TnT/CodeNet4Repair huggingface dataset:
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):
n,m = map(int,input().split())
a = [0 for _ in range(m)]
b = [0 for _ in range(m)]

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

ans = [0 for _ in range(n-1)]
looked = [1 for _ in range(n)]
queue = [0]
looked[0] = 0
can = [[] for _ in range(n)]

for i in range(m):
    can[a[i]].append(b[i])
    can[b[i]].append(a[i])

while(len(queue)):
    now = queue.pop(0)
    for i in can[now]:
        if looked[i]:
            queue.append(i)
            looked[i] = 0
            ans[i-1] = now+1
print("\n".join(list(map(str,ans))))

Fixed, accepted submission:
n,m = map(int,input().split())
a = [0 for _ in range(m)]
b = [0 for _ in range(m)]

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

ans = [-1 for _ in range(n-1)]
looked = [1 for _ in range(n)]
queue = [0]
looked[0] = 0
can = [[] for _ in range(n)]

for i in range(m):
    can[a[i]].append(b[i])
    can[b[i]].append(a[i])

while(len(queue)):
    now = queue.pop(0)
    for i in can[now]:
        if looked[i]:
            queue.append(i)
            looked[i] = 0
            ans[i-1] = now+1
for i in ans:
    if ans == -1:
        print("No")
        exit
print("Yes")
for i in ans:
    print(i)
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 = list(map(int,input().split()))
set =[]
s=int(N*(N+1)/2)
count = 0
for i in range(M):
    set.append(list(map(int,input().split())))

def BFS(L):
    start = L.pop(0)
    visited.append(start)
    for item in lis[start-1]:
        if (item not in visited) and (item not in L):
            L.append(item)
    if L:
        BFS(L)
    else:
        global check
        if sum(visited) == s:
            check = 1
        else:
            check = 0



for num in range(N):
    visited = []
    lis = [[] for i in range(N)]
    que = [1]
    for i,item in enumerate(set):
        if i == num:
            continue
        x,y = item
        lis[x-1].append(y)
        lis[y-1].append(x)
    check = -1
    BFS(que)
    print (check)
    if check == 0:
        count += 1
print (count)


Fixed, accepted submission:
N,M = list(map(int,input().split()))
lis_s =[]
count = 0
for _ in range(M):
  lis_s.append(tuple(map(int,input().split())))

graph = [[False for i in range(N+1)] for k in range(N+1)]

for tp in lis_s:
  graph[tp[0]][tp[1]] = True
  graph[tp[1]][tp[0]] = True


visited = [False]*(N+1)
bridge = False

def dfs(v):
  visited[v] = True
  for vs in range(1,N+1):
    if (graph[v][vs] == True) and (visited[vs] == False):
      dfs(vs)

ans = 0

for num in range(M):
  graph[lis_s[num][0]][lis_s[num][1]] = False
  graph[lis_s[num][1]][lis_s[num][0]] = False
  for i in range(N+1):
    visited[i] = False
  
  dfs(1)
  
  bridge = False
  for check in range(1,N+1):
    if visited[check] == False:
      bridge = True
  
  if bridge:
    ans += 1
  
  graph[lis_s[num][0]][lis_s[num][1]] = True 
  graph[lis_s[num][1]][lis_s[num][0]] = True
  
  
print(ans)
Problem:
Snuke stands on a number line. He has L ears, and he will walk along the line continuously under the following conditions:    * He never visits a point with coordinate less than 0, or a point with coordinate greater than L.   * He starts walking at a point with integer coordinate, and also finishes walking at a point with integer coordinate.   * He only changes direction at a point with integer coordinate.  Each time when Snuke passes a point with coordinate i-0.5, where i is an integer, he put a stone in his i-th ear.  After Snuke finishes walking, Ringo will repeat the following operations in some order so that, for each i, Snuke's i-th ear contains A_i stones:    * Put a stone in one of Snuke's ears.   * Remove a stone from one of Snuke's ears.  Find the minimum number of operations required when Ringo can freely decide how Snuke walks.

Incorrect submission (verdict: Wrong Answer):
a = [int(input()) for _ in range(int(input()))]

dp = [[0] * 5 for _ in range(len(a) + 1)]

for i, x in enumerate(a):
    y = x % 2 + (x==0) * 2
    z = (x + 1) % 2
    
    dp[i+1][0] = dp[i][0] + x
    dp[i+1][1] = min(dp[i][0] + y, dp[i][1] + y)
    dp[i+1][2] = min(dp[i][1] + z, dp[i][0] + z, dp[i][2] + z)
    dp[i+1][3] = min(dp[i][2] + y, dp[i][1] + y, dp[i][0] + y, dp[i][3] + y)
    dp[i+1][4] = min(dp[i][3] + x, dp[i+1][3])

print(dp[-1][4])

Fixed, accepted submission:
a = [int(input()) for _ in range(int(input()))]

dp = [[0] * 5 for _ in range(len(a) + 1)]

for i, x in enumerate(a):
    m = x % 2
    y = m + (not x) * 2
    z = 1 - m
    
    dp[i+1][0] = dp[i][0] + x
    dp[i+1][1] = min(dp[i][:2]) + y
    dp[i+1][2] = min(dp[i][:3]) + z
    dp[i+1][3] = min(dp[i][:4]) + y
    dp[i+1][4] = min(dp[i]) + x

print(min(dp[-1]))
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())

t = 0
m = 100

while m < X:
  m = math.ceil(m * 1.01)
  t += 1

print(t)

Fixed, accepted submission:
import math

X = int(input())

t = 0
m = 100

while m < X:
  m = math.floor(m * 1.01)
  t += 1

print(t)
Problem:
Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten. He remembers that the remainder of a divided by b was greater than or equal to K. Find the number of possible pairs that he may have had.

Incorrect submission (verdict: Wrong Answer):
n,m = map(int,input().split())
ans = 0
for i in range(1,n+1):
    print(ans)
    ans += n // i * max(i-m,0) + max(n%i-m+1,0)
if m == 0:
    ans -= n
print(ans)

Fixed, accepted submission:
n,m = map(int,input().split())
ans = 0
for i in range(1,n+1):
    ans += n // i * max(i-m,0) + max(n%i-m+1,0)
if m == 0:
    ans -= n
print(ans)
Problem:
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.  You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.

Incorrect submission (verdict: Wrong Answer):
#!/usr/bin/env python3

def main():
    O = input()
    E = input()
    s = ['x']*(len(O) + len(E))
    for i in range(len(O)//2):
        s[i*2] = O[i]
        s[i*2+1] = E[i]
    if len(O) % 2:
        s[-1] = O[-1]
    print(''.join(s))


if __name__ == '__main__':
    main()

Fixed, accepted submission:
#!/usr/bin/env python3

def main():
    O = input()
    E = input()
    s = []
    for i in range(len(E)):
        s.append(O[i])
        s.append(E[i])
    if len(O) - len(E):
        s.append(O[-1])
    print(''.join(s))


if __name__ == '__main__':
    main()
Problem:
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions:    * Question 1: Are you subscribing to Newspaper X?   * Question 2: Are you subscribing to Newspaper Y?  As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2.  What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y?  Write a program to answer this question.

Incorrect submission (verdict: Wrong Answer):






# 1 <= N < 100
# 0 <= A <= N
# 0 <= B <= N


inputstr = input()
inputary = inputstr.split(" ")
print(inputary)
N = int(inputary[0])
A = int(inputary[1])
B = int(inputary[2])

suball = N - A - B
if suball > 0:
    suball = 0
else:
    suball = abs(suball)

if A >= B:
    print(str(B) + " " + str(suball))
elif A < B:
    print(str(A) + " " + str(suball))


Fixed, accepted submission:






# 1 <= N < 100
# 0 <= A <= N
# 0 <= B <= N


inputstr = input()
inputary = inputstr.split(" ")

N = int(inputary[0])
A = int(inputary[1])
B = int(inputary[2])

suball = N - A - B
if suball > 0:
    suball = 0
else:
    suball = abs(suball)

if A >= B:
    print(str(B) + " " + str(suball))
elif A < B:
    print(str(A) + " " + str(suball))
Problem:
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i.  Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0.  Now, the following Q operations will be performed:    * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j.  Find the value of the counter on each vertex after all operations.

Incorrect submission (verdict: Wrong Answer):
N, Q = map(int, input().split())
lst = []
cnt = []
ans = [0]*N
N = N - 1
for i in range(N):
    lst.append(list(map(int, input().split())))
for j in range(Q):
    cnt.append(list(map(int, input().split())))
for k in range(Q):
    ans[cnt[k][0]-1] += cnt[k][1]
for i in range(N):
    ans[lst[i][1]-1] += ans[lst[i][0]-1]
for a in ans:
    print(a, end=" ")

Fixed, accepted submission:
import collections

N, Q = map(int, input().split())
lst = [[] for _ in range(N)]
ans = [0]*N
for i in range(N-1):
    a, b = map(int, input().split())
    lst[b-1].append(a-1)
    lst[a-1].append(b-1)
for j in range(Q):
    p, x = map(int, input().split())
    p -= 1
    ans[p] += x
               
q=collections.deque()
q.append(0)
flag = [0]*N
while q:
    y = q.pop()
    flag[y] = 1
    for k in lst[y]:
        if flag[k] == 1:
            continue
        ans[k] += ans[y]
        q.append(k)
print(*ans)
Problem:
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

Incorrect submission (verdict: Wrong Answer):
N = int(input())
cnt = 0
for i in range(1, N + 1):
  cnt += 1
print(cnt)

Fixed, accepted submission:
N = int(input())
cnt = 0
for i in range(1, N + 1):
  cnt += i
print(cnt)
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())
maxdiv = 0
for i in range(1, N + 1):
    d = i
    count = 0
    while d & 1 == 0:
        d >>= 1
        count += 1
    maxdiv = max(maxdiv, count)
print(count)

Fixed, accepted submission:
N = int(input())
maxdiv = 0
ans = 1
for i in range(1, N + 1):
    d = i
    count = 0
    while d % 2 == 0:
        d //= 2
        count += 1
    if count > maxdiv:
        ans = i
        maxdiv = count
print(ans)
Problem:
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N).  For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it.  We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied:    * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2.  Does there exist a good set of choices? If the answer is yes, find one good set of choices.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = list(map(int,input().split()))
box = [0]*n
for i in range(n,0,-1):
  k = a[i-1]
  c = 0
  for j in range(1,n//i+1):
    c += box[i*j-1]
  if c%2 != k:
    box[i-1] = 1
ans = []
w = 0
for m in range(n):
  if box[m] == 1:
    w += 1
    ans.append(str(i+1))
print(w)
print(' '.join(ans))

Fixed, accepted submission:
n = int(input())
a = list(map(int,input().split()))
box = [0]*n
for i in range(n,0,-1):
  k = a[i-1]
  c = 0
  for j in range(1,n//i+1):
    c += box[i*j-1]
  if c%2 != k:
    box[i-1] = 1
ans = []
w = 0
for m in range(n):
  if box[m] == 1:
    w += 1
    ans.append(str(m+1))
print(w)
print(' '.join(ans))
Problem:
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.  After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.

Incorrect submission (verdict: Wrong Answer):
S = input()
t = '2018' + S[4:]

Fixed, accepted submission:
S = input()
t = '2018' + S[4:]
print(t)
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):
import collections

N = int(input())
A = [int(i) for i in input().split()]

stick_num = collections.Counter(A)
flag = True
ans = 0
for i in sorted(stick_num, reverse=True):
    if stick_num[i] % 4 == 0:
        ans = i * i
    elif stick_num[i] % 2 >= 0:
        if flag:
            a = i
            flag = False
        else:
            ans = a * i
            break

print(ans)

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

flag_h = True
flag_v = True
h, v = 0, 0

for i in A:
    if flag_h:
        if h == i:
            flag_h = False
        else:
            h = i
    elif flag_v:
        if v == i:
            flag_v = False
        else:
            v = i

if not (flag_v and flag_h):
    print(h * v)
else:
    print(0)
Problem:
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs.  Determine if it is possible that there are exactly X cats among these A + B animals.

Incorrect submission (verdict: Wrong Answer):
a, b, x = list(map(int, input().split()))
if a + b >= x and a <= x:
    print('Yes')
else:
    print('No')

Fixed, accepted submission:
a, b, x = list(map(int, input().split()))
if a + b >= x and a <= x:
    print('YES')
else:
    print('NO')
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=input().split()
if sorted(x) == x:
  s='='
elif sorted(x)[0] == x[0]:
  s='<'
else:
  s='>'
print(s)

Fixed, accepted submission:
s=list(input().split())
ss=sorted(s)
if s[0]==s[1]:
  r='='
elif s==ss:
  r='<'
else:
  r='>'
print(r)
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]:
  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:
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs.  Determine if it is possible that there are exactly X cats among these A + B animals.

Incorrect submission (verdict: Wrong Answer):


def main():
    A,B,X = map(int, input().split(" "))

    if A > X or A + B < X:
        print("No")
    else:
        print("Yes")

if __name__ == '__main__':
    main()


Fixed, accepted submission:


def main():
    A,B,X = map(int, input().split(" "))

    if A > X or A + B < X:
        print("NO")
    else:
        print("YES")

if __name__ == '__main__':
    main()
Problem:
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements?  Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds:    * r_1 + c_1 = r_2 + c_2   * r_1 - c_1 = r_2 - c_2  For example, in the following figure, the bishop can move to any of the red squares in one move:  

Incorrect submission (verdict: Wrong Answer):
num = input().split(" ")
ki = 0
gu = 0
sum_kaku = 0
for a in range(int(num[1])):
  if a % 2 == 0:
    gu += 1
  else:
    ki += 1
a = 0
while a < int(num[1]):
  if a % 2 == 0:
    sum_kaku += ki
  else:
    sum_kaku += gu
  a += 1
print(sum_kaku)
  

Fixed, accepted submission:
num = input().split(" ")
if "1" in num:
  print(1)
else:
	gu = int(num[1]) // 2
	ki = int(num[1]) - int(gu)
	gu2 = int(num[0]) // 2
	ki2 = int(num[0]) - int(gu2)
	print(gu * gu2 + ki * ki2)
Problem:
Print the K-th element of the following sequence of length 32:                1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51

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

Fixed, accepted submission:
a = [1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
print(a[int(input()) - 1])
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]:
  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 a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.

Incorrect submission (verdict: Wrong Answer):
s = input()
if set([s[0], s[1], s[2]]) == set(['a', 'b', 'c']):
    print('YES')
else:
    print('NO')

Fixed, accepted submission:
s = input()
if set([s[0], s[1], s[2]]) == set(['a', 'b', 'c']):
    print('Yes')
else:
    print('No')
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):
row = int(input())
s = [input() for i in range(row)]
ac = 0
wa = 0
tle = 0
re = 0
for j in s:
    if j == "AC":
        ac += 1
    elif j == "WA":
        wa += 1
    elif j == "TLE":
        tle += 1
    elif j == "re":
        re += 1
ac = str(ac)
wa = str(wa)
tle = str(tle)
re = str(re)
print("AC × " + ac)
print("WA × " + wa)
print("TLE × " + tle)
print("RE × " + re)

Fixed, accepted submission:
row = int(input())
s = [input() for i in range(row)]
ac = 0
wa = 0
tle = 0
re = 0
for j in s:
    if j == "AC":
        ac += 1
    elif j == "WA":
        wa += 1
    elif j == "TLE":
        tle += 1
    elif j == "RE":
        re += 1
ac = str(ac)
wa = str(wa)
tle = str(tle)
re = str(re)
print("AC x " + ac)
print("WA x " + wa)
print("TLE x " + tle)
print("RE x " + re)
Problem:
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R.  Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan.  Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.

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

print('Right' if a+b > c+d else 'left' if a+b < c+d else 'Balanced')

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

print('Right' if a+b < c+d else 'Left' if a+b > c+d else 'Balanced')
Problem:
You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them.  For example, if K = 3, print `ACLACLACL`.

Incorrect submission (verdict: Wrong Answer):
k = int(input())
ans = 'ACL' * k
print(k)

Fixed, accepted submission:
k = int(input())
ans = 'ACL' * k
print(ans)