TnT/CodeNet4Repair huggingface dataset:
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())

k=0
m=n

while m%2==0:
  k+=1
  m=m/2
  
print(k)
  

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

k=0


while n//2>=1:
  k+=1
  n=n//2
 
print(2**k)
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):
import sys
sys.setrecursionlimit(10**6)


def main():
    def dfs(now, prev=-1):
        for next in edge[now]:
            if next == prev:
                continue
            score[next] += score[now]
            dfs(next, now)

    N, Q = map(int, input().split())
    edge = [[] for _ in range(N+1)]
    for _ in range(N-1):
        u, v = map(int, input().split())
        edge[u].append(v)
        edge[v].append(u)
    p = []
    x = []
    score = [0] * (N+1)
    for _ in range(Q):
        p, x = map(int, input().split())
        score[p] += x
    dfs(1)
    print(score)


if __name__ == "__main__":
    main()


Fixed, accepted submission:
import sys
sys.setrecursionlimit(10 ** 6)


def input():
    return sys.stdin.readline()[:-1]


def main():
    def dfs(now, prev=-1):
        for next in edge[now]:
            if next == prev:
                continue
            score[next] += score[now]
            dfs(next, now)

    N, Q = map(int, input().split())
    edge = [[] for _ in range(N+1)]
    for _ in range(N-1):
        u, v = map(int, input().split())
        edge[u].append(v)
        edge[v].append(u)
    p = []
    x = []
    score = [0] * (N+1)
    for _ in range(Q):
        p, x = map(int, input().split())
        score[p] += x
    dfs(1)
    print(*score[1:])


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):
def main():
    H, W = list(map(int, input().split()))

    return min(H, W)


print(main())


Fixed, accepted submission:
def main():
    H, W = list(map(int, input().split()))
    if min(H, W) == 1:
        return 1

    return (H * W + 1) // 2


print(main())
Problem:
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.  He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.

Incorrect submission (verdict: Wrong Answer):
A,V = map(int,input().split())
B,W = map(int,input().split())
T = int(input())
if abs(A + V*T) >= abs(B + W*T) :
    print('Yes')
else :
    print('No')


Fixed, accepted submission:
A,V = map(int,input().split())
B,W = map(int,input().split())
T = int(input())
if B > A :
    if A + V*T >= B + W*T :
        print('YES')
    else :
        print('NO')
elif A > B :
    if B - W*T >= A - V*T :
        print('YES')
    else :
        print('NO')
Problem:
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?

Incorrect submission (verdict: Wrong Answer):
n=int(input())
if n%2==0:
  print(n/2-1)
else:
  print(int(n/2))

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

Incorrect submission (verdict: Wrong Answer):
S = list(input().split())
print(S.count('1'))

Fixed, accepted submission:
s=input()
print(s.count('1'))
Problem:
There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities.  We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly.  For each city, find the number of the cities connected to that city by both roads and railways.

Incorrect submission (verdict: Wrong Answer):
print('1 2 2 1')
print('1 2 2 1')

Fixed, accepted submission:
from collections import defaultdict

class UnionFind():

    def __init__(self, n):
        self.par = [i for i in range(n)]

    def root(self, x):
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.root(self.par[x])
            return self.par[x]

    def same(self, x, y):
        return self.root(x) == self.root(y)

    def unite(self, x, y):
        x = self.root(x)
        y = self.root(y)

        if x == y:
            return
        
        self.par[x] = y
        return


def main():
    N, K, L = map(int, input().split())
    
    uf1 = UnionFind(N)
    uf2 = UnionFind(N)

    
    for _ in range(K):
        p, q = map(int, input().split())
        uf1.unite(p-1, q-1)
    
    # O(L)
    for _ in range(L):
        r, s = map(int, input().split())
        uf2.unite(r-1, s-1)

    # O(N) x O(log_N)
    
    
    cnts = defaultdict(int)
    for i in range(N):
        pos = (uf1.root(i), uf2.root(i))
        cnts[pos] += 1

    ans = []
    for i in range(N):
        pos = (uf1.root(i), uf2.root(i))
        ans.append(cnts[pos])

    print(*ans)


if __name__ == '__main__':
    main()
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):
import numpy as np

N, Q = map(int, input().split())
tree = [[0] * N for i in range(N)]
for i in range(N-1):
    a, b = map(int, input().split())
    print(a, b)
    tree[a-1][b-1] = 1

tree_np = np.array(tree)

for i, _ in enumerate(tree):
    for node, _ in enumerate(tree[i]):
        if tree[i][node] == 1:
            tree[i] = np.array(tree[i]) + np.array(tree[node])

print(tree)

ans = np.array([0] * N)
for q in range(Q):
    p, x = map(int, input().split())
    ans += np.array(np.array(tree[p-1]) * x)
    
print(ans)



Fixed, accepted submission:
from collections import deque

N, Q = map(int, input().split())
edge = [[] for i in range(N)]
ops = []
ans = [0] * N

for _ in range(N-1):
    a, b = map(int, input().split())
    edge[a-1].append(b-1)
    edge[b-1].append(a-1)
for _ in range(Q):
    ops.append(tuple(map(int, input().split())))

for op in ops:
    node = op[0]-1
    pnt = op[1]
    ans[node] += pnt

stack = deque()
stack.append((0, -1))

while stack:
    st = stack.pop()
    node, parent = st[0], st[1]
    children = edge[node]
    for child in children:
        if child == parent:
            continue
        ans[child] += ans[node]
        stack.append((child, node))

print(*ans)
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())
C = min(C, A+B)

ans = 0
ans += C*min(X, Y)
if X > Y:
  ans += min(C, A) * abs(X-Y)
else:
  ans + min(C, B) * abs(X-Y)
print(ans)

Fixed, accepted submission:
A, B, C, X, Y = map(int, input().split())
C = min(2*C, A+B)

ans = 0
ans += C*min(X, Y)
if X > Y:
  ans += min(C, A) * abs(X-Y)
else:
  ans += min(C, B) * abs(X-Y)
print(ans)
Problem:
M-kun has the following three cards:    * A red card with the integer A.   * A green card with the integer B.   * A blue card with the integer C.  He is a genius magician who can do the following operation at most K times:    * Choose one of the three cards and multiply the written integer by 2.  His magic is successful if both of the following conditions are satisfied after the operations:    * The integer on the green card is **strictly** greater than the integer on the red card.   * The integer on the blue card is **strictly** greater than the integer on the green card.  Determine whether the magic can be successful.

Incorrect submission (verdict: Wrong Answer):
#!/usr/bin/env python
# -*- coding: utf-8 -*-


def main():
    A, B, C = map(int, input().split())
    K = int(input())
    print(A, B, C, K)

    for _ in range(K):
        if A > B:
            B *= 2
        elif B > C:
            C *= 2
    if A < B < C:
        print("Yes")
    else:
        print("No")


if __name__ == "__main__":
    main()


Fixed, accepted submission:
#!/usr/bin/env python
# -*- coding: utf-8 -*-


def main():
    A, B, C = map(int, input().split())
    K = int(input())

    for _ in range(K):
        if not A < B:
            B *= 2
        elif not B < C:
            C *= 2
    if A < B < C:
        print("Yes")
    else:
        print("No")


if __name__ == "__main__":
    main()
Problem:
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

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

n,a,b = [int(x) for x in stdin.readline().rstrip().split()]
sum = 0



for i in range(n+1):
    i_sum = 0
    i_num = i
    print("i_num:"+str(i_num))
    while(True):
        if(i_num//10 ==0):
            i_sum += (i_num%10)
            break
        else:
            i_sum += (i_num%10)
            i_num = i_num//10
    if(a<= i_sum <= b):
        print(i_sum)
        sum += i
print("-------------------")
print(sum)




Fixed, accepted submission:
from sys import stdin

n,a,b = [int(x) for x in stdin.readline().rstrip().split()]
sum = 0



for i in range(n+1):
        #print(i)
        if (a<= i//10000 + (i-i//10000*10000)//1000 + (i-i//1000*1000)//100 + (i-i//100*100)//10 + i%10 <=b):
            sum += i


print(sum)
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()

cnt = 0
j = len(S) - 1

for i in range(len(S)-1, -1, -1):
    if j-1 > len("dreamer"):
        break
    for dd in ("dream", "dreamer", "erase", "eraser"):
        if S[i:j+1] == dd:
            j = i-1
            cnt += len(dd)

print(("NO", "YES")[len(S) == cnt])


Fixed, accepted submission:
S = input()

cnt = 0
j = len(S) - 1

for i in range(len(S) - 1, -1, -1):
    if j - i > len("dreamer"):
        break
    for dd in ("dream", "dreamer", "erase", "eraser"):
        if S[i:j + 1] == dd:
            j = i - 1
            cnt += len(dd)

print(("NO", "YES")[len(S) == cnt])
Problem:
You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.

Incorrect submission (verdict: Wrong Answer):
S = list(str(input()))
if len(S) == 2:
  print(S[0],S[1])
else:
  print(S[2],S[1],S[0])

Fixed, accepted submission:
S = str(input())
if len(S) == 2:
  print(S)
else:
  print(S[2]+S[1]+S[0])
Problem:
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.

Incorrect submission (verdict: Wrong Answer):
import math
X = int(input())
b = 1
for i in range(33,1,-1):
    a = math.floor(math.log(X, i))
    b = max(i**a,b)
print(b)

Fixed, accepted submission:
x=int(input())
c=1
for b in range(1,x):
    for p in range(2,x):
        if b**p<=x:c=max(c,b**p)
        else:break
print(c)
Problem:
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately.  He has very high awareness of safety, and decides to buy two bells, one for each hand.  The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.

Incorrect submission (verdict: Wrong Answer):
price = input().split(" ")
price = sorted(price)
print(price[0] + price[1])

Fixed, accepted submission:
price = list(map(int, input().split(" ")))
print(sum(price) - max(price))
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 = int(input())
b = int(input())
c = int(input())
X = int(input())
count = 0
for i in range(1,a+1):
    x = 500 * i
    for j in range(1,b+1):
        y = 100 * j
        for k in range(1,c+1):
            z = 50 * k
            if x+y+z == X :
                count += 1
print(count)

Fixed, accepted submission:
a = int(input())
b = int(input())
c = int(input())
X = int(input())
count = 0
for i in range(a+1):
    x = 500 * i
    for j in range(b+1):
        y = 100 * j
        for k in range(c+1):
            z = 50 * k
            if x+y+z == X :
                count += 1
print(count)
Problem:
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i.  Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.

Incorrect submission (verdict: Wrong Answer):
print(list(input()))

Fixed, accepted submission:
print(list(input()).count('1'))
Problem:
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis.  You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0.  However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned.  For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.

Incorrect submission (verdict: Wrong Answer):
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits

def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7

N = INT()
A = LIST()

A.insert(0, 0)
A.append(0)
print(A)

ans = 0
for i in range(N+1):
	ans += abs(A[i+1]-A[i])

for i in range(N):
	if A[i]<=A[i+1]<=A[i+2] or A[i+2]<=A[i+1]<=A[i]:  
		print(ans) 
	else:
		print(ans-2*
			min(
				abs(A[i+2]-A[i+1]), 
				abs(A[i+1]-A[i])
				)
			)


Fixed, accepted submission:
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits

def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7

N = INT()
A = LIST()

A.insert(0, 0)
A.append(0)

ans = 0
for i in range(N+1):
	ans += abs(A[i+1]-A[i])

for i in range(N):
	if A[i]<=A[i+1]<=A[i+2] or A[i+2]<=A[i+1]<=A[i]:  
		print(ans) 
	else:
		print(ans-2*
			min(
				abs(A[i+2]-A[i+1]), 
				abs(A[i+1]-A[i])
				)
			)
Problem:
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.  The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.  How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?

Incorrect submission (verdict: Wrong Answer):

n=int(input())
h=int(input())
w=int(input())
h -= (n-1)
w -= (n-1)
print(h*w)

Fixed, accepted submission:

n=int(input())
h=int(input())
w=int(input())
s = n-h+1
t = n-w+1
print(s*t)
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):
s = input()
if s.replace("eraser"," ").replace("erase"," ").replace("dreamer"," ").replace("dream"," ").replace(" ", ""):
    print("NO")
else:
    print("YES")

Fixed, accepted submission:
l = input()
def vowel(l):
    return "vowel" if l in "aeiou" else "consonant"
print(vowel(l))
Problem:
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

Incorrect submission (verdict: Wrong Answer):
a,b=map(int,input().split())
c=a*b
if(c%2==0):
  print("even")
else:
  print("odd")
        
        
        

Fixed, accepted submission:
a,b=map(int,input().split())
c=a*b
if(c%2==0):
  print("Even")
else:
  print("Odd")
Problem:
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets.  Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.

Incorrect submission (verdict: Wrong Answer):
x=1
while True:
    i=input()
    print('case {}: {}'.format(x,i))
    x+=1
    if x==999:
        break


Fixed, accepted submission:
import sys
 
i=1
for s in sys.stdin:
  n = int(s)
  if n == 0:
    break
  print("Case ",i,": ",n,sep="")
  i += 1
Problem:
Let's play Amidakuji.  In the following example, there are five vertical lines and four horizontal lines. The horizontal lines can intersect (jump across) the vertical lines.      In the starting points (top of the figure), numbers are assigned to vertical lines in ascending order from left to right. At the first step, 2 and 4 are swaped by the first horizontal line which connects second and fourth vertical lines (we call this operation (2, 4)). Likewise, we perform (3, 5), (1, 2) and (3, 4), then obtain "4 1 2 5 3" in the bottom.  Your task is to write a program which reads the number of vertical lines w and configurations of horizontal lines and prints the final state of the Amidakuji. In the starting pints, numbers 1, 2, 3, ..., w are assigne to the vertical lines from left to right.

Incorrect submission (verdict: Wrong Answer):
def main():
    LIST=[]
    for i in range(int(input())):
        LIST.append(i+1)

    for j in range(int(input())):
        a,b=map(int,input().split(","))
        LIST[a-1],LIST[b-1]=LIST[b-1],LIST[a-1]

    for i in LIST:
        print(i)

Fixed, accepted submission:
def main():
    LIST=[]
    for i in range(int(input())):
        LIST.append(i+1)

    for j in range(int(input())):
        a,b=map(int,input().split(","))
        LIST[a-1],LIST[b-1]=LIST[b-1],LIST[a-1]

    for i in LIST:
        print(str(i))
            
    
if __name__ == '__main__':
    main()
Problem:
You are given two positive integers A and B. Compare the magnitudes of these numbers.

Incorrect submission (verdict: Wrong Answer):
A = int(input())
B = int(input())

print(A)

if A > B:
  print("GREATER")
elif A<B:
  print("LESS")
else:
  print("EQUAL")

Fixed, accepted submission:
A = int(input())
B = int(input())

if A > B:
  print("GREATER")
elif A<B:
  print("LESS")
else:
  print("EQUAL")
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):
s = list(input())
t = set(s)
print(s)
print(t)
if len(s) == len(t):
	print('yes')
else:
	print('no')

Fixed, accepted submission:
s = list(input())
t = set(s)
if len(s) == len(t):
	print('yes')
else:
	print('no')