TnT/CodeNet4Repair huggingface dataset:
Problem:
Snuke has N dogs and M monkeys. He wants them to line up in a row.  As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys.  How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.

Incorrect submission (verdict: Wrong Answer):
N,M = map(int,input().split())
MOD = 10 ** 9 + 7

if abs(N-M) >= 2:
  print(0)
  exit()
  
N,M = max(N,M),min(N,M)
fctr_M = 1
for i in range(1,M+1):
  fctr_M *= i
  fctr_M %= MOD

ans = fctr_M * fctr_M
ans %= MOD
ans *= N
if N == M:
  ans *= 2
  ans %= MOD
else:
  ans *= N
  ans %= MOD

print(ans)
  
  

Fixed, accepted submission:
N,M = map(int,input().split())
MOD = 10 ** 9 + 7

if abs(N-M) >= 2:
  print(0)
  exit()
  
N,M = max(N,M),min(N,M)
fctr_M = 1
for i in range(1,M+1):
  fctr_M *= i
  fctr_M %= MOD

ans = fctr_M * fctr_M
ans %= MOD
if N == M:
  ans *= 2
  ans %= MOD
else:
  ans *= N
  ans %= MOD

print(ans)
Problem:
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?

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

print ( 24 - m )

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

print ( 48 - m )
Problem:
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i.  The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere.  When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0.  What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?

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


ans = sum(A) - min(A)

Fixed, accepted submission:
N = int(input())
A = sorted(list(map(int, input().split())), reverse=True)

ans = -A[0]
for i in range(N):
    ans += A[i//2]
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 = map(int, input().split())
flag = None

for i in range(A, B*A):
    if C == i % B:
        flag = True

if flag:
    print('Yes')
else:
    print('No')


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

for i in range(B):
    if C == i * A % B:
        flag = True

if flag:
    print('YES')
else:
    print('NO')
Problem:
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.  Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)  For example, when n=7, it becomes 0 after two operations, as follows:    * \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.   * \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.  You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).

Incorrect submission (verdict: Wrong Answer):
import sys
input = sys.stdin.buffer.readline

def binToInt(X):
    ret = 0
    for s in X:
        ret <<= 1
        if s == '1':
            ret += 1
    return ret

def cnt(n):
    ret = 0
    while n > 0:
        n, r = divmod(n, 2)
        ret += r
    return ret

def f(n):
    ret = 0
    while n > 0:
        ret += 1
        n %= cnt(n)
    return ret

N = int(input())
X = binToInt(input().decode())

for i in range(N)[::-1]:
    print(f(X ^ (1 << i)))


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

cnt = X.count('1')
if cnt == 1:
    for x in X[:-1]:
        if x == '0':
            print(1)
        else:
            print(0)
    if X[-1] == '1':
        print(0)
    else:
        print(2)
    exit()

M = 0
for x in X:
    M <<= 1
    if x == '1':
        M += 1

def solve(n):
    ret = 1
    while n > 0:
        ret += 1
        n %= bin(n).count('1')
    return ret

ans = []
p = M % (cnt + 1)
m = M % (cnt - 1)

for i in range(N):
    if X[i] == '0':
        ans.append(solve((p + pow(2, N - i - 1, cnt + 1)) % (cnt + 1)))
    else:
        ans.append(solve((m - pow(2, N - i - 1, cnt - 1)) % (cnt - 1)))
print(*ans, sep='\n')
Problem:
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it.  They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card.  Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.

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

CW = 0

for s in S:
    if s == '.' : CW += 1
    
CB = 0
ans = 2**30
for s in S:
    if s == '.' : CW -= 1
    if s == '#' : CB += 1
    ans = min(ans,CB+CW)
    
print(ans)
    


Fixed, accepted submission:
import numpy as np
N = int(input())
a = list(map(int,input().split()))
tmp = a[0]
S = sum(a)
ans = np.abs(S-2*tmp)
for i in range(1,N-1):
    tmp += a[i]
    ans = min(ans,np.abs(S-2*tmp))
    
print(ans)
Problem:
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times:    * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n.  What is the largest possible sum of the integers written on the blackboard after K operations?

Incorrect submission (verdict: Wrong Answer):
a = [int(i) for i in input().split()]
k = int(input())
a.sort()
print(a[len(a)-1]**8+sum(a)-a[len(a)-1])

Fixed, accepted submission:
a = [int(i) for i in input().split()]
k = int(input())
a.sort()
print(a[len(a)-1]*2**k+sum(a)-a[len(a)-1])
Problem:
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive.  Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.

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

colors = input().split()

bef = 0
count = 0
colors.sort()

for c in colors:
    print(c)
    if bef == 0:
        count += 1
        bef = c
    else:
        if c > bef:
            count += 1
            bef = c

print(count)


Fixed, accepted submission:
# coding: utf-8

colors = input().split()

bef = 0
count = 0
colors.sort()

for c in colors:
    if bef == 0:
        count += 1
        bef = c
    else:
        if c > bef:
            count += 1
            bef = c

print(count)
Problem:
百人一首の札を使った遊戯の１つに、「坊主めくり」というものがあります。絵札だけを使う簡単な遊戯なので広く楽しまれています。きまりには様々な派生型がありますが、ここで考える坊主めくりはN人の参加者で、以下のようなルールで行います。    * ６４枚の「男」、１５枚の「坊主」、２１枚の「姫」、計１００枚の札を使う。   * 絵が見えないように札を裏がえしにしてよく混ぜ、「札山」をつくる。   * 参加者の一人目から順番に１枚ずつ札山の札を引く。N人目の次は、また一人目から繰り返す。      * 引いた札が男なら、引いた人はその札を手に入れる。     * 引いた札が坊主なら、引いた人はその札を含め、持っている札をすべて「場」に出す。     * 引いた札が姫なら、引いた人はその札を含め、場にある札をすべて手に入れる。   * 札山の札がなくなったら終了で、一番多くの札を持っている人の勝ち。  参加者数と札山に積まれた札の順番が与えられたとき、遊戯が終了した時点で各参加者が持っている札数を昇順で並べたものと、場に残っている札数を出力するプログラムを作成してください。

Incorrect submission (verdict: Wrong Answer):
def ans(n,s):
    p = [[] for i in range(n+1)]
    i = 0
    while True:
        if s == []:
            break
        c = s.pop(0)
        if c == 'M':
            if p[i] == []:
                p[i] = [c]
            else:
                p[i].append(c)
        elif c == 'L':
            if p[i] == []:
                p[i] = [c]
            else:
                p[i].append(c)
            if p[n] != []:
                p[i] += p[n]
            p[n] = []
        elif c == 'S':
            if p[n] == []:
                p[n] = [c]
            else:
                p[n].append(c)
            if p[i] != []:
                p[n] += p[i]
            p[i] = []
        
        i = (i+1) % n
    return p

while True:
    n = int(input())

    if n == 0:
        break
    s = list(input())
    o = ans(n,s)
    oo= list(map(lambda x: str(len(x)),o))
    print(" ".join(oo))

Fixed, accepted submission:
def ans(n,s):
    p = [[] for i in range(n)]
    h = []
    i = 0
    while True:
        if s == []:
            break
        c = s.pop(0)
        if c == 'M':
            if p[i] == []:
                p[i] = [c]
            else:
                p[i].append(c)
        elif c == 'L':
            if p[i] == []:
                p[i] = [c]
            else:
                p[i].append(c)
            if h != []:
                p[i] += h
            h = []
        elif c == 'S':
            if h == []:
                h = [c]
            else:
                h.append(c)
            if p[i] != []:
                h += p[i]
            p[i] = []
        
        i = (i+1) % n
    pp = list(sorted(map(len,p)))
    hh = len(h)
    pp.append(hh)
    
    return " ".join(map(str,pp))

while True:
    n = int(input())
    if n == 0:
        break
    s = list(input())
    o = ans(n,s)
    print(o)
Problem:
Write a program which calculates the cube of a given integer x.

Incorrect submission (verdict: Wrong Answer):
def f(x):
    return x*x*x

Fixed, accepted submission:
def cubic(x):
	return x*x*x

print(cubic(int(input())))
Problem:
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.  The person numbered i says:    * "In front of me, exactly A_i people are wearing hats with the same color as mine."  Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.  Since the count can be enormous, compute it modulo 1000000007.

Incorrect submission (verdict: Wrong Answer):
print(0)

Fixed, accepted submission:
#%%
n = int(input())
s = list(map(int, input().split()))
inf =  1000000007

a, b, c = 0, 0, 0
ans = 1

for i in range(n):
	tmp = [a, b, c].count(s[i])
	ans *= tmp
	ans %= 1000000007
	if s[i] == a:
		a += 1
	elif s[i] == b:
		b += 1
	else:
		c += 1
	

print(ans)
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):
def main():
    N = int(input())
    AC = 0
    WA = 0
    TLE = 0
    RE = 0
    for i in range(N):
        St = input()
        if St == 'AC':
            AC += 1
        elif St == 'WA':
            WA += 1
        elif St == 'TLE':
            TLE += 1
        elif St == 'RE':
            RE += 1
    print('AC x ' + str(AC))
    print('AC x ' + str(WA))
    print('AC x ' + str(TLE))
    print('AC x ' + str(RE))
main()

Fixed, accepted submission:
def main():
    N = int(input())
    AC = 0
    WA = 0
    TLE = 0
    RE = 0
    for i in range(N):
        St = input()
        if St == 'AC':
            AC += 1
        elif St == 'WA':
            WA += 1
        elif St == 'TLE':
            TLE += 1
        elif St == 'RE':
            RE += 1
        else:
            Er += 1
    print('AC x', AC)
    print('WA x', WA)
    print('TLE x', TLE)
    print('RE x', RE)
main()
Problem:
Write a program which reads an integer n and prints the factorial of n. You can assume that n ≤ 20\.

Incorrect submission (verdict: Wrong Answer):
a=1
for i in range(1,int(input())+1):a*i
print(a)


Fixed, accepted submission:
def f(n):
 if n==1:return 1
 return f(n-1)*n
print(f(int(input())))
Problem:
We have N sticks with negligible thickness. The length of the i-th stick is A_i.  Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle.

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

if len(a) > len(A)-2:
    print(0)
    exit()

A.sort(reverse=True)
ans = []

for i in range(len(A)-1):
    if A[i] !=A[i+1]:
        continue
    else:
        ans.append(A[i])
        i += 1

z = list(set(A))
z.reverse()
print(z[0]*z[1])



Fixed, accepted submission:
from collections import Counter

N = int(input())
A = list(map(int,input().split()))
c = Counter(A)
sticks = []
for i in c:
    s = c[i]
    if s >= 4:
        sticks.append(i)
    if s >= 2:
        sticks.append(i)
if len(sticks) < 2:
    print(0)
else:
    sticks.sort(reverse=True)
    print(sticks[0]*sticks[1])
Problem:
The goal of the matrix-chain multiplication problem is to find the most efficient way to multiply given $n$ matrices $M_1, M_2, M_3,...,M_n$.  Write a program which reads dimensions of $M_i$, and finds the minimum number of scalar multiplications to compute the maxrix-chain multiplication $M_1M_2...M_n$.

Incorrect submission (verdict: Wrong Answer):
import sys

p=[0 for i in range (101)]
m=[[0 for i in range(100)]for i in range(100)]

def MatrixChain(n) : 
	for i in range(1,n+1) : m[i][i] = 0;
	for l in range(2,n+1) :
		for i in range(1,n-l+2) : 
			j = i + l - 1
			m[i][j] = float('Infinity')
			for k in range(i,j) :
				q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j]
				m[i][j] = m[i][j] if m[i][j] <= q else q

n = int(input())
for i in range(n) :
	input_num = input()
	input_num = input_num.split(' ')
	p[i] = int(input_num[0][0])
	p[i+1] = int(input_num[1])

MatrixChain(n)
print(m[1][n])

Fixed, accepted submission:
import sys

p=[0 for i in range (101)]
m=[[0 for i in range(100)]for i in range(100)]

def MatrixChain(n) : 
	for i in range(1,n+1) : m[i][i] = 0;
	for l in range(2,n+1) :
		for i in range(1,n-l+2) : 
			j = i + l - 1
			m[i][j] = float('Infinity')
			for k in range(i,j) :
				q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j]
				m[i][j] = m[i][j] if m[i][j] <= q else q

n = int(input())
for i in range(n) :
	p[i], p[i+1] = map(int, input().split())

MatrixChain(n)
print(m[1][n])
Problem:
Takahashi is a user of a site that hosts programming contests.   When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows:    * Let the current rating of the user be a.   * Suppose that the performance of the user in the contest is b.   * Then, the new rating of the user will be the avarage of a and b.  For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000.  Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest.   Find the performance required to achieve it.

Incorrect submission (verdict: Wrong Answer):
x = int(input())
y = int(input())
print(y**2-x)


Fixed, accepted submission:
x = int(input())
y = int(input())
print(y*2-x)
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):
def solve():
    A, B, C, D = input()
    op = '+-'
    
    for op1 in op:
        for op2 in op:
            for op3 in op:
                left_hand = A + op1 + B + op2 + C + op3 + D
                if eval(left_hand) == 7:
                    return left_hand + '=7'

solve()

Fixed, accepted submission:
def solve():
    A, B, C, D = input()
    op = '+-'
    
    for op1 in op:
        for op2 in op:
            for op3 in op:
                left_hand = A + op1 + B + op2 + C + op3 + D
                if eval(left_hand) == 7:
                    return left_hand + '=7'

print(solve())
Problem:
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

Incorrect submission (verdict: Wrong Answer):
a,b = map(int, input().split())
x = (a+b)/2
print(round(x+1)/2)

Fixed, accepted submission:
import math

a,b = map(int, input().split())
x = (a+b)/2
print(math.ceil(x))
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[-1] == B[0] and B[-1] == C[0]:
    ans = "YES"
else:
    print('NO')


Fixed, accepted submission:
A, B, C = input().split()
if A[-1] == B[0] and B[-1] == C[0]:
    ans = "YES"
    print(ans)
else:
    print('NO')
Problem:
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.

Incorrect submission (verdict: Wrong Answer):
a, b = input().split()
c = int(a + b)
t = c ** 0.5
d = int(t) ** 2
print(c, d)
if c == d:
    ans = 'Yes'
else:
    ans = 'No'
print(ans)


Fixed, accepted submission:
a, b = input().split()
c = int(a + b)
t = c ** 0.5
d = int(t) ** 2
if c == d:
    ans = 'Yes'
else:
    ans = 'No'
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):
s = input()
opl = ['+', '-']

flag = [0,0,0]
for i in range(0, 0b111+1):
    for j in range(3):
        #print(i)
        flag[j] = (i>>(j+1) & 1)
        ans = s[0] + opl[flag[0]] + s[1] + opl[flag[1]] + s[2] + opl[flag[2]] + s[3]
        if(eval(ans) == 7):
            print(s[0] + opl[flag[0]] + s[1] + opl[flag[1]] + s[2] + opl[flag[2]] + s[3])
            break
    else:
        continue
    break
            
        


Fixed, accepted submission:
s = input()
opl = ['+', '-']

flag = [0,0,0]
for i in range(0, 0b111+1):
    for j in range(3):
        #print(i)
        flag[j] = (i>>(j) & 1)
        ans = s[0] + opl[flag[0]] + s[1] + opl[flag[1]] + s[2] + opl[flag[2]] + s[3]
        if(eval(ans) == 7):
            print(ans + '=7')
            break
    else:
        continue
    break
Problem:
You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
lst = [int(input()) for _ in range(n)]
lst.sort()
print(lst)
if n%2==0:
    lower = lst[:n//2]
    upper = lst[n//2:]
    x = 2*sum(upper) - upper[0]
    y = 2*sum(lower) - lower[-1]
    print(x-y)
else:
    lower = lst[:n//2]
    upper = lst[n//2+1:]
    mid = lst[n//2]
    x = 2*sum(upper) - 2*sum(lower) - mid + lower[-1]
    y = 2*sum(upper) - 2*sum(lower) + mid - upper[0]
    print(max(x,y))

Fixed, accepted submission:
n = int(input())
lst = [int(input()) for _ in range(n)]
lst.sort()
if n%2==0:
    lower = lst[:n//2]
    upper = lst[n//2:]
    x = 2*sum(upper) - upper[0]
    y = 2*sum(lower) - lower[-1]
    print(x-y)
else:
    lower = lst[:n//2]
    upper = lst[n//2+1:]
    mid = lst[n//2]
    x = 2*sum(upper) - 2*sum(lower) - mid + lower[-1]
    y = 2*sum(upper) - 2*sum(lower) + mid - upper[0]
    print(max(x,y))
Problem:
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these.  According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.

Incorrect submission (verdict: Wrong Answer):
a,sums= list(map(int,input().split()))
print(a)
yuki = sums//10000
taishi = (sums%10000) //5000
noguchi = (sums%5000) //1000
while a<yuki+taishi+noguchi:
  if a%(yuki+taishi+noguchi) == 9:
    yuki-=1
    noguchi+=10
  if a%(yuki+taishi+noguchi) == 4:
    taishi-=1
    noguchi+=5
  else:
    yuki-=1
    taishi+=2
print(yuki,taishi,noguchi)

Fixed, accepted submission:
a,sums= list(map(int,input().split()))
yuki = sums//10000
taishi = (sums%10000) //5000
noguchi = (sums%5000) //1000
while a>yuki+taishi+noguchi:
  if ((a-(yuki+taishi+noguchi))%4==0)&(taishi>0):
    taishi-=1
    noguchi+=5
  elif yuki>0:
    yuki-=1
    taishi+=2
  else:
    break
if a==yuki+taishi+noguchi:
  print(yuki,taishi,noguchi)
else:
  print("-1 -1 -1")
Problem:
An elementary school student Takahashi has come to a variety store.  He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it?  Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

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

Fixed, accepted submission:
A, B, C = map(int, input().split())
if A + B >= C:
  print('Yes')
else:
  print('No')