TnT/CodeNet4Repair huggingface dataset:
Problem:
You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**.  Here, an sequence b is a **good sequence** when the following condition holds true:    * For each element x in b, the value x occurs exactly x times in b.  For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not.  Find the minimum number of elements that needs to be removed so that a will be a good sequence.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = list(map(int,input().split()))
dic = {}
for i in range(n):
    if a[i] in dic:
        dic[a[i]] += 1
    else:
        dic[a[i]] = 1
ans = 0
for i in dic:
    if dic[i] < i:
        ans += dic[i]
    else:
        ans += i-dic[i]
print(ans)

Fixed, accepted submission:
n = int(input())
a = list(map(int,input().split()))
dic = {}
for i in range(n):
    if a[i] in dic:
        dic[a[i]] += 1
    else:
        dic[a[i]] = 1
ans = 0
for i in dic:
    if dic[i] < i:
        ans += dic[i]
    else:
        ans += dic[i]-i
print(ans)
Problem:
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.

Incorrect submission (verdict: Wrong Answer):
import math
import itertools

a, b = map(str, input().split())
C = a+b
c = int(C)

I=c
s=0
R=int(I)
L=[]
while s==0:
    for i in range(2,R+1):
        if I%i==0:
            I=I/i
            if I==1:
                s=1
            L.append(i)
            break
index =0
sameCCount =0
flag =0
while index < len(L)-1:
    if L[index] == L[index+1]:
        sameCCount+=1
        print(sameCCount, index)
    else:
        if sameCCount%2 ==0 or sameCCount==0:
            flag=1
            break
        else:
            sameCCount=0
    index+=1

if flag ==0:
    print("Yes")
else:
    print("No")

Fixed, accepted submission:
import math
import itertools

a, b = map(str, input().split())
C = a+b
c = int(C)

I=c
s=0
R=int(I)
L=[]
while s==0:
    for i in range(2,R+1):
        if I%i==0:
            I=I/i
            if I==1:
                s=1
            L.append(i)
            break
beforeC =""
count=0
index=0
flag=0
while index < len(L):
    if beforeC =="":
        beforeC = L[index]
        count+=1
    else:
        if L[index] == beforeC:
            count+=1
        else:
            if count%2==0:
                count =1
                beforeC = L[index]
            else:
                flag=1
                break
    index+=1
if count==1:
    flag=1

if flag ==1:
    print("No")
else:
    print("Yes")
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 = 10 ** 6

for a in A:
  c = 0
  while a % 2 == 0:
    a //= 2
  ans = min(c,ans)
  
print(ans)

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

ans = 10 ** 6

for a in A:
  c = 0
  while a % 2 == 0:
    a //= 2
    c += 1
  ans = min(c,ans)
  
print(ans)
Problem:
Write a program which reads a $ n \times m$ matrix $A$ and a $m \times 1$ vector $b$, and prints their product $Ab$.  A column vector with m elements is represented by the following equation.  \\[ b = \left( \begin{array}{c} b_1 \\\ b_2 \\\ : \\\ b_m \\\ \end{array} \right) \\]  A $n \times m$ matrix with $m$ column vectors, each of which consists of $n$ elements, is represented by the following equation.  \\[ A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1m} \\\ a_{21} & a_{22} & ... & a_{2m} \\\ : & : & : & : \\\ a_{n1} & a_{n2} & ... & a_{nm} \\\ \end{array} \right) \\]  $i$-th element of a $m \times 1$ column vector $b$ is represented by $b_i$ ($i = 1, 2, ..., m$), and the element in $i$-th row and $j$-th column of a matrix $A$ is represented by $a_{ij}$ ($i = 1, 2, ..., n,$ $j = 1, 2, ..., m$).  The product of a $n \times m$ matrix $A$ and a $m \times 1$ column vector $b$ is a $n \times 1$ column vector $c$, and $c_i$ is obtained by the following formula:  \\[ c_i = \sum_{j=1}^m a_{ij}b_j = a_{i1}b_1 + a_{i2}b_2 + ... + a_{im}b_m \\]

Incorrect submission (verdict: Wrong Answer):
n, m = map(int, input().split())
M1 = []
M2 = []
for i in range(n):
 M1.append(list(map(int, input().split())))

for i in range(m):
 M2.append(int(input()))

for i in range(n):
 c = 0
 for s in range(m):
  c += M1[i][s] + M2[s]
 print(c)

Fixed, accepted submission:
n, m = map(int, input().split())
M1 = []
M2 = []
for i in range(n):
 M1.append(list(map(int, input().split())))

for i in range(m):
 M2.append(int(input()))

for i in range(n):
 c = 0
 for s in range(m):
  c += M1[i][s] * M2[s]
 print(c)
Problem:
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value:    * Operation A: The displayed value is doubled.   * Operation B: The displayed value increases by K.  Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
K = int(input())

def ans(i,x):
    print(i)
    if i == 0:
        return min(2*x, x+K)
    else:
        return min(ans(i-1, 2*x), ans(i-1, x+K))

print(ans(N-1, 1))


Fixed, accepted submission:
N = int(input())
K = int(input())

def ans(i,x):
    if i == 0:
        return min(2*x, x+K)
    else:
        return min(ans(i-1, 2*x), ans(i-1, x+K))

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

Incorrect submission (verdict: Wrong Answer):
import sys

def dfs(s):
    color = [0]*n
    visited = [False]*n
    visited[s] = True
    q = [s]
    while q:
        p = q.pop()
        visited[p] = True
        for d,v in edge[p]:
            if visited[v]:
                continue
            if d%2 != 0:
                color[v] = 1
            q.append(v)
    return color

if __name__ == "__main__":
    n = int(input())
    edge = [[] for i in range(n)]
    for i in range(n-1):
        u,v,w = map(int,input().split())
        u-=1
        v-=1
        edge[u].append([w,v])
        edge[v].append([w,u])
    ans = dfs(0)
    print(*ans)

Fixed, accepted submission:
import sys
INF = 10**18
import sys
from heapq import heapify,heappop,heappush,heappushpop


class PriorityQueue:
    
    
    def __init__(self,heap):
        self.heap = heap
        heapify(self.heap)
    
    
    def push(self,item):
        heappush(self.heap,item)
    
    
    def pop(self):
        return heappop(self.heap)
    
    
    def pushpop(self,item):
        return heappushpop(self.heap,item)
    
    
    def __call__(self):
        return self.heap
    
    
    def __len__(self):
        return len(self.heap)

def dijkstra(s):
    
    d = [INF] * n
    used = [True] * n 
    d[s] = 0
    used[s] = False
    edgelist = []
    q = PriorityQueue(edgelist)
    for e in edge[s]:
        q.push(e)
    while len(edgelist):
        minedge = q.pop()
        
        if not used[minedge[1]]:
            continue
        v = minedge[1]
        d[v] = minedge[0]
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                q.push([e[0]+d[v],e[1]])
    return d

if __name__ == "__main__":
    n = int(input())
    edge = [[] for i in range(n)]
    for i in range(n-1):
        u,v,w = map(int,input().split())
        u-=1
        v-=1
        edge[u].append([w,v])
        edge[v].append([w,u])
    ans = dijkstra(0)
    c_ans = [0]*n
    for i in range(n):
        if ans[i]%2 == 0:
            c_ans[i] = 0
        else:
            c_ans[i] = 1
    for i in range(n):
        print(c_ans[i])
Problem:
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

Incorrect submission (verdict: Wrong Answer):
S=list(map(str,input().split()))
s=S[::2]
print(s)

Fixed, accepted submission:
S=input()
print(S[::2])
Problem:
In this problem, we use the 24-hour clock.  Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?

Incorrect submission (verdict: Wrong Answer):
#A - Study Scheduling

def answ():
    h1, m1, h2,m2,k = map(int, input().split())
    h = h2 - h1
    m = m2 - m1
    if m < 0:
        m = m + 60
        h = h - 1
    time = h*60 + m 
    return time - k
answ()

Fixed, accepted submission:
h1, m1, h2,m2,k = map(int, input().split())
if h1 > h2:
    h2 +=24
h = h2 - h1
m = m2 - m1
if m < 0:
    m = m + 60
    h = h - 1
time = h*60 + m 
print(time - k)
Problem:
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right.  Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N.  Find the minimum cost incurred before reaching the goal.

Incorrect submission (verdict: Wrong Answer):
N,M,X = map(int,input().split())
A = list(map(int,input().split()))
for i,x in enumerate(A):
  if x > X:
    t = i
    break

print(t,M-t)

Fixed, accepted submission:
N,M,X = map(int,input().split())
A = list(map(int,input().split()))
for i,x in enumerate(A):
  if x > X:
    t = i
    break

print(min(t,M-t))
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):
import math
n = int(input())
print(int(math.sqrt(2*n))+1)

Fixed, accepted submission:
n = int(input())
num = 0
for i in range(n+1):
	num += i
	if n<=num:
		print(i)
		break
Problem:
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b.  The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.

Incorrect submission (verdict: Wrong Answer):
while True :
    i = input().split()
    r = int(i[0])
    op = i[1]
    l = int(i[2])
    if op == '?':
        break
    else:
        if op == '+':
            print(str(r+l))
        elif op == '-':
            print(str(r-l))
        elif op == '*':
            print(str(r*l))
        else :
            print(str(r/l))
        

Fixed, accepted submission:
while True :
    i = input().split()
    r = int(i[0])
    op = i[1]
    l = int(i[2])
    if op == '?':
        break
    else:
        if op == '+':
            print(str(r+l))
        elif op == '-':
            print(str(r-l))
        elif op == '*':
            print(str(r*l))
        else :
            print(str(r//l))
Problem:
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives.  You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative.  For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins.  If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?

Incorrect submission (verdict: Wrong Answer):
a=int(input())

res = float(0.0)

for i in range(a):
  b=input().split()
  if b[1] == "JPY":
    res += float(b[0])
  if b[1] == "BTC":
    res += float(b[0])*38000

print(res)

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

res = float(0.0)

for i in range(a):
  b=input().split()
  if b[1] == "JPY":
    res += float(b[0])
  if b[1] == "BTC":
    res += float(b[0])*380000

print(res)
Problem:
You are given three integers A, B and C.  Determine if there exists an equilateral triangle whose sides have lengths A, B and C.

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

if A == B == C:
    print("YES")
else:
    print("NO")

Fixed, accepted submission:
A, B, C = map(int, input().split())

if A == B == C:
    print("Yes")
else:
    print("No")
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 = int(input())
b = int(input())
h = int(input())

print((a+b) * h / 2)

Fixed, accepted submission:
a = int(input())
b = int(input())
h = int(input())

print(int((a+b) * h / 2))
Problem:
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key.  To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:    * The `0` key: a letter `0` will be inserted to the right of the string.   * The `1` key: a letter `1` will be inserted to the right of the string.   * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.  Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?

Incorrect submission (verdict: Wrong Answer):
import queue
q = queue.LifoQueue()
for s in input():
    if s =="B":
        if q.empty():
            continue
        q.get()
    else:q.put(s)
while not q.empty():
    print(q.get(),end="")

Fixed, accepted submission:
a = []
for s in input():
    if "B" in s:
        if not a:continue
        else:a.pop()
    else:a.append(s)
print(*a,sep="")
Problem:
A bipartite graph G = (V, E) is a graph in which the vertex set V can be divided into two disjoint subsets X and Y such that every edge e ∈ E has one end point in X and the other end point in Y.  A matching M is a subset of edges such that each node in V appears in at most one edge in M.  Given a bipartite graph, find the size of the matching which has the largest size.

Incorrect submission (verdict: Wrong Answer):
from collections import defaultdict
def dfs(here):
    used[here] = 1
    for near in connect[here]:
        if not match[near] or (not used[near] and dfs(near)):
            match[here] = near
            match[near] = here
            return 1
        return 0
    
a_num, b_num, e_num = (int(n) for n in input().split(" "))
connect = defaultdict(list)
all_v = a_num + b_num
match = [0 for n in range(all_v)]
used = [0 for n in range(all_v)]
for _ in range(e_num):
    v1, v2 = (int(n) for n in input().split(" "))
    v2 += a_num + 1
    connect[v1].append(v2)
    connect[v2].append(v1)
answer = 0
for i in range(all_v):
    if not match[i] and dfs(i):
        answer += 1
print(answer)

Fixed, accepted submission:
from collections import defaultdict
def dfs(here,used,match,connect):
    used[here] = 1
    for near in connect[here]:
        m = match[near]
        if m < 0 or (not used[m] and dfs(m,used,match,connect)):
            match[here] = near
            match[near] = here
            return 1
    return 0

a_num, b_num, e_num = (int(n) for n in input().split(" "))
connect = defaultdict(list)
all_v = a_num + b_num
match = [-1 for n in range(all_v)]
for _ in range(e_num):
    v1, v2 = (int(n) for n in input().split(" "))
    v2 += a_num
    connect[v1].append(v2)
    connect[v2].append(v1)
answer = 0
for i in range(all_v):
    if match[i] < 0:
        used = [0 for n in range(all_v)]
        if dfs(i,used,match,connect):
            answer += 1
print(answer)
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=int(input())
b=int(input())
h=int(input())
s=(a+b)*h/2
print(s)

Fixed, accepted submission:
a=int(input())
b=int(input())
h=int(input())
s=(a+b)*h/2
s=int(s)
print(s)
Problem:
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime.  You are given Q queries.  In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

Incorrect submission (verdict: Wrong Answer):
Q = int(input())
lr = list(list(map(int,input().split())for i in range(Q)))
pn = [0,0]+[1]*(10**5-2)
O = 0
for i in range(2,317):
    for j in range(i*2,10**5,i):
        pn[j] = 0
p = pn.copy()
for i in range(7,10**5):
    if p[i]==1:
        if(pn[int((i+1)/2)]==0):
            p[i] = 0
tmp = 0
p[2] =0
for i in range(10**5):
    p[i] = tmp+p[i]
    tmp = p[i]
print(p[:30])
for l,r in lr:
    print(p[r]-p[l-1])

Fixed, accepted submission:
Q = int(input())
lr = list(list(map(int,input().split())for i in range(Q)))
pn = [0,0]+[1]*(10**5-2)
O = 0
for i in range(2,317):
    for j in range(i*2,10**5,i):
        pn[j] = 0
p = pn.copy()
for i in range(7,10**5):
    if p[i]==1:
        if(pn[int((i+1)/2)]==0):
            p[i] = 0
tmp = 0
p[2] =0
for i in range(10**5):
    p[i] = tmp+p[i]
    tmp = p[i]
for l,r in lr:
    print(p[r]-p[l-1])
Problem:
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`.  You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Incorrect submission (verdict: Wrong Answer):
x,y=map(int, input().split())
S=input()
if len(S)==x+y and S[x]=='-':
  print('Yes')
else:
  print('No')

Fixed, accepted submission:
x,y=map(int, input().split())
S=input()
if len(S)==x+y+1 and S[x]=='-' and S.count('-')==1:
  print('Yes')
else:
  print('No')
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()))

cnt=0
a=0

if n==1:print(0)
else:
  for i in range(n-1):
    for j in range(i+1,n):
      if A[i]>A[j]:
        cnt+=1

  for i in range(n):
    for j in range(n):
      if A[i]>A[j]:
        a+=1

      
print((cnt*k+a*k*(k-1)/2)%(10**9+7))


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

cnt=0
a=0

if n==1:print(0)
else:
  for i in range(n-1):
    for j in range(i+1,n):
      if A[i]>A[j]:
        cnt+=1

  for i in range(n):
    for j in range(n):
      if A[i]>A[j]:
        a+=1

      
  print(int(cnt*k+a*(k*(k-1)//2))%(10**9+7))
Problem:
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode:                 BubbleSort(A) 1 for i = 0 to A.length-1 2     for j = A.length-1 downto i+1 3         if A[j] < A[j-1] 4             swap A[j] and A[j-1]   Note that, indices for array elements are based on 0-origin.  Your program should also print the number of swap operations defined in line 4 of the pseudocode.

Incorrect submission (verdict: Wrong Answer):
def bubbleSort(A,N):
    count = 0
    flag = 1
    while flag:
        flag = 0
        for i in range(1,N):
            if A[N-i] < A[N-i-1]:
                A[N-i] , A[N-i-1] = A[N-i-1] , A[N-i]
                flag = 1
                count += 1
    return (count)

n = int(input())
data = [int(i) for i in input().split()]
print(bubbleSort(data,n))

Fixed, accepted submission:
def bubbleSort(A,N):
    count = 0
    flag = 1
    while flag:
        flag = 0
        for i in range(1,N):
            if A[N-i] < A[N-i-1]:
                A[N-i] , A[N-i-1] = A[N-i-1] , A[N-i]
                flag = 1
                count += 1
    return (count)

n = int(input())
data = [int(i) for i in input().split()]
a = bubbleSort(data,n)
print(" ".join(map(str,data)))
print(a)