TnT/CodeNet4Repair huggingface dataset:
Problem:
You start with the number 0 and you want to reach the number N.  You can change the number, paying a certain amount of coins, with the following operations:    * Multiply the number by 2, paying A coins.   * Multiply the number by 3, paying B coins.   * Multiply the number by 5, paying C coins.   * Increase or decrease the number by 1, paying D coins.  You can perform these operations in arbitrary order and an arbitrary number of times.  What is the minimum number of coins you need to reach N?  **You have to solve T testcases.**

Incorrect submission (verdict: Wrong Answer):
import sys
check=dict()
N,a,b,c,d=0,0,0,0,0


def dfs_sub(n,ttf,cha):
    global check,d
    if n%ttf<=(ttf//2):
        if n//ttf==0:
            if n//ttf in check:
                if check[n//ttf]>check[n]+(n%ttf)*d:
                    check[n//ttf]=check[n]+(n%ttf)*d
                    dfs(n//ttf,check[n]+(n%ttf)*d)
            else:
                check[n//ttf]=check[n]+(n%ttf)*d
                dfs(n//ttf,check[n]+(n%ttf)*d)
        else:
            if n//ttf in check:
                if check[n//ttf]>check[n]+(n%ttf)*d+cha:
                    check[n//ttf]=check[n]+(n%ttf)*d+cha
                    dfs(n//ttf,check[n]+(n%ttf)*d+cha)
            else:
                check[n//ttf]=check[n]+(n%ttf)*d+cha
                dfs(n//ttf,check[n]+(n%ttf)*d+cha)
    if n%ttf>(ttf//2) or (ttf==2 and n%ttf==1):
        if 1+n//ttf in check:
            if check[1+n//ttf]>check[n]+(ttf-n%ttf)*d+cha:
                check[1+n//ttf]=check[n]+(ttf-n%ttf)*d+cha
                dfs(1+n//ttf,check[n]+(ttf-n%ttf)*d+cha)
        else:
            check[1+n//ttf]=check[n]+(ttf-n%ttf)*d+cha
            dfs(1+n//ttf,check[n]+(ttf-n%ttf)*d+cha)


def dfs(n,x):
    global check,a,b,c,d
    if n==1:
        if n-1 in check:
            if check[n-1]>x+d:
                check[n-1]=x+d
        else:
            check[n-1]=x+d
    if n==0:
        return
    dfs_sub(n,2,a)
    dfs_sub(n,3,b)
    dfs_sub(n,5,c)

t=int(input())
ans=[0]*t
for i in range(t):
    N,a,b,c,d=map(int,input().split())
    check=dict()
    check[N]=0
    dfs(N,0)
    ans[i]=check[0]
for i in range(t):
    print(ans[i])

Fixed, accepted submission:
def f(n):
 if (m:=s.get(n,-1))<0:s[n]=m=min([d*n]+[abs(p*k-n)*d+c+f(k)for p,c in zip((2,3,5),b)for k in(n//p,0--n//p)])
 return m
exec("n,*b,d=map(int,input().split());s={0:0,1:d};print(f(n));"*int(input()))
Problem:
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?

Incorrect submission (verdict: Wrong Answer):
n, m = map(int, input().split())
ab = [input().split() for _ in range(m)]
ab = [item for ab in ab for item in ab]
print(ab)

for x in range(1, n+1):
    print(ab.count(str(x)))

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

for x in range(1, n+1):
    print(ab.count(str(x)))
Problem:
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year.   For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year.  In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi.   How many days from 2018-1-1 through 2018-a-b are Takahashi?

Incorrect submission (verdict: Wrong Answer):
a, b = map(int, input().split())
print(a if a<b else a-1)

Fixed, accepted submission:
a, b = map(int, input().split())
print(a if a<=b else a-1)
Problem:
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}.  Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.

Incorrect submission (verdict: Wrong Answer):

c = input('')
d = input('')

if c == d[::-1]:
    print('Yes')
else:
    print('No')

Fixed, accepted submission:

c = input('')
d = input('')

if c == d[::-1]:
    print('YES')
else:
    print('NO')
Problem:
We have a rectangular grid of squares with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left. On this grid, there is a piece, which is initially placed at square (s_r,s_c).  Takahashi and Aoki will play a game, where each player has a string of length N. Takahashi's string is S, and Aoki's string is T. S and T both consist of four kinds of letters: `L`, `R`, `U` and `D`.  The game consists of N steps. The i-th step proceeds as follows:    * First, Takahashi performs a move. He either moves the piece in the direction of S_i, or does not move the piece.   * Second, Aoki performs a move. He either moves the piece in the direction of T_i, or does not move the piece.  Here, to move the piece in the direction of `L`, `R`, `U` and `D`, is to move the piece from square (r,c) to square (r,c-1), (r,c+1), (r-1,c) and (r+1,c), respectively. If the destination square does not exist, the piece is removed from the grid, and the game ends, even if less than N steps are done.  Takahashi wants to remove the piece from the grid in one of the N steps. Aoki, on the other hand, wants to finish the N steps with the piece remaining on the grid. Determine if the piece will remain on the grid at the end of the game when both players play optimally.

Incorrect submission (verdict: Wrong Answer):
import sys

input = sys.stdin.readline


def main():
    h, w, n = map(int, input().split())
    y, x = map(int, input().split())
    y -= 1
    x -= 1
    s, t = input(), input()

    u, d, l, r = 0, h - 1, 0, w - 1

    for ss, tt in zip(s[::-1], t[::-1]):
        if tt == 'U':
            d = min(d + 1, h - 1)
        elif tt == 'D':
            u = max(u - 1, 0)
        elif tt == 'L':
            r = min(r + 1, w - 1)
        else:
            l = max(l - 1, 0)  

        if ss == 'U':
            u += 1
        elif ss == 'D':
            d -= 1
        elif ss == 'L':
            l += 1
        else:
            r -= 1  # ss == 'R'

        if l > r or u > d:
            print('NO')
            return

    if l <= x <= r and u <= y <= d:
        print('YES')
        return
    else:
        print('NO')
        return


if __name__ == '__main__':
    main()

Fixed, accepted submission:
def main():
    h, w, n = map(int, input().split())
    y, x = map(int, input().split())
    s, t = input(), input()

    u, d, l, r = 1, h, 1, w

    for ss, tt in zip(s[::-1], t[::-1]):
        if tt == 'U':
            d = min(d + 1, h)
        elif tt == 'D':
            u = max(u - 1, 1)
        elif tt == 'L':
            r = min(r + 1, w)
        else:
            l = max(l - 1, 1)  

        if ss == 'U':
            u += 1
        elif ss == 'D':
            d -= 1
        elif ss == 'L':
            l += 1
        else:
            r -= 1  # ss == 'R'

        if l > r or u > d:
            print('NO')
            return

    if l <= x <= r and u <= y <= d:
        print('YES')
        return
    else:
        print('NO')
        return


if __name__ == '__main__':
    main()
Problem:
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.

Incorrect submission (verdict: Wrong Answer):
s=input()
all="abcdefghijklmnopqrstuvwxyz"
for i in all:
    if i not in s:
        print(i)
    elif i=="z":
        print("None")

Fixed, accepted submission:
s=input()
ans=1
all="abcdefghijklmnopqrstuvwxyz"
for i in all:
    if i not in s:
        print(i)
        ans=0
        break
if ans==1:
    print("None")
Problem:
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j).   Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`.   However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.   Determine if square1001 can achieve his objective.

Incorrect submission (verdict: Wrong Answer):
H,W=map(int,input().split())
S = [list(str(input())) for i in range(H)]
judge=True
for i in range(H):
    for j in range(W):
        if S[i][j]=='#':
            judge1=True
            judge2=True
            judge3=True
            judge4=True
            
            try:S[i][j-1]
            except IndexError:judge1=False
            else:
                if S[i][j-1]=='.':
                    judge1=False
                    
            try:S[i][j+1]
            except IndexError:judge2=False
            else:
                if S[i][j+1]=='.':
                    judge2=False
                    
            try:S[i+1][j]
            except IndexError:judge3=False
            else:
                if S[i+1][j]=='.':
                    judge3=False
                    
            try:S[i-1][j]
            except IndexError:judge4=False
            else:
                if S[i-1][j]=='.':
                    judge4=False

            if judge1==False and judge2==False and judge3==False and judge4==False:
                judge=False
                break

    if judge==False:break


print(judge)

Fixed, accepted submission:
H,W=map(int,input().split())
S = [list(str(input())) for i in range(H)]
judge=True
for i in range(H):
    for j in range(W):
        if S[i][j]=='#':
            judge1=True
            judge2=True
            judge3=True
            judge4=True
            
            try:S[i][j-1]
            except IndexError:judge1=False
            else:
                if S[i][j-1]=='.':
                    judge1=False
                    
            try:S[i][j+1]
            except IndexError:judge2=False
            else:
                if S[i][j+1]=='.':
                    judge2=False
                    
            try:S[i+1][j]
            except IndexError:judge3=False
            else:
                if S[i+1][j]=='.':
                    judge3=False
                    
            try:S[i-1][j]
            except IndexError:judge4=False
            else:
                if S[i-1][j]=='.':
                    judge4=False

            if judge1==False and judge2==False and judge3==False and judge4==False:
                judge=False
                break

    if judge==False:break


if judge:print('Yes')
else:print('No')
Problem:
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.

Incorrect submission (verdict: Wrong Answer):
class Judge():
    @classmethod
    def judge1(cls,s,t):
        if len(s) == len(t):
            return False
        judge = True
        n = min(len(s),len(t))
        for i in range(n):
            judge = judge and s[i]==t[i]
        return judge

    @classmethod
    def judge2(cls,s,t):
        n = min(len(s),len(t))
        for l in range(1,n):
            judge = True
            for i in range(l-1):
                judge = judge and s[i]==t[i]
            judge = judge and s[l] < t[l]
            if judge == True:
                break

        return judge

s = list(input())
s.sort()
t = list(input())
t.sort()

judge = Judge.judge1(s,t) or Judge.judge2(s,t)

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


Fixed, accepted submission:
s = list(input())
s.sort()
s = ''.join(s)
t= list(input())
t.sort()
t = ''.join(reversed(list(t)))

if s < t:
    print('Yes')
else:
    print('No')
Problem:
You are given two integers A and B. Find the largest value among A+B, A-B and A \times B.

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

print(max(a+b,a-b,a*b))

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

print(max(a+b,a-b,a*b))
Problem:
You start with the number 0 and you want to reach the number N.  You can change the number, paying a certain amount of coins, with the following operations:    * Multiply the number by 2, paying A coins.   * Multiply the number by 3, paying B coins.   * Multiply the number by 5, paying C coins.   * Increase or decrease the number by 1, paying D coins.  You can perform these operations in arbitrary order and an arbitrary number of times.  What is the minimum number of coins you need to reach N?  **You have to solve T testcases.**

Incorrect submission (verdict: Wrong Answer):
t=int(input())
INF=10**19
def dfs(k):
    if k==1:return d
    if k==0:return 0
    if k in dic:return dic[k]
    ans=INF
    ans=min(ans,dfs(k//2)+a+k%2*d,dfs((k+1)//2)+a+-k%2*d,dfs(k//3)+b+k%3*d,dfs((k+2)//3)+b+-k%3*d,dfs(k//5)+c+k%5*d,dfs((k+4)//5)+c+-k%5*d)
    dic[k]=ans
    return ans
for _ in range(t):
    dic={}
    n,a,b,c,d=map(int,input().split())
    print(dfs(n))

Fixed, accepted submission:
def f(k):
    if k==1:return d
    if k==0:return 0
    if k in m:return m[k]
    s=min(k*d,f(k//2)+a+k%2*d,f(-(-k//2))+a+-k%2*d,f(k//3)+b+k%3*d,f(-(-k//3))+b+-k%3*d,f(k//5)+c+k%5*d,f(-(-k//5))+c+-k%5*d)
    m[k]=s
    return s
for _ in range(int(input())):
    m={}
    n,a,b,c,d=map(int,input().split())
    print(f(n))
Problem:
AtCoDeer has three cards, one red, one green and one blue.   An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card.   We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer.   Is this integer a multiple of 4?

Incorrect submission (verdict: Wrong Answer):
r, g, b = map(int, input().split())
print('Yes' if (r*100+g*10+b) % 4 == 0 else 'No')

Fixed, accepted submission:
r, g, b = map(int, input().split())
print('YES' if (r*100+g*10+b) % 4 == 0 else 'NO')
Problem:
Write a program which reads a list of student test scores and evaluates the performance for each student.  The test scores for a student include scores of the midterm examination m (out of 50), the final examination f (out of 50) and the makeup examination r (out of 100). If the student does not take the examination, the score is indicated by -1.  The final performance of a student is evaluated by the following procedure:    * If the student does not take the midterm or final examination, the student's grade shall be F.   * If the total score of the midterm and final examination is greater than or equal to 80, the student's grade shall be A.   * If the total score of the midterm and final examination is greater than or equal to 65 and less than 80, the student's grade shall be B.   * If the total score of the midterm and final examination is greater than or equal to 50 and less than 65, the student's grade shall be C.   * If the total score of the midterm and final examination is greater than or equal to 30 and less than 50, the student's grade shall be D. However, if the score of the makeup examination is greater than or equal to 50, the grade shall be C.   * If the total score of the midterm and final examination is less than 30, the student's grade shall be F.

Incorrect submission (verdict: Wrong Answer):
while True:
    m, f, r = map(int, input().split())
    A = m + f + r
    if m == f == r == -1:
        break
    elif m == 0 or f == 0:
        print('F')
    elif A >= 80:
        print('A')
    elif 65 <= A < 80:
        print('B')
    elif 50 <= A < 65:
        print('C')
    elif 30 <= A < 50:
        print('D')
    else:
        print('F')

Fixed, accepted submission:
while True:
    m, f, r = map(int, input().split())
    A = m + f
    if m == f == r == -1:
        break
    elif m == -1 or f == -1:
        print('F')
    elif A >= 80:
        print('A')
    elif 65 <= A < 80:
        print('B')
    elif 50 <= A < 65:
        print('C')
    elif 30 <= A < 50:
        if 50 <= r:
            print('C')
        else:
            print('D')
    else:
        print('F')
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):
n = int(input())
a = sorted([list(map(int, input().split())) for i in range(n - 1)])
color = [-1 for i in range(n)]
color[0] = 1
for i in a:
    u, v, w = i
    u -= 1
    v -= 1
    if w % 2 == 0:
        if color[u] == 0:
            color[v] = 0
        if color[u] == 1:
            color[v] = 1
        if color[v] == 0:
            color[u] = 0
        if color[v] == 1:
            color[u] = 1
    else:
        if color[u] == 0:
            color[v] = 1
        if color[u] == 1:
            color[v] = 0
        if color[v] == 0:
            color[u] = 1
        if color[v] == 1:
            color[u] = 0
for i in color:
    if i == -1:
        print(0)
    else:
        print(i)

Fixed, accepted submission:
import heapq

def dijkstra(graph, node, start):
    dist = [float("inf") for _ in range(node)]

    dist[start] = 0
    q = []
    heapq.heappush(q, (0, start))

    while q:
        cost, cur_node = heapq.heappop(q)

        if dist[cur_node] < cost:
            continue

        for nex_cost, nex_node in graph[cur_node]:
            dist_cand = dist[cur_node] + nex_cost
            if dist_cand < dist[nex_node]:
                dist[nex_node] = dist_cand
                heapq.heappush(q, (dist[nex_node], nex_node))

    return dist


n = int(input())
adj = [[] for _ in range(n)]
for i in range(n - 1):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    adj[a].append((c, b))
    adj[b].append((c, a))
shortest = dijkstra(adj, n, 0)
for i in shortest:
    print(i % 2)
Problem:
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.  Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.  Find the inversion number of B, modulo 10^9 + 7.  Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

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

A = list(map(int, input().split()))

mod = 1000000007


B = A * 2
syokiti_A = 0
syokiti_B = 0
for i in range(N):
    for j in range(i, N):
        if A[i] > A[j]:
            syokiti_A += 1

for i in range(N*2):
    for j in range(i, N*2):
        if B[i] > B[j]:
            syokiti_B += 1

ans = int((syokiti_A + syokiti_B*K*(K-1)/2)) % mod

print(int(ans))



Fixed, accepted submission:
N, K = map(int, input().split())
A = list(map(int, input().split()))
mod = 10**9+7
ans = 0


normal = 0
for i in range(N):
    for j in range(i+1, N):
        if A[i] > A[j]:
            normal += 1
un_normal = 0
for i in range(N):
    for j in range(N):
        if A[i] > A[j]:
            un_normal += 1
print((normal * K + un_normal * K * (K-1) // 2) % mod)
Problem:
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower.  He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i.  To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar.  How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

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

a = sorted(a)
b = sorted(b)
c = sorted(c)

def bs(li, num):
    """return boundary index"""
    l = 0
    r = len(li) - 1
    if li[l] >= num:
        return -1
    elif li[r] < num:
        return r
    while r-l > 1:
        mid = (l+r) // 2
        if li[mid] < num:
            l = mid
        else:
            r = mid
    return l

def bs_1(li, num):
    """return boundary index"""
    l = 0
    r = len(li) - 1
    if li[l] > num:
        return r
    elif li[r] <= num:
        return -1
    while r-l > 1:
        mid = (l+r) // 2
        if li[mid] <= num:
            l = mid
        else:
            r = mid
    return r


ans = 0
for i in range(len(b)):

    c_i = bs_1(c, b[i])
    a_i = bs(a, b[i])
    ans += (n-c_i) * (a_i+1)

print(ans)



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

cnt = [0] * len(b)
a = sorted(a)
c = sorted(c)

for i in range(len(b)):
    target = b[i]

    
    l = 0
    r = len(a) - 1
    if a[l] >= target:
        cnt[i] += 0
    elif a[r] < target: 
        cnt[i] += len(a)
    else:
        while r - l > 1:
            mid = (l + r) // 2
            if a[mid] >= target:
                r = mid
            else:
                l = mid
        cnt[i] += l + 1

    # decide c cnt
    l = 0
    r = len(c) - 1
    if c[l] > target:
        cnt[i] *= len(c)
    elif c[r] <= target: 
        cnt[i] *= 0
    else:
        while r - l > 1:
            mid = (l + r) // 2
            if c[mid] > target:
                r = mid
            else:
                l = mid
        cnt[i] *= len(c) - (l + 1)
print(sum(cnt))
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):
# coding: utf-8
x,a,b=map(int,input().split())
print('A' if abs(x-a)>abs(x-b) else 'B')

Fixed, accepted submission:
# coding: utf-8
x,a,b=map(int,input().split())
print('A' if abs(x-a)<abs(x-b) else 'B')
Problem:
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions:    * 1 \leq x,y,z   * x^2 + y^2 + z^2 + xy + yz + zx = n  Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

Incorrect submission (verdict: Wrong Answer):
n = int(input())
li = [0]*n
for i in range(1,32):
    for j in range(1,32):
        for k in range(1,32): 
            if i*2 + j*2 + k*2 + i*j + j*k + k*i <= n:
                li[i*2 + j*2 + k*2 + i*j + j*k + k*i-1] += 1
for i in range(n):
    print(li[i])

Fixed, accepted submission:
n = int(input())
li = [0]*n
for i in range(1,100):
    for j in range(1,100):
        for k in range(1,100): 
            if i**2 + j**2 + k**2 + i*j + j*k + k*i <= n:
                li[i**2 + j**2 + k**2 + i*j + j*k + k*i -1] += 1
for i in range(n):
    print(li[i])