TnT/CodeNet4Repair huggingface dataset:
Problem:
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.   The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i.  An edge whose removal disconnects the graph is called a _bridge_.   Find the number of the edges that are bridges among the M edges.

Incorrect submission (verdict: Wrong Answer):
n,m=map(int,input().split())
graph=[[False]*n for i in range(n)] 
node=[None for i in range(m)]
visited=[False]*n 
for i in range(m):
    a,b=map(int,input().split())
    graph[a-1][b-1],graph[b-1][a-1]=True,True
    node[i]=[a-1,b-1]
def dfs(v):
    visited[v]=True
    for i in range(n):
        if graph[v][i]==False:continue
        if visited[i]==True:continue
        dfs(i)
ans=0
for i in range(m):
    graph[node[i][0]][node[i][1]],graph[node[i][1]][node[i][0]]=False,False 
    for j in range(n):visited[j]=False
    dfs(0)
    for h in range(n):
        if visited[h]==False:
            ans+=1
            break
print(ans)

Fixed, accepted submission:
n,m=map(int,input().split())
graph=[[False]*n for i in range(n)] 
node=[None for i in range(m)]
visited=[False]*n 
for i in range(m):
    a,b=map(int,input().split())
    graph[a-1][b-1],graph[b-1][a-1]=True,True
    node[i]=[a-1,b-1]
def dfs(v):
    visited[v]=True
    for i in range(n):
        if graph[v][i]==False:continue
        if visited[i]==True:continue
        dfs(i)
ans=0
for i in range(m):
    graph[node[i][0]][node[i][1]],graph[node[i][1]][node[i][0]]=False,False 
    for j in range(n):visited[j]=False
    dfs(0)
    for h in range(n):
        if visited[h]==False:
            ans+=1
            break
    graph[node[i][0]][node[i][1]],graph[node[i][1]][node[i][0]]=True,True
print(ans)
Problem:
Write a program which reads two integers a and b, and calculates the following values:    * a ÷ b: d (in integer)   * remainder of a ÷ b: r (in integer)   * a ÷ b: f (in real number)

Incorrect submission (verdict: Wrong Answer):
values = input()
a, b = [int(x) for x in values.split()]

print(a // b)
print(a % b)
print(a / b)

Fixed, accepted submission:
values = input()
a, b = [int(x) for x in values.split()]

print('{0} {1} {2:.5f}'.format(a // b, a % b, a / b))
Problem:
We have a deck consisting of N cards. Each card has an integer written on it. The integer on the i-th card from the top is a_i.  Two people X and Y will play a game using this deck. Initially, X has a card with Z written on it in his hand, and Y has a card with W written on it in his hand. Then, starting from X, they will alternately perform the following action:    * Draw some number of cards from the top of the deck. Then, discard the card in his hand and keep the last drawn card instead. Here, at least one card must be drawn.  The game ends when there is no more card in the deck. The score of the game is the absolute difference of the integers written on the cards in the two players' hand.  X will play the game so that the score will be maximized, and Y will play the game so that the score will be minimized. What will be the score of the game?

Incorrect submission (verdict: Wrong Answer):
N, Z, W = map(int, input().split())
a = list(map(int, input().split()))
a = a[::-1]
print(a)
if N == 1:
    print(a[0]-W)
elif abs(a[0]-a[1]) > abs(a[0] - W):
    print(abs(a[0]-a[1]))
else:
    print(abs(a[0] - W))

Fixed, accepted submission:
N, Z, W = map(int, input().split())
a = list(map(int, input().split()))
a = a[::-1]
if N == 1:
    print(abs(a[0]-W))
elif abs(a[0]-a[1]) > abs(a[0] - W):
    print(abs(a[0]-a[1]))
else:
    print(abs(a[0] - W))
Problem:
Write a program which performs a sequence of commands to a given string $str$. The command is one of:    * print a b: print from the a-th character to the b-th character of $str$   * reverse a b: reverse from the a-th character to the b-th character of $str$   * replace a b p: replace from the a-th character to the b-th character of $str$ with p  Note that the indices of $str$ start with 0.

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

s = list(input())
q = int(input())

for i in range(q):
    n = list(input().split())
    if 'print' in n:
        for i in range(int(n[1]), int(n[2])):
            print(str(s[i]), end='')
        print()
    
    elif 'reverse' in n:
        for i in range(int(int(n[2]) / 2)):
            s[int(n[1]) + i], s[int(n[2]) - i] = s[int(n[2]) - i], s[int(n[1]) + i]
    
    elif 'replace' in n:
        for i in range(int(n[1]), int(n[2]) + 1):
            s.pop(int(n[1]))
    
        count  = int(n[1])
    
        for j in list(n[3]):
            s.insert(count, j)
            count += 1


Fixed, accepted submission:
# -*- coding: utf-8 -*-

import math

s = list(input())
q = int(input())

for i in range(q):
    n = list(input().split())
    if 'print' in n:
        for j in range(int(n[1]), int(n[2]) + 1):
            print(str(s[j]), end='')
        print()

    elif 'reverse' in n:
        for k in range(math.floor((int(n[2]) - int(n[1]) + 1) / 2)):
            s[int(n[1]) + k], s[int(n[2]) - k] = s[int(n[2]) - k], s[int(n[1]) + k]

    elif 'replace' in n:
        for l in range(int(n[1]), int(n[2]) + 1):
            s.pop(int(n[1]))

        count  = int(n[1])

        for m in list(n[3]):
            s.insert(count, m)
            count += 1
Problem:
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied:    * For any two vertices painted in the same color, the distance between them is an even number.  Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.

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


n = int(input())
es = [[] for _ in range(n)]
for _ in range(n-1):
    u, v, w = map(int, input().split())
    u -= 1
    v -= 1
    es[u].append((v, w))
    es[v].append((u, w))
print(es)
clr = [None]*n
clr[0] = 0
dq = deque()
dq.append(0)
while dq:
    v = dq.popleft()
    for nv, w in es[v]:
        if clr[nv] is not None:
            continue
        clr[nv] = (clr[v] + w)%2
        dq.append(nv)
print(*clr, sep='\n')

Fixed, accepted submission:
import sys
sys.setrecursionlimit(10**5)
n = int(input())
t = {i:{} for i in range(n)}
ans = [-1]*n
for _ in range(n-1):
    u ,v, w = map(int, input().split())
    t[u-1][v-1] = w
    t[v-1][u-1] = w


def dfs(x):
    for y in t[x]:
        del t[y][x]
        ans[y] = ans[x] if t[x][y] % 2 == 0 else 1 - ans[x]
        if len(t[y]) > 0:
            dfs(y)


ans[0] = 0
dfs(0)
for i in range(n):
    print(ans[i])
Problem:
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K.  She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change).  However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money.  Find the amount of money that she will hand to the cashier.

Incorrect submission (verdict: Time Limit Exceeded):
N, K = map(int, input().split())
D = list(map(int, input().split()))

L=[]
for i in range(0, 10):
    if i not in D:
        L.append("i")    

i = N
while True:
    if set(list(str((i)))) <= set(L):
        print(i)
        break
    i += 1   

Fixed, accepted submission:
N, K = map(int, input().split())
D = list(map(int, input().split()))

L=[]
for i in range(0, 10):
    if i not in D:
        L.append(str(i))    

L.sort()

i = N
while True:
    if set(list(str((i)))) <= set(L):
        print(i)
        break
    i += 1
Problem:
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".

Incorrect submission (verdict: Wrong Answer):
clist = list(map(int,input().split()))
clist.sort()
ans = "No"
if min(clist) == 1 and max(clist) == 9 :
  clist.pop(0)
  clist.pop()
  if min(clist) == 4 and max(clist) == 7 :
    ans = "Yes"
  else:
    ans = "No"
else:
  ans = "No"

print(ans)

Fixed, accepted submission:
clist = list(map(int,input().split()))
clist.sort()
if clist == [1,4,7,9] :
  print("YES")
else:
  print("NO")
Problem:
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`.  You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.

Incorrect submission (verdict: Wrong Answer):
s = input()
a = len(s)-2
print(a)

Fixed, accepted submission:
s = input()
a = len(s)-2
print(s[0] + str(a) + s[len(s) - 1])
Problem:
テンアイスクリームという名前のアイスクリーム屋さんがあります。このお店では常に 10 種類のアイスクリームが店頭に並ぶようにしています。お店の店長は商品開発の参考にするために、アイスクリームの売れ具合を表すグラフを毎日作成しています。  そんな店長のために、あなたは各アイスクリームの販売数をグラフで表示するプログラムを作成することになりました。  一日に販売されるアイスクリームの総数と売れたアイスクリームの番号を入力とし、アイスクリームの種類ごとに販売した数だけ * (半角アスタリスク) を出力するプログラムを作成してください。ただし、アイスクリームの種類を 0 から 9 までの整数で表わします。また、販売個数がゼロのアイスクリームは、- (半角ハイフン) をひとつ出力します。

Incorrect submission (verdict: Wrong Answer):
while True:
    n = int(input())
    if n == 0:
        break
        
    d = {k: 0 for k in range(n)}
    for _ in range(n):
        x = int(input())
        d[x] += 1

    for k in range(n):
        print("*"*d[k] + "-"*(d[k]==0))

Fixed, accepted submission:
while True:
    n = int(input())
    if n == 0:
        break

    ice = [int(input()) for _ in range(n)]
    for i in range(10):
        nums = ice.count(i)
        print("*"*nums + "-"*(nums==0))
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):
from collections import deque

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

graph = [[] for _ in range(n + 1)]
for i in range(m):
    a = ab[i][0]
    b = ab[i][1]
    graph[a].append(b)
    graph[b].append(a)

inf = 100100100
ans = [[inf, inf] for _ in range(n + 1)]
ans[0] = [0, 0]
ans[1] = [0, 0]

q = deque([1])
while q:
    now = q.popleft()
    depth = ans[now][1]
    for x in graph[now]:
        if ans[x][0] == inf:
            q.append(x)
            if ans[x][1] > depth + 1:
                ans[x][0] = now
                ans[x][1] = depth + 1

for i in range(2, n + 1):
    print(ans[i][0])


Fixed, accepted submission:
from collections import deque

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

graph = [[] for _ in range(n + 1)]
for i in range(m):
    a = ab[i][0]
    b = ab[i][1]
    graph[a].append(b)
    graph[b].append(a)

inf = 100100100
ans = [[inf, inf] for _ in range(n + 1)]
ans[0] = [0, 0]
ans[1] = [0, 0]

q = deque([1])
while q:
    now = q.popleft()
    depth = ans[now][1]
    for x in graph[now]:
        if ans[x][0] == inf:
            q.append(x)
            if ans[x][1] > depth + 1:
                ans[x][0] = now
                ans[x][1] = depth + 1

print('Yes')
for i in range(2, n + 1):
    print(ans[i][0])
Problem:
E869120 found a chest which is likely to contain treasure.   However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.   He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`.  One more thing he found is a sheet of paper with the following facts written on it:    * Condition 1: The string S contains a string T as a contiguous substring.   * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.  Print the string S.   If such a string does not exist, print `UNRESTORABLE`.

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

ans = []
for i in range(len(S) - len(T) + 1):
    c = ""
    for j in range(len(T)):
        if S[i+j] != "?" and S[i+j] != T[j]:
            break
        c += T[j]
    if len(c) == len(T):
        print(S[:i], c, S[i+len(T):])
        ans.append((S[:i] + c + S[i+len(T):]).replace("?", "a"))
print(min(ans) if ans else "UNRESTORABLE")


Fixed, accepted submission:
S = input()
T = input()

ans = []
for i in range(len(S) - len(T) + 1):
    c = ""
    for j in range(len(T)):
        if S[i+j] != "?" and S[i+j] != T[j]:
            break
        c += T[j]
    if len(c) == len(T):
        ans.append((S[:i] + c + S[i+len(T):]).replace("?", "a"))
print(min(ans) if ans else "UNRESTORABLE")
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):
# coding: utf-8
# Your code here!
S = input()
flag = False
st_match = ["erase",'dream']
while len(S)>0:
    for st in st_match:
        if(S.startswith(st)):
            S=S[len(st):]
            if(S.startswith('er') and S.startswith('era')==False):
                S=S[2:]
            elif(S.startswith('r')):
                S=S[2:]
            flag=True
            break
    if(flag):
        flag=False
        continue
    else:
        break
if(len(S)==0):
    print('Yes')
else:
    print('NO')
    
        

Fixed, accepted submission:
def rev (s):
    _re=''.join(list(reversed(s)))
    return _re

name = ['dream','dreamer','eraser','erase']
S = input()
S_re = rev(S)
i=0
can = True
while i < len(S_re):
    for div in name:
        can2 = False
        if len(S_re[i:]) >= len(div):
           if S_re[i:i+len(div)]==rev(div) :
                i+=len(div)
                can2 = True
                break
    if not can2 :
        can = False
        break


if can :
    print('YES')
else :
    print('NO')
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):
h, w = [int(x) for x in input().split()]
s = "*" * (w + 2)
print(s)
for _ in range(h):
    print("*" + input() + "*")
print(s)

Fixed, accepted submission:
h, w = [int(x) for x in input().split()]
c = "#"
s = c * (w + 2)
print(s)
for _ in range(h):
    print(c + input() + c)
print(s)
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):
def C():
    A, B, C, nA, nB = map(int, input().split())
    minAB = min(nA, nB)
    # 3, 2 --> 1, 0
    ns = [nA - minAB, nB - minAB]
    print(ns)
    amounts = [
        A * nA + B * nB,
        minAB * 2 * C + ns[0] * A + ns[1] * B,
        max(nA, nB) * 2 * C
    ]
    return min(amounts)

print(C())

Fixed, accepted submission:
def C():
    A, B, C, nA, nB = map(int, input().split())
    minAB = min(nA, nB)
    # 3, 2 --> 1, 0
    ns = [nA - minAB, nB - minAB]
    amounts = [
        A * nA + B * nB,
        minAB * 2 * C + ns[0] * A + ns[1] * B,
        max(nA, nB) * 2 * C
    ]
    return min(amounts)

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

Incorrect submission (verdict: Wrong Answer):
s = input()
data =[]
for i in range(len(s)):
  data.append(s[i])
array = []
for i in range(1, len(data)):
  if data[i] == "0":
    array.append("0")
  elif data[i] == "1":
    array.append("1")
  elif (len(array) != 0 and data[i]== "B"):
    array.pop(-1)
  else:
    pass

mojiretu = ''

for x in array:
    mojiretu += x

print(mojiretu)
      
      

Fixed, accepted submission:
s = input()
data =[]
for i in range(len(s)):
  data.append(s[i])
array = []
for i in range(len(data)):
  if data[i] == "0":
    array.append("0")
  elif data[i] == "1":
    array.append("1")
  elif (len(array) != 0 and data[i]== "B"):
    array.pop(-1)
  else:
    pass

mojiretu = ''

for x in array:
    mojiretu += x

print(mojiretu)
Problem:
You are given a sequence of _n_ integers S and a sequence of different _q_ integers T. Write a program which outputs C, the number of integers in T which are also in the set S.

Incorrect submission (verdict: Wrong Answer):
def BinarySearch(A, c):
	left = 0
	right = len(A)-1
	while(left < right):
		mid = (left + right) // 2
		print(mid)
		
		if A[mid] == c:
			return mid
		elif (A[mid] < c):
			left = mid + 1
		else:
			right = mid 
	else:
		return 0
	
	
	

N = int(input())
S = list(map(int, input().split()))
Q = int(input())
T = list(map(int, input().split()))

ctr = 0

n = N

for i in range(Q):
	if BinarySearch(S, T[i]):
		ctr += 1

print(ctr)

Fixed, accepted submission:
def BinarySearch(A, c):
	left = 0
	right = len(A)
	while(left < right):
		mid = (left + right) // 2
		if A[mid] == c:
			return True
		elif (A[mid] < c):
			left = mid + 1
		else:
			right = mid 
	else:
		return False
	
	
	

N = int(input())
S = list(map(int, input().split()))
Q = int(input())
T = list(map(int, input().split()))

ctr = 0



for i in range(Q):
	if BinarySearch(S, T[i]):
		ctr += 1

print(ctr)
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 _ in range(4)]
print([500*i+100*j+50*k for i in range(a) for j in range(b) for k in range(c)].count(x))

Fixed, accepted submission:
a,b,c,x=[int(input()) for _ in range(4)]
print([500*i+100*j+50*k for i in range(a+1) for j in range(b+1) for k in range(c+1)].count(x))
Problem:
It is September 9 in Japan now.  You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?

Incorrect submission (verdict: Wrong Answer):
print('YNeos'[input().count('9')>0::2])

Fixed, accepted submission:
print('YNeos'[input().count('9')==0::2])
Problem:
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up.   Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1.   Here, both the x\- and y-coordinates before and after each movement must be integers.   He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy).   Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty).   Under this condition, find a shortest path for him.

Incorrect submission (verdict: Wrong Answer):
sx,sy,tx,ty = map(int,input().split())

yl = ty - sy
xl = tx - sx

ans = ''
ans += 'U'*yl
ans += 'R'*xl
ans += 'D'*yl
ans += 'L'*xl
ans += 'R'*(yl+1)
ans += 'R'*(xl+1)
ans += 'DR'
ans += 'D'*(yl+1)
ans += 'L'*(xl+1)
ans += 'U'

print(ans)

Fixed, accepted submission:
sx,sy,tx,ty = map(int,input().split())

yl = ty - sy
xl = tx - sx

ans = ''
ans += 'U'*yl
ans += 'R'*xl
ans += 'D'*yl
ans += 'L'*(xl+1)
ans += 'U'*(yl+1)
ans += 'R'*(xl+1)
ans += 'DR'
ans += 'D'*(yl+1)
ans += 'L'*(xl+1)
ans += 'U'

print(ans)
Problem:
We have a grid with H rows and W columns, where all the squares are initially white.  You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions:    * Choose one row, then paint all the squares in that row black.   * Choose one column, then paint all the squares in that column black.  At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.

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

count = n//h



bit = w - count

while count<n:
  if count>=n:
    print(count)
    break
  count += bit



Fixed, accepted submission:
h = int(input())
w = int(input())
n = int(input())

num = n//max(h,w) if (n//max(h,w)+1)==n else (n//max(h,w)+1)

if (n%max(h,w)==0):
  print((n//max(h,w)))
else:
  print((n//max(h,w)+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):
a,b,c=map(int,input().split())
ans=1
if a==b:
	ans+=1
if b==c:
	ans+=1
if a==c:
	ans+=1
if ans==4:
	ans=3
print(ans)


Fixed, accepted submission:
a,b,c=input().split()
if a!=b and b!=c and c!=a:
	print(3)
elif a!=b or b!=c or c!=a:
	print(2)
else:
	print(1)
Problem:
Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string.

Incorrect submission (verdict: Wrong Answer):
sentence=input()
print(sentence.upper())

Fixed, accepted submission:
print(input().swapcase())