TnT/CodeNet4Repair huggingface dataset:
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())
ans=n/(max(h,w))
print(int(ans))

Fixed, accepted submission:
import math
h=int(input())
w=int(input())
n=int(input())
ans=math.ceil(n/(max(h,w)))
print(int(ans))
Problem:
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.  To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Incorrect submission (verdict: Wrong Answer):
A = sorted(map(int,input().split()))
print('Yes' if A[0]==5 and A[1]==5 and A[2]==7 else 'No')


Fixed, accepted submission:
A = sorted(input().split())
print('YES' if A[0]=='5' and A[1]=='5' and A[2]=='7' else 'NO')
Problem:
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.  The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.  One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.  If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.    * X < Z \leq Y   * x_1, x_2, ..., x_N < Z   * y_1, y_2, ..., y_M \geq Z  Determine if war will break out.

Incorrect submission (verdict: Wrong Answer):


N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))

x.sort()
y.sort()

if x[-1] < y[0]:
    for i in range(x[-1], y[0]):
        if X < i <= Y:
            print("X({}) < i({}) <= Y({})".format(X, i, Y))
            print("No War")
            exit()

print("War")


Fixed, accepted submission:


N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))

x.sort()
y.sort()

if x[-1] < y[0]:
    for i in range(x[-1]+1, y[0]+1):
        if X < i <= Y:
            
            print("No War")
            exit()

print("War")
Problem:
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.

Incorrect submission (verdict: Wrong Answer):
X = int(input())
count = 0
step = 0
while X > step:
  count += 1
  step += count
  print(count)

Fixed, accepted submission:
X = int(input())
 
i = 0
while i * (i + 1) / 2 < X:
    i += 1
 
print(i)
Problem:
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.

Incorrect submission (verdict: Wrong Answer):
tmp_lst = input().split()

A, B, C = int(tmp_lst[0]), int(tmp_lst[1]), int(tmp_lst[2])

if A < C and C < B:
    print("YES")
elif B < C and C < A:
    print("YES")
else:
    print("NO")



Fixed, accepted submission:
tmp_lst = input().split()

A, B, C = int(tmp_lst[0]), int(tmp_lst[1]), int(tmp_lst[2])

if A < C and C < B:
    print("Yes")
elif B < C and C < A:
    print("Yes")
else:
    print("No")
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: Wrong Answer):
import sys
N, K = map(int, input().split())
D = list(input().split())
for i in range(N, 100 * N):
    for d in D:
        if d not in str(i):
            print(i)
            sys.exit()

Fixed, accepted submission:
import sys
N, K = map(int, input().split())
D = list(input().split())
D = set(D)
for i in range(N, 100 * N):
    if D & set(str(i)):
        continue
    else:
        print(i)
        sys.exit()
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):
def main():
  S = input()
  while len(S)>=5:
    if len(S)>=7 and S[:-7]=="dreamer":
      S = S[-7:]
      continue
    
    elif len(S)>=6 and S[:-6]=="eraser":
      S = S[-6:]
      continue
    elif S[:-5]=="dream" or S[:-5]=="erase":
      S = S[-5:]
      continue
    else:
      break
  if len(S)==0:
    print('Yes')
  else:
    print('No')
    
main()
    

Fixed, accepted submission:
def main():
  S = input()
  while len(S)>=5:
    if len(S)>=7 and S[-7:]=="dreamer":
      S = S[:-7]
      continue
    
    elif len(S)>=6 and S[-6:]=="eraser":
      S = S[:-6]
      continue
    elif S[-5:]=="dream" or S[-5:]=="erase":
      S = S[:-5]
      continue
    else:
      break
  if len(S)==0:
    print('YES')
  else:
    print('NO')
    
main()
Problem:
Snuke has a favorite restaurant.  The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer.  So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
ans=n*800-150*(n//15)
print(ans)

Fixed, accepted submission:
n=int(input())
ans=int(n*800-200*(n//15))
print(ans)
Problem:
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019.  Integers M_1, D_1, M_2, and D_2 will be given as input.   It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1.   Determine whether the date 2019-M_1-D_1 is the last day of a month.

Incorrect submission (verdict: Wrong Answer):
_,s=open(0)
c=0
for i in range(1000):
 p=0;F=1
 for j in str(i).zfill(3):
  q=s[p:].find(j)
  if q<0:F=0;break
  p+=q+1
 c+=F
print(c)

Fixed, accepted submission:
print(max(2-int(open(0).read().split()[3]), 0))
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 = [int(x) for x in input().split()]

if a*b//2==0:
    print("Even")
else:
    print("Odd")

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

if (a*b)%2==0:
    print("Even")
else:
    print("Odd")
Problem:
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery.  Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence.  Here, the distance between two points s and t on a number line is represented by |s-t|.

Incorrect submission (verdict: Wrong Answer):
a, b, c= map(int, input().split())
if abs(a-b) >= abs(a-c):
  print("A")
else:
  print("B")

Fixed, accepted submission:
x, a, b = map(int,input().split())
print("A" if abs(x-a) < abs(x-b) else "B")
Problem:
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively.    An example of a trapezoid  Find the area of this trapezoid.

Incorrect submission (verdict: Wrong Answer):
a,b,h=[int(input()) for i in range(3)]
print((a+b)*h/2)

Fixed, accepted submission:
a,b,h=[int(input()) for i in range(3)]
print(int((a+b)*h/2))
Problem:
There is a data which provides heights (in meter) of mountains. The data is only for ten mountains.  Write a program which prints heights of the top three mountains in descending order.

Incorrect submission (verdict: Wrong Answer):
mt=[]
for i in range(10):
    mt.append(int(input()))

high=[0,0,0]
for i in range(10):
    if  mt[i] > high[0]:
        high=[mt[i]]+high[:2]
    elif mt[i] > high[1]:
        high=[high[0]]+[mt[1]]+[high[1]]
    elif mt[i] > high[2]:
        high=high[:2]+[mt[i]]
print(high,sep='\n')

Fixed, accepted submission:

mt=[]
high=[0,0,0]
for i in range(10):
    mt.append(int(input()))
    if  mt[i] > high[0]:
        high=[mt[i]]+high[:2]
    elif mt[i] > high[1]:
        high=[high[0]]+[mt[i]]+[high[1]]
    elif mt[i] > high[2]:
        high=high[:2]+[mt[i]]
print ('\n'.join(map(str,high)))
Problem:
N persons are standing in a row. The height of the i-th person from the front is A_i.  We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person:  Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool.  Find the minimum total height of the stools needed to meet this goal.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
a_list=list(map(int,input().split()))
b_list=[]
for i in range(n-1):
  d=a_list[i]-a_list[i+1]
  if(d>=0):
    b_list.append(d)
    a_list[i+1]+=d
    
ans=sum(b_list)
print(b_list)
print(ans)

Fixed, accepted submission:
n=int(input())
a_list=list(map(int,input().split()))
b_list=[]
for i in range(n-1):
  d=a_list[i]-a_list[i+1]
  if(d>=0):
    b_list.append(d)
    a_list[i+1]+=d
    
ans=sum(b_list)
print(ans)
Problem:
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively.    An example of a trapezoid  Find the area of this trapezoid.

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

a = int(input())

b = int(input())

h = int(input())



print("{}".format((a+b)*h/2))

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

a = int(input())

b = int(input())

h = int(input())



print("{:.0f}".format((a+b)*h/2))
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())
ret = 0
if 2 * c <= a + b:
  m = min(x, y)
  ret += m * c
  x -= m
  y -= m
  ret += min(2 * c, a) * x
  ret += min(2 * c, b) * y
else:
  ret += a * x
  ret += b * y
print(ret)

Fixed, accepted submission:
a, b, c, x, y = map(int, input().split())
ret = 0
if 2 * c <= a + b:
  m = min(x, y)
  ret += m * 2 * c
  x -= m
  y -= m
ret += min(2 * c, a) * x
ret += min(2 * c, b) * y
print(ret)
Problem:
N persons are standing in a row. The height of the i-th person from the front is A_i.  We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person:  Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool.  Find the minimum total height of the stools needed to meet this goal.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = list(map(int, input().split()))
m = 0
ma = a[0]
for i in range(1, n):
    ma = max(ma, a[i])
    m = max(m, ma-a[i])
print(m)

Fixed, accepted submission:
n = int(input())
a = list(map(int, input().split()))
m = 0
ma = a[0]
for i in range(1, n):
    ma = max(ma, a[i])
    m += ma-a[i]
print(m)
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):
# coding: utf-8
# Here your code !

i = 0

while i < 1000:
    num = input().rstrip()
    if int(num) > 0:
        print("Case i: " + num)
    else:
        print("Case i: " + num)
        break

Fixed, accepted submission:
# coding: utf-8
# Here your code !

i = 0

while i < 10000:
    num = input().rstrip()
    if int(num) > 0:
        i += 1
        print("Case " + str(i) + ": " + str(num))
    else:
        break
Problem:
Print all the integers that satisfies the following in ascending order:    * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

Incorrect submission (verdict: Wrong Answer):
a,b,k=map(int,input().split())
l=[i for i in range(1,b+1)]
l=l[a-1:]
l=l[:k]+l[-k:]
l=set(l)
for i in l:
    print(i)

Fixed, accepted submission:
a,b,k=map(int,input().split())
l=[i for i in range(a,min(a+k,b+1))]
s=[i for i in range(max(a,b-k+1),b+1)]
l=l+s
l=list(set(l))
l.sort()
for i in l:
    print(i)
Problem:
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand.  How many grams of sand will the upper bulb contains after t seconds?

Incorrect submission (verdict: Wrong Answer):
a, b = map(int, input().split())
if b-a > 0:
    print(b-a)
else:
    print(0)


Fixed, accepted submission:
a, b = map(int, input().split())
if a-b > 0:
    print(a-b)
else:
    print(0)
Problem:
There are N positive integers written on a blackboard: A_1, ..., A_N.  Snuke can perform the following operation when all integers on the blackboard are even:    * Replace each integer X on the blackboard by X divided by 2.  Find the maximum possible number of operations that Snuke can perform.

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

ans = 0

for a in A:
    if a % 2 == 0:
        ans += 1
    else:
        print(0)
        exit()
        
print(ans)

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

ans = 0
flag = True

while flag == True:
    for i in range(N):
        if A[i] % 2 == 0:
            A[i] = A[i]//2
        else:
            flag = False
            break
    if flag == True:
        ans += 1
        
print(ans)
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):
n,y = map(int, input().split())

for i in range(n+1):
    for j in range(n+1-i):
        for k in range(n+1-i-j):
            sum = 10000*i + 5000*j + 1000*k
            if sum == y:
                print(i,j,k)
                break
print(-1,-1,-1)

Fixed, accepted submission:
n,y = map(int, input().split())
a = -1
b = -1
c = -1
for i in range(n+1):
    for j in range(n+1-i):
        k = n-i-j
        sum = 10000*i + 5000*j + 1000*k
        if sum == y:
            a = i
            b = j
            c = k
                
print(a,b,c)
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):
from collections import deque

N, Q = map(int, input().split())
tree = [[] for _ in range(N)] 
counters = deque([0] * N)
answer = [0] * N

for _ in range(N-1):
    index, child = map(int, input().split())
    tree[index-1].append(child-1)

for _ in range(Q):
    index, count = map(int, input().split())
    counters[index-1] += count

for i in range(N):
    count = counters.popleft()
    answer[i] += count

    for j in range(len(tree[i])):
        answer[tree[i][j]] += answer[i]

for i in answer:
    print(i, end=' ')

Fixed, accepted submission:
from collections import deque

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

counts = [0] * (N+1)
for _ in range(Q):
    p, x  = map(int, input().split())
    counts[p] += x
visited = [-1] * (N+1)

q = deque()
q.append(1)
visited[1] = 1
while q:
    node = q.pop()

    next_nodes = graph[node]
    for next in next_nodes:
        if visited[next] != -1:
            continue
        q.append(next)
        visited[next] = 1
        counts[next] += counts[node]

print(*counts[1:])
Problem:
Depth-first search (DFS) follows the strategy to search ”deeper” in the graph whenever possible. In DFS, edges are recursively explored out of the most recently discovered vertex $v$ that still has unexplored edges leaving it. When all of $v$'s edges have been explored, the search ”backtracks” to explore edges leaving the vertex from which $v$ was discovered.  This process continues until all the vertices that are reachable from the original source vertex have been discovered. If any undiscovered vertices remain, then one of them is selected as a new source and the search is repeated from that source.  DFS timestamps each vertex as follows:    * $d[v]$ records when $v$ is first discovered.   * $f[v]$ records when the search finishes examining $v$’s adjacency list.  Write a program which reads a directed graph $G = (V, E)$ and demonstrates DFS on the graph based on the following rules:    * $G$ is given in an adjacency-list. Vertices are identified by IDs $1, 2,... n$ respectively.   * IDs in the adjacency list are arranged in ascending order.   * The program should report the discover time and the finish time for each vertex.   * When there are several candidates to visit during DFS, the algorithm should select the vertex with the smallest ID.   * The timestamp starts with 1.

Incorrect submission (verdict: Wrong Answer):
G = {}
for _ in range(int(input())):
  x = list(map(int, input().split()))
  G[x[0]] = sorted(x[2:])

t = 1
stack = [1]
timestamp = {1:[t]}

while len(stack) > 0:
  t += 1
  v = stack[-1]

  while len(G[v]) > 0:
    v = G[v].pop(0)
    if v in timestamp: continue
    stack.append(v)
    timestamp[v] = [t]
    break
  else:
    v = stack.pop()
    timestamp[v].append(t)

for i in timestamp:
  print(i, *timestamp[i])

Fixed, accepted submission:
def get_unvisted_child(v, timestamp):
  while len(G[v]) > 0:
    c = G[v].pop(0)
    if not c in timestamp: return c
  return -1

def get_unvisited_node(G, timestamp):
  for v in sorted(G):
    if not v in timestamp: return v
  return -1

def dept_first_search(G):
  t = 1
  timestamp = {}
  stack = []

  v = get_unvisited_node(G, timestamp)
  while v > 0:
    stack.append(v)
    timestamp[v] = [t]
    t += 1

    while len(stack) > 0:
      v = stack[-1]
      c = get_unvisted_child(v, timestamp)
      if c > 0:
        #c = G[v].pop(0)
        stack.append(c)
        timestamp[c] = [t]
      else:
        timestamp[v].append(t)
        stack.pop()
      t += 1

    v = get_unvisited_node(G, timestamp)

  for v in sorted(timestamp):
    print(v, *timestamp[v])

G = {}
for i in range(int(input())):
  x = list(map(int, input().split()))
  G[x[0]] = x[2:]

dept_first_search(G)
Problem:
Snuke built an online judge to hold a programming contest.  When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)  Determine whether the judge can return the string `AC` as the verdict to a program.

Incorrect submission (verdict: Wrong Answer):
s=input()
c=s.replace("x","")
def insert(s,n,b):
    c=s[n]
    a=s.split(c)
    h=a[0]+b+c+a[1]
    return h

def kaibun(s):
    n=0
    for i in range(int(len(s)/2)):
        if s[i]!=s[len(s)-1-i]:
            n=1
    return n
    
center=int(len(c)/2)
num=0
j=1
f=c[:center+1]
if kaibun(c)==1:
    print(-1)
else:
    for ch in f:
        i=num
        while(1):
            if ch==s[i]:
                break
            i+=1
            num=i
        print(len(s)-num)
    
            
        
            












            

Fixed, accepted submission:
s=input()
if "AC" in s:
    print("Yes")
else:
    print("No")
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):
from bisect import bisect_right, bisect_left

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

A.sort()
B = set(A)

ans = []

for i in B:
    p = bisect_right(A, i) - bisect_left(A, i)
    if p > 2:
        ans.append(i)

ans.sort(reverse=True)

if len(ans) >= 2:
    print(ans[0] * ans[1])
else:
    print(0)


Fixed, accepted submission:
from bisect import bisect_right, bisect_left

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

A.sort()
B = set(A)

ans = []

for i in B:
    p = bisect_right(A, i) - bisect_left(A, i)
    if 4 > p >= 2:
        ans.append(i)
    elif p >= 4:
        ans.append(i)
        ans.append(i)
        

ans.sort(reverse=True)

if len(ans) >= 2:
    print(ans[0] * ans[1])
else:
    print(0)
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())
if a*b // 2 == 0:
  print('Even')
else:
  print('Odd')

Fixed, accepted submission:
a,b = map(int, input().split())
if a * b % 2 == 0:
  print('Even')
else:
  print('Odd')
Problem:
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10.  Given an integer N, determine whether it is a Harshad number.

Incorrect submission (verdict: Wrong Answer):
use = int(input())
if sum([int(i) for i in str(use)]) % use == 0:
    print("YES")
else: print("NO")

Fixed, accepted submission:
use = int(input())
if use % sum([int(i) for i in str(use)]) == 0:
    print("Yes")
else: print("No")