacent_list: [[int]]):
    cost = 0
    next = 0
    for l in adjacent_list:
        l[next] = MAX
    while True:
        tree.append(next)
        min_list = list()
        for t in tree:
            min_list.append(min([(v, i) for i, v in enumerate(adjacent_list[t])]))
        min_cost, i = min(min_list)
        if min_cost == MAX:
            return cost
        cost += min_cost
        for l in adjacent_list:
            l[i] = MAX
        print(adjacent_list)
        next = i


if __name__ == '__main__':
    N = int(input())
    adjacent_list = [[int(e) if int(e) != -1 else MAX for e in input().rstrip().split()] for _ in range(N)]
    test_list = []
    print(prim(test_list, adjacent_list))
    print(test_list)

Fixed, accepted submission:
MAX = 2001


def prim(tree: [int], adjacent_list: [[int]]):
    cost = 0
    next = 0
    for l in adjacent_list:
        l[next] = MAX
    while True:
        tree.append(next)
        min_list = list()
        for t in tree:
            min_list.append(min([(v, i) for i, v in enumerate(adjacent_list[t])]))
        min_cost, i = min(min_list)
        if min_cost == MAX:
            return cost
        cost += min_cost
        for l in adjacent_list:
            l[i] = MAX
        next = i


if __name__ == '__main__':
    N = int(input())
    adjacent_list = [[int(e) if int(e) != -1 else MAX for e in input().rstrip().split()] for _ in range(N)]
    test_list = []
    print(prim(test_list, adjacent_list))
Problem:
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.

Incorrect submission (verdict: Wrong Answer):
s = input()
A = 'abcdefghijklmnopqrstuvwxyz'
for c in A:
    print(c+':{}'.format(s.count(c)))

Fixed, accepted submission:
import sys
s = []
for line in sys.stdin:
    s += line.lower()
A = 'abcdefghijklmnopqrstuvwxyz'
for c in A:
    print(c+' : {}'.format(s.count(c)))
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 V <= W:
  print("No")
else:
  if (max(A, B) - min(A, B)) / (V - W) <= T:
    print("Yes")
  else:
    print("No")


Fixed, accepted submission:
A ,V = map(int, input().split())
B ,W = map(int, input().split())
T = int(input())

if V <= W:
  print("NO")
else:
  if (max(A, B) - min(A, B)) / (V - W) <= T:
    print("YES")
  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())
#a,b = map(int,input().split())
N = n
cnt = 0
num = 0
while N:
    cnt += n//5**num
    num += 1
    N = N//5
print(cnt)

Fixed, accepted submission:
n = int(input())
if n%2 ==1:
    print(0)
    exit()
cnt = 0
i = 1

if n%2 == 0:
    while n//(2*(5**i)) > 0:
        cnt += n//(2*(5**i))
        i +=1
print(cnt)
Problem:
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

Incorrect submission (verdict: Wrong Answer):
S = input()
for i in range(len(S)):
  if i % 2 == 0:
    print(S[i])


Fixed, accepted submission:
S = input()
k = ""
for i in range(len(S)):
  if i % 2 == 0:
    k += S[i]
print(k)
Problem:
Takahashi has N balls. Initially, an integer A_i is written on the i-th ball.  He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls.  Find the minimum number of balls that Takahashi needs to rewrite the integers on them.

Incorrect submission (verdict: Wrong Answer):
N, K = map(int, input().split())
A = list(map(int, input().split()))
d = {}
for i in set(A):
  d[i] = A.count(i)
ans = K
for i, (k, v) in enumerate(sorted(d.items())):
  if i < K -1:
    ans -= v

print(ans)

Fixed, accepted submission:
from collections import Counter

n, k = map(int, input().split())
a = list(map(int, input().split()))

c = Counter(a)

arr = c.most_common()[k:]

print(sum(x[1] for x in arr))
Problem:
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it.  The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).  Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer.  If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
for X in range(int(N/1.1), N):
    if int(X * 10.8) == N:
        print(X)
        exit()
print(":(")


Fixed, accepted submission:
N = int(input())
for X in [-int(-N/1.08), int((N+1)/1.08)]:
    if int(X * 1.08) == N:
        print(X)
        exit()
print(":(")
Problem:
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively.    An example of a trapezoid  Find the area of this trapezoid.

Incorrect submission (verdict: Wrong Answer):
a = int(input())
b = int(input())
h = int(input())
print ((a+b)/h)

Fixed, accepted submission:
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
Problem:
Given are N integers A_1,\ldots,A_N.  We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.  Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Incorrect submission (verdict: Wrong Answer):
N, K = map(int, input().split())
A = list(map(int, input().split()))

MOD = 10**9 + 7

A_abs = [abs(a) for a in A]
A_is_negative = [a < 0 for a in A]


A_abs, A_is_negative, A = zip(*sorted(zip(A_abs, A_is_negative, A), key=lambda x: x[0], reverse=True))

ans = 1
ans_is_negative = 0

for i in range(K-1):
    ans *= A[i]
    if A_is_negative[i]:
        
        ans_is_negative = 1 - ans_is_negative
        last_negative = i


if ans_is_negative and A_is_negative[K-1]:
    ans *= A[K-1]
    ans %= MOD
    print(ans)
elif ans_is_negative and not A_is_negative[K-1]:
    
    for i in range(K, N):
        if A_is_negative[i]:
            ans *= A[K-1]
            ans %= MOD
            print(ans)
            exit()
    
    
    tar = A[last_negative]
    tar_inv = pow(tar, MOD-2, MOD)
    ans *= tar_inv
    ans %= MOD
    ans *= A[K-1]
    ans %= MOD
    ans *= A[K]
    ans %= MOD
    print(ans)
elif not ans_is_negative and A_is_negative[K-1]:
    
    for i in range(K, N):
        if not A_is_negative[i]:
            ans *= A[K-1]
            ans %= MOD
            print(ans)
            exit()
    .
    
    ans = 1
    for i in range(1, K+1):
        ans *= A[-i]
        ans %= MOD
    print(ans)
else:
    
    ans *= A[K-1]
    ans %= MOD
    print(ans)


Fixed, accepted submission:
from heapq import heappush, heappop


N, K = map(int, input().split())
A = list(map(int, input().split()))
A = sorted(A, reverse=True)

MOD = 10**9 + 7


if N == K:
    ans = 1
    for i in range(K):
        ans *= A[i]
        ans %= MOD
    print(ans)

elif max(A) < 0 and K % 2 == 1:
    ans = 1
    for i in range(K):
        ans *= A[i]
        ans %= MOD
    print(ans)
else:
    

    
    

    if K%2 == 1:
        ans = A.pop(0)  
        n = (K-1)//2
    else:
        ans = 1
        n = K//2

    
    A = sorted(A, key=lambda x: abs(x), reverse=True)

    
    cand = []  

    BIGINT = 10**9 + 1  
    tmp1 = BIGINT
    tmp2 = BIGINT

    for a in A:
        if a >= 0:
            if tmp1 == BIGINT:
                tmp1 = a
            else:
                heappush(cand, -tmp1*a)
                tmp1 = BIGINT
        else:
            if tmp2 == BIGINT:
                tmp2 = a
            else:
                heappush(cand, -tmp2*a)
                tmp2 = BIGINT

    
    for i in range(n):
        ans *= -heappop(cand)
        ans %= MOD
    print(ans)
Problem:
We will say that two integer sequences of length N, x_1, x_2, ..., x_N and y_1, y_2, ..., y_N, are _similar_ when |x_i - y_i| \leq 1 holds for all i (1 \leq i \leq N).  In particular, any integer sequence is similar to itself.  You are given an integer N and an integer sequence of length N, A_1, A_2, ..., A_N.  How many integer sequences b_1, b_2, ..., b_N are there such that b_1, b_2, ..., b_N is similar to A and the product of all elements, b_1 b_2 ... b_N, is even?

Incorrect submission (verdict: Wrong Answer):
def get_result(A):
    res = 1
    total = 1
    for i, ai in enumerate(A):
        print(ai, res, total)
        res *= 2 if ai % 2 == 0 else 1
        total *= 3 if ai > 1 else 2
    return total - res

A = [int(ai) for ai in input().split()]
print(get_result(A))

Fixed, accepted submission:
def get_result(A):
    res = 1
    total = 3 ** len(A)
    for i, ai in enumerate(A):
        res *= 2 if ai % 2 == 0 else 1
    return total - res
n = input()
A = [int(ai) for ai in input().split()]
print(get_result(A))
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):
s = input()
#import sys
#s = '1222'
n = len(s)
is7 = False
for i in range(2**(n-1)):
  #print(i)
  f = ''
  b = format(i, 'b').zfill(n)
  #print(b)
  for j in range(n):
    f += s[j]
    if b[j] == '0':
      f += '-'
    else:
      f += '+'
  f = f[:len(f)-1]
  if eval(f) == 7:
    print(f)
    break

Fixed, accepted submission:
s = input()
#import sys
#s = '1222'
n = len(s)
for i in range(2**(n-1)):
  #print(i)
  f = ''
  b = format(i, 'b').zfill(n-1) + '0'
  #print(b)
  for j in range(n):
    f += s[j]
    if b[j] == '0':
      f += '-'
    else:
      f += '+'
  f = f[:len(f)-1]
  #print(f)
  if eval(f) == 7:
    print(f+'=7')
    break
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=input().split()
N,K=int(N),int(K)
A=input().split()

A=[int(N) for N in A]

prev=0
foll=0
for i in range(N):
    print('Ai',A[i])
    for j in range(i):
        if A[i]>A[j]:
            prev+=1
    for j in range(i,N):
        if A[i]>A[j]:
            foll+=1
    print(foll)

a = (K*(K-1)//2)%(10**9+7)
b = foll*K%(10**9+7)
u= (prev+foll)*a+b
print(u%(10**9+7))

Fixed, accepted submission:
N,K=input().split()
N,K=int(N),int(K)
A=input().split()

A=[int(N) for N in A]

prev=0
foll=0
for i in range(N):
    for j in range(i):
        if A[i]>A[j]:
            prev+=1
    for j in range(i,N):
        if A[i]>A[j]:
            foll+=1

a = (K*(K-1)//2)%(10**9+7)
b = foll*K%(10**9+7)
u= (prev+foll)*a+b
print(u%(10**9+7))
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):
A, B, C = map(int, input().split())

if (A ==7) and (B==C==5):
    print("Yes")
elif (A==C==5) and (B==7):
    print("Yes")
elif (A==B==5) and (C==7):
    print("Yes")
else:
    print("NO")

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

if A==7:
    if B==5:
        if C==5:
            print("YES")
        else:
            print("NO")
    else:
        print("NO")
elif A==5:
    if B==7:
        if C==5:
            print("YES")
        else:
            print("NO")
    elif B==5:
        if C==7:
            print("YES")
        else:
            print("NO")
    else:
        print("NO")
else:
    print("NO")
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):
a = tuple(map(int, input().split()))

a5 = 0
a7 = 0
b = 0
i = 0
while i < 3:
    if a[i] == 5:
        a5 += 1
    elif a[i] == 7:
        a7 += 1
    i += 1

if a5 == 2 and a7 == 1:
    print("Yes")
else:
    print("No")


Fixed, accepted submission:
a = tuple(map(int, input().split()))

a5 = 0
a7 = 0
b = 0
i = 0
while i < 3:
    if a[i] == 5:
        a5 += 1
    elif a[i] == 7:
        a7 += 1
    i += 1

if a5 == 2 and a7 == 1:
    print("YES")
else:
    print("NO")
Problem:
The ICPC committee would like to have its meeting as soon as possible to address every little issue of the next contest. However, members of the committee are so busy maniacally developing (possibly useless) programs that it is very difficult to arrange their schedules for the meeting. So, in order to settle the meeting date, the chairperson requested every member to send back a list of convenient dates by E-mail. Your mission is to help the chairperson, who is now dedicated to other issues of the contest, by writing a program that chooses the best date from the submitted lists. Your program should find the date convenient for the most members. If there is more than one such day, the earliest is the best.

Incorrect submission (verdict: Wrong Answer):
while True:
    n, q = map(int, input().split())
    if n == 0: break
    ls = []
    for i in range(n):
        ls.append(list(map(int, input().split())))

    for i in range(1, 101):
        count = 0
        for j in range(len(ls)):
            if i in ls[j][1:]:
                count += 1
                if count == q:
                    break
        else: continue
        print(i)
        break
    else:
        print(0)

Fixed, accepted submission:
while True:
    n, q = map(int, input().split())
    if n == 0: break
    ls = []
    for i in range(n):
        ls.append(list(map(int, input().split())))

    max_q = 0
    date = 0
    for i in range(1, 101):
        count = 0
        for j in range(len(ls)):
            if i in ls[j][1:]:
                count += 1
                if count > max_q:
                    max_q = count
                    date = i
    if max_q < q:
        print(0)
    else:
        print(date)
Problem:
Takahashi loves numbers divisible by 2.  You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.  Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.  For example,    * 6 can be divided by 2 once: 6 -> 3.   * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.   * 3 can be divided by 2 zero times.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
ans = 0
while True:
    if n < pow(2, ans + 1):
        break
    else:
        ans += 1
print(ans)


Fixed, accepted submission:
n = int(input())
ans = 0
while True:
    if n < pow(2, ans + 1):
        break
    else:
        ans += 1
print(pow(2, ans))
Problem:
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i.  Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.

Incorrect submission (verdict: Wrong Answer):
print(list(input().split()).count("1"))

Fixed, accepted submission:
print(list(input()).count("1"))
Problem:
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".

Incorrect submission (verdict: Wrong Answer):
from sys import stdin

a, b, c = (int(n) for n in stdin.readline().rsplit())

answer = "YES" if a < b < c else "NO"

print(answer)



Fixed, accepted submission:
from sys import stdin

a, b, c = (int(n) for n in stdin.readline().rsplit())

answer = "Yes" if a < b < c else "No"

print(answer)
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 detect_odd_and_even(a, b):
    if (a * b % 2 == 0):
        return 'Odd'
    else:
        return 'Even'

Fixed, accepted submission:
def detect_odd_and_even(a, b):
    res = a * b
    if (res % 2 == 0):
        return 'Even'
    else:
        return 'Odd'

if __name__ == '__main__':
    a, b = input().split()
    print(detect_odd_and_even(int(a), int(b)))
Problem:
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers.  You are given an integer N. Find the N-th Lucas number.  Here, the i-th Lucas number L_i is defined as follows:    * L_0=2   * L_1=1   * L_i=L_{i-1}+L_{i-2} (i≥2)

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

answer = []
for p in range(2, 10):

    for i in range(1, 32):
        if i**p > x:
            answer.append((i-1)**p)
            break
        elif pow(i, p) == x:
            answer.append(i**p)
            break

print(answer)
print(max(answer))

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

L = []
L.append(2)
L.append(1)
if N == 1:
    print(L[1])
else:
    for i in range(2, N+1):
        L.append(L[i-1] + L[i-2])
    print(L[N])
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, B, C, X = [int(input()) for i in range(4)]

res = 0
for a in range(A):
	for b in range(B):
		for c in range(C):
			total = 500 * a + 100 * b + 50 * c
			if total == X:
				res += 1
print(res)

Fixed, accepted submission:
A, B, C, X = [int(input()) for i in range(4)]

res = 0
for a in range(A + 1):
	for b in range(B + 1):
		for c in range(C + 1):
			total = 500 * a + 100 * b + 50 * c
			if total == X:
				res += 1
print(res)
Problem:
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it.  The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).  Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer.  If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

Incorrect submission (verdict: Wrong Answer):
import numpy

N = int(input())
X0 = N / 1.08
X1 = (N + 1) / 1.08

if numpy.floor(X1) - numpy.floor(X0) == 0:
    print(':(')
else:
    print(int(numpy.floor(X0) + 1))

Fixed, accepted submission:
import numpy

N = int(input())
X0 = N / 1.08
X1 = (N + 1) / 1.08

if numpy.ceil(X1) - numpy.ceil(X0) == 0:
    print(':(')
else:
    print(int(numpy.ceil(X0)))
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())
S=[input() for _ in range(1,N+1)]

C0=0
C1=0
C2=0
C3=0

for s in S:
    if s=='AC':
        C0+=1
    elif s=='WA':
        C1+=1
    elif s=='TLE':
        C2+=1
    elif s=='RE':
        C3+=1
print('AC x'+str(C0))
print('WA x'+str(C1))
print('TLE x'+str(C2))
print('RE x'+str(C3))


Fixed, accepted submission:
N = int(input())
s = [input() for i in range(N)]
for v in ['AC', 'WA', 'TLE', 'RE']:
    print('{0} x {1}'.format(v, s.count(v)))
Problem:
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}.  Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Incorrect submission (verdict: Wrong Answer):
for i in range(3):
  print(input()[i])
  

Fixed, accepted submission:
s=''
for i in range(3):
  temp=input()
  s=s+temp[i]
print(s)