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):
def main():
    from itertools import accumulate as ac
    n,c=map(int,input().split())
    xv=[list(map(int,input().split())) for _ in [0]*n]
    rmemo=[0]*n
    lmemo=[0]*n
    rmemo[0]=xv[0][1]-xv[0][0]
    lmemo[n-1]=xv[n-1][1]-(c-xv[n-1][0])
    for i in range(1,n):
        rmemo[i]=rmemo[i-1]+xv[i][1]-xv[i][0]+xv[i-1][0]
    for i in range(n-2,-1,-1):
        lmemo[i]=lmemo[i+1]+xv[i][1]+xv[i][0]-xv[i+1][0]

    temp=rmemo[0]
    rmemo2=[temp]*(n+1)
    rmemo2[-1]=0
    temp=lmemo[-1]
    lmemo2=[temp]*(n+1)
    lmemo2[-1]=0
    for i in range(1,n):
        rmemo2[i]=max(rmemo[i],rmemo2[i-1])
    for i in range(n-2,-1,-1):
        lmemo2[i]=max(lmemo[i],lmemo2[i+1])
    #print(rmemo2)
    #print(lmemo)
    ans=max(rmemo+lmemo+[0])
    for i in range(n):
        ans=max(ans,rmemo[i]+lmemo2[i+1]-xv[i][0])
        ans=max(ans,lmemo[i]+rmemo2[i+1]+xv[i][0]-c)
    print(ans)
main()

Fixed, accepted submission:
def main():
    from itertools import accumulate as ac
    n,c=map(int,input().split())
    xv=[list(map(int,input().split())) for _ in [0]*n]
    rmemo=[0]*n
    lmemo=[0]*n
    rmemo[0]=xv[0][1]-xv[0][0]
    lmemo[n-1]=xv[n-1][1]-(c-xv[n-1][0])
    for i in range(1,n):
        rmemo[i]=rmemo[i-1]+xv[i][1]-xv[i][0]+xv[i-1][0]
    for i in range(n-2,-1,-1):
        lmemo[i]=lmemo[i+1]+xv[i][1]+xv[i][0]-xv[i+1][0]
    temp=rmemo[0]
    rmemo2=[temp]*(n)
    temp=lmemo[-1]
    lmemo2=[temp]*(n+1)
    lmemo2[-1]=0
    for i in range(1,n):
        rmemo2[i]=max(rmemo[i],rmemo2[i-1])
    for i in range(n-2,-1,-1):
        lmemo2[i]=max(lmemo[i],lmemo2[i+1])
    rmemo2=[0]+rmemo2
    ans=max(rmemo+lmemo+[0])
    for i in range(n):
        ans=max(ans,rmemo[i]+lmemo2[i+1]-xv[i][0])
        ans=max(ans,lmemo[i]+rmemo2[i]+xv[i][0]-c)
    print(ans)
main()
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())
a1=min(h,w)
a2=10**6
a3=10**6
for i in range(1,h):
  mx=max(w*i,(w//2)*(h-i),((w+1)//2)*(h-i))
  mn=min(w*i,(w//2)*(h-i),((w+1)//2)*(h-i))
  a2=min(a2,mx-mn)
for j in range(1,w):
  mx=max(h*i,(h//2)*(w-i),((h+1)//2)*(w-i))
  mn=min(h*i,(h//2)*(w-i),((h+1)//2)*(w-i))
  a3=min(a3,mx-mn)
print(min(a1,a2,a3))


Fixed, accepted submission:
h,w=map(int,input().split())
a1=min(h*(w%3!=0),w*(h%3!=0))
a2=10**6
a3=10**6
for i in range(1,h):
  mx=max(w*i,(w//2)*(h-i),((w+1)//2)*(h-i))
  mn=min(w*i,(w//2)*(h-i),((w+1)//2)*(h-i))
  a2=min(a2,mx-mn)
for i in range(1,w):
  mx=max(h*i,(h//2)*(w-i),((h+1)//2)*(w-i))
  mn=min(h*i,(h//2)*(w-i),((h+1)//2)*(w-i))
  a3=min(a3,mx-mn)
print(min(a1,a2,a3))
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())
price = n % 1000
print(price)

Fixed, accepted submission:
n = int(input())
price = n % 1000
if price == 0:
  print(0)
else:
  print(1000 - price)
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):
N=int(input())
K=int(input())

ans=2**N
tmp=1
for i in range(2**N - 1):
  for j in range(N):
    if i>>j %2==1:
      tmp*=2
    else:
      tmp+=K
      
  ans=min(ans,tmp)
print(ans)

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

ans=2**N

for i in range(2**N):
  tmp=1
  for j in range(N):
    if (i>>j) %2==1:
      tmp*=2
    else:
      tmp+=K     
  ans=min(ans,tmp)
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):
import sys
read = sys.stdin.read
readline = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 8)
INF = float('inf')
MOD = 10 ** 9 + 7

def main():
    N = int(readline())
    ans = 1000-N
    print(ans)

if __name__ == '__main__':
    main()


Fixed, accepted submission:
import sys
read = sys.stdin.read
readline = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 8)
INF = float('inf')
MOD = 10 ** 9 + 7

def main():
    N = int(readline())
    ans = 1000-N%1000
    if ans ==1000:
        print(0)
    else:
        print(ans)

if __name__ == '__main__':
    main()
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):
def main():
    s = input()
    ans = 'no'
    if len(set(s)) == len(s):
        ans = 'yes'
    print(ans)

Fixed, accepted submission:
def main():
    s = input()
    ans = 'no'
    if len(set(s)) == len(s):
        ans = 'yes'
    print(ans)



if __name__ == '__main__':
    main()
Problem:
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room.  For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left.  Assume that initially no person lives in the building.

Incorrect submission (verdict: Wrong Answer):
roomsOfBuilding = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]

n = int(input())
for l in range(n):
    
    b, f, r, v = map(int, input().split())
    roomsOfBuilding[b - 1][f - 1][r - 1] += v

for building in roomsOfBuilding:
    for rooms in building:
        for room in rooms:
            print(room, end=' ')
        print('')
    print('#' * 20)



Fixed, accepted submission:
roomsOfBuilding = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]

n = int(input())
for l in range(n):
    
    b, f, r, v = map(int, input().split())
    roomsOfBuilding[b - 1][f - 1][r - 1] += v
counter = 0
for building in roomsOfBuilding:
    counter += 1
    for rooms in building:
        for room in rooms:
            print(' ' + str(room), end='')
        print('')
    if counter < 4:
        print('#' * 20)
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):
print("YNeos"[sorted(map(int,input().split()))!=[5,5,7]::2])

Fixed, accepted submission:
print("YNEOS"[sorted(map(int,input().split()))!=[5,5,7]::2])
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=list(map(int,list(input())))
for i in range(2**3):
    op=[0]*3
    for j in range(3):
        op[j]=i%2
        i//=2
    val=ABCD[0]
    for j in range(3):
        if op[j]==1:
            val+=ABCD[j+1]
        else:
            val-=ABCD[j+1]
    if val==7:
        ans=str(ABCD[0])
        for j in range(3):
            if op[j]==1:
                ans+="+"
            else:
                ans+="-"
            ans+=str(ABCD[j+1])
        print(ans)
        exit()

Fixed, accepted submission:
ABCD=list(map(int,list(input())))
for i in range(2**3):
    op=[0]*3
    for j in range(3):
        op[j]=i%2
        i//=2
    val=ABCD[0]
    for j in range(3):
        if op[j]==1:
            val+=ABCD[j+1]
        else:
            val-=ABCD[j+1]
    if val==7:
        ans=str(ABCD[0])
        for j in range(3):
            if op[j]==1:
                ans+="+"
            else:
                ans+="-"
            ans+=str(ABCD[j+1])
        print(ans+"=7")
        exit()
Problem:
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.

Incorrect submission (verdict: Wrong Answer):

if __name__ == "__main__":
    c = input()
    if c == "a" or c == "i" or c == "u" or c == "e" or c == "o":
        print("voewl")
    else:
        print("consonant")


Fixed, accepted submission:

if __name__ == "__main__":
    c = input()
    if c == "a" or c == "i" or c == "u" or c == "e" or c == "o":
        print("vowel")
    else:
        print("consonant")
Problem:
Takahashi is organizing a party.  At the party, each guest will receive one or more snack pieces.  Takahashi predicts that the number of guests at this party will be A or B.  Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.  We assume that a piece cannot be divided and distributed to multiple guests.

Incorrect submission (verdict: Wrong Answer):
import math
a,b=map(int,input().split())
print(math.gcd(a, b))

Fixed, accepted submission:
import math
a,b=map(int,input().split())
def lcm(x, y):
    return (x * y) // math.gcd(x, y)
print(lcm(a,b))
Problem:
You are given three strings A, B and C. Check whether they form a _word chain_.  More formally, determine whether both of the following are true:    * The last character in A and the initial character in B are the same.   * The last character in B and the initial character in C are the same.  If both are true, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
a, b, c = input().split()

if (a[len(a) - 1] == b[0]) and (b[len(b) - 1] == c[0]):
    print("Yes")
else:
    print("No")

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

if a[-1] == b[0] and b[-1] == c[0]:
    print("YES")
else:
    print("NO")
Problem:
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?

Incorrect submission (verdict: Wrong Answer):
a,b,c = map(int,input().split())

print(a//c-b//c)

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

if a%c == 0:
    print(b//c-a//c+1)
else:
    print(b//c-a//c)
Problem:
Today is August 24, one of the five Product Days in a year.  A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day):    * d_1 \geq 2   * d_{10} \geq 2   * d_1 \times d_{10} = m  Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D.  In Takahashi Calendar, how many Product Days does a year have?

Incorrect submission (verdict: Wrong Answer):
m,d=map(int,input().split())
x=int(d//10)
a,b,c=list(range(1,10)),list(range(1,x+1)),list(range(1,m+1))
d=sum([i*b for i in a],[])
e=len(set(c)&set(d))
print(e)

Fixed, accepted submission:
def main():
    import sys
    import bisect
    from itertools import accumulate

    m, d = map(int, input().split())
    if d < 22:
        print(0)
        return

    count = 0
    for i in range(22, d + 1):
        d1 = int(str(i)[1])
        d10 = int(str(i)[0])

        x = d1 * d10
        if 1 <= x <= m and d1 >= 2:
            count += 1
    print(count)


if __name__ == '__main__':
    main()
Problem:
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied:    * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds.  Find all favorite numbers and print the sum of those.

Incorrect submission (verdict: Wrong Answer):
import math

N = int(input())

total = 0


for i in range(1,int(math.sqrt(N))):
    amari = i
    if N % amari == 0:
        total += N//amari - 1
        #print(amari)

print(total)

Fixed, accepted submission:
import math

N = int(input())

total = 0
   
if N != 1:        
    for i in range(1,int(math.sqrt(N)+1)):
        amari = i
        if N % amari == 0:
            k = N // amari - 1
            if amari == N % k:
                total += N//amari - 1
        #print(amari)

print(total)
Problem:
In this problem, we use the 24-hour clock.  Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?

Incorrect submission (verdict: Wrong Answer):
t = list(input())
a = 0
b = 0

for i in range(len(t)):
    if i == 0 and t[i] == '?':
        t[i] = 'D'
    
    if 0 < i < len(t) - 1 and t[i] == '?':
        if t[i-1] == 'P':
            t[i] = 'D'
        else:
            t[i] = 'P'

    if i == len(t) -1 and t[i] == '?':
        t[i] = 'D'

print(''.join(t))

Fixed, accepted submission:
H1,M1,H2,M2,K = map(int,input().split())
A = (H1 * 60) + M1
B = (H2 * 60) + M2
C = B - A
print(C-K)
Problem:
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once.  Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.

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

for i in range(8):
    print(i, Ss[:i] + Ss[-7+i:])
    if Ss[:i] + Ss[-7+i:] == 'keyence':
        print('YES')
        break
else:
    print('NO')


Fixed, accepted submission:
Ss = input()

for i in range(8):
    if Ss[:i] + Ss[len(Ss)-7+i:] == 'keyence':
        print('YES')
        break
else:
    print('NO')
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):
import math

x = int(input())
r = 0.01
i = 0
s = 100

while True:
    i += 1
    s = int(s * (1 + r))
    if s > x:
        print(i)
        break

Fixed, accepted submission:
import math

x = int(input())
r = 0.01
i = 0
s = 100

while True:
    i += 1
    s = math.floor(s * (1 + r))
    if s >= x:
        print(i)
        break
Problem:
There are N children, numbered 1, 2, \ldots, N.  They have decided to share K candies among themselves. Here, for each i (1 \leq i \leq N), Child i must receive between 0 and a_i candies (inclusive). Also, no candies should be left over.  Find the number of ways for them to share candies, modulo 10^9 + 7. Here, two ways are said to be different when there exists a child who receives a different number of candies.

Incorrect submission (verdict: Wrong Answer):
def main():
    N, K = map(int, input().split())
    A = list(map(int, input().split()))
    u = [1] + [0] * K
    v = [1] + [0] * K
    s = 0
    for i in A:
        u, v = v, u
        s += i
        t = 0
        for k in range(0, i):
            t += u[k]
            v[k] = t
        for k in range(i, min(s, K) + 1):
            t += u[i]
            v[k] = t
            t -= u[k - i]
    print(v[K] % (10 ** 9 + 7))

main()


Fixed, accepted submission:
def main():
    N, K = map(int, input().split())
    A = map(int, input().split())
    u = [1] + [0] * K
    v = [1] + [0] * K
    s = 1
    for a in A:
        u, v = v, u
        s += a
        c = 0
        for i, t in enumerate(u[:a + 1]):
            c += t
            v[i] = c
        for i, t, tt in zip(range(a + 1, K + 1), u[a + 1:s], u):
            c += t - tt
            v[i] = c
    print(v[K] % (10 ** 9 + 7))

main()
Problem:
Two students of AtCoder Kindergarten are fighting over candy packs.  There are three candy packs, each of which contains a, b, and c candies, respectively.  Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.  Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

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

Fixed, accepted submission:
a, b, c = map(int,input().split())
print("Yes" if (a+b) == c or (b+c) == a or (c+a) == b else "No")
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())
n = max(x,y)
kar = a*x + b*y
for i in range(1,n+1):
    if kar > a*(max(0,x-i)) + b*(max(0,y-i)) +c*i:
        kar = a*(max(0,x-i)) + b*(max(0,y-i)) +2*c*i
print(kar)

Fixed, accepted submission:
a , b , c , x , y = map(int,input().split())
n = max(x,y)
kar = a*x + b*y
for i in range(1,n+1):
    if kar > a*(max(0,x-i)) + b*(max(0,y-i)) +2*c*i:
        kar = a*(max(0,x-i)) + b*(max(0,y-i)) +2*c*i
print(kar)
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 = map(int, [input() for i in range(4)])
cnt = 0
for a in range(a+1):
    for b in range(b+1):
        for c in range(c+1):
            if 500*a + 100*b + 50*c == x:
                cnt += 1
                print(cnt)

Fixed, accepted submission:
a, b, c, x = map(int, [input() for i in range(4)])
cnt = 0
for i in range(a+1):
    for j in range(b+1):
        for k in range(c+1):
            if 500*i + 100*j + 50*k == x:
                cnt += 1
print(cnt)
Problem:
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times:    * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

Incorrect submission (verdict: Wrong Answer):
s = input()
while(1):
    if s.endswith('dream'):
        s = s.rstrip('dream')
    elif s.endswith('dreamer'):
        s = s.rstrip('dreamer')
    elif s.endswith('eraser'):
        s = s.rstrip('eraser')
    elif s.endswith('erase'):
        s = s.rstrip('erase')
    else:
        print("NO")
        quit()
    if s == '':
        break
print("YES")

                   

Fixed, accepted submission:
s = input()
while(1):
    if s.endswith('dream'):
        s = s[:-5]
    elif s.endswith('dreamer'):
        s = s[:-7]
    elif s.endswith('eraser'):
        s = s[:-6]
    elif s.endswith('erase'):
        s = s[:-5]
    else:
        print("NO")
        quit()
    if s == '':
        break
print("YES")
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())/2
print((a+b)*h)

Fixed, accepted submission:
a=int(input())
b=int(input())
h=int(input())//2
print((a+b)*h)
Problem:
In programming, hexadecimal notation is often used.  In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.  In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.  When X and Y are seen as hexadecimal numbers, which is larger?

Incorrect submission (verdict: Wrong Answer):
x, y = input().split()
x = int(x, 16)
y = int(y, 16)
if x > y:
  print('<')
elif x < y:
  print('>')
else:
  print('=')

Fixed, accepted submission:
x, y = input().split()
x = int(x, 16)
y = int(y, 16)
if x > y:
  print('>')
elif x < y:
  print('<')
else:
  print('=')
Problem:
Write a program which reverses a given string str.

Incorrect submission (verdict: Wrong Answer):
print(input()[-1])


Fixed, accepted submission:
print(input()[::-1])
Problem:
半角アルファベット文字列からなる、1 行あたり 100 文字以内のデータがあります。いくつかの行は対称（左端から読んでも右端から読んでも同じ）です。このデータを読み込んで、その中の対称な文字列の数を出力するプログラムを作成してください。なお、１文字だけからなる行は対称であるとします。

Incorrect submission (verdict: Wrong Answer):
import sys
print(e[:-1]==e[-2::-1]for e in sys.stdin)


Fixed, accepted submission:
import sys
print(sum(e[:-1]==e[-2::-1]for e in sys.stdin))