TnT/CodeNet4Repair huggingface dataset:
Problem:
A railroad running from west to east in Atcoder Kingdom is now complete.  There are N stations on the railroad, numbered 1 through N from west to east.  Tomorrow, the opening ceremony of the railroad will take place.  On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated.  The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds.  Here, it is guaranteed that F_i divides S_i.  That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains.  For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
nums = []
ans = []
for v in range(n-1):
    c, t, f = map(int,input().split())
    te = [c, t, f]
    nums.append(te)

print(nums)

for q in range(0,n-1):
    if(q==n-2):
        time = 0
        time += nums[q][0]
        time += nums[q][1]
        ans.append(time)
        break;
    time = 0
    time += nums[q][0]
    time += nums[q][1]
    for t in range(q+1,n-1):
        p = nums[t][1];
        while(time > p):
            p += nums[t][2]
        p += nums[t][0]
        time = p
    ans.append(p)

ans.append(0)
print(ans)

Fixed, accepted submission:
n = int(input())
c=[]
t=[]
f=[]
for v in range(n-1):
    cn, tn, fn = map(int,input().split())
    c.append(cn)
    t.append(tn)
    f.append(fn)

for q in range(n):
    time=0
    for p in range(q,n-1):
        if time>t[p]:
            while time%f[p]>0:
                time+=1
        else:
            time=t[p]
        time+=c[p]
    print(time)
Problem:
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters.  We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person.  At most how many people can sit on the seat?

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

x,y,z=map(int,input().split())
cnt=z
hum=0
while 1:
    if cnt>x:
        break
    else:
        cnt+=y+z
    hum+=1
print(hum)

Fixed, accepted submission:
# coding: utf-8
x,y,z=map(int,input().split())
cnt=z
hum=0
while 1:
    cnt+=y+z#isu to haba tasu
    if cnt>x:
        break
    hum+=1#koetenakya hairu
print(hum)
Problem:
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes:    * Hit his pocket, which magically increases the number of biscuits by one.   * Exchange A biscuits to 1 yen.   * Exchange 1 yen to B biscuits.  Find the maximum possible number of biscuits in Snuke's pocket after K operations.

Incorrect submission (verdict: Wrong Answer):
k, a, b = map(int, input().split())
biscuit = 1
yen = 0
if b <= a or k < a or (b - a) / 2 < 1:
    print(k + 1)
    exit()
else:
    biscuit += (a - 1)
    k -= (a - 1)

print((b - a) * (k // 2) + biscuit)

Fixed, accepted submission:
k, a, b = map(int, input().split())
biscuit = 1
tmp = 1

if b <= a or k < a or (b - a) / 2 < 1:
    print(k + 1)
    exit()
else:
    biscuit += (a - 1)
    k -= (a - 1)

print((b - a) * (k // 2) + biscuit + (k%2==1 &1))
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())
if (a+b)<=2*c:
  print(a*x+b*y)
else:
  val=2*c*(min(x,y)//2)
  X=x-2*(min(x,y)//2)
  Y=y-2*(min(x,y)//2)
  print(min(val+2*c,val+c+(X-1)*a+(Y-1)*b,val+X*a+Y*b))
    
  
  

Fixed, accepted submission:
import sys
a,b,c,x,y=map(int,input().split())
val=0
if a+b<=2*c:
  print(a*x+b*y)
  sys.exit()
else:
  m=min(x,y)
  val+=m*c*2
  x=x-m
  y=y-m
  if x==0:
    if y==0:
      print(val)
      sys.exit()
    val+=min(b*y,y*2*c)
  if y==0:
    val+=min(a*x,x*2*c)
  print(val)
  sys.exit()
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())
if A>B:
    print("GRATER")
elif A==B:
    print("EQUAL")
else:
    print("LESS")

Fixed, accepted submission:
A = int(input())
B = int(input())
if A>B:
    print("GREATER")
elif A==B:
    print("EQUAL")
else:
    print("LESS")
Problem:
_n_ _q_   _name 1 time1_   _name 2 time2_   ...   _name n timen_    In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space.  In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.

Incorrect submission (verdict: Wrong Answer):
import collections
import sys

n,q = map(int,input().split())
data = [[i for i in input().split()]for i in range(n)]
time = 0
while data:
    task = data[0]
    del data[0]
    if int(task[1]) < q:
        time += int(task[1])
        print(task[0],time)
      
    else:
        time += q
        task[1] = str(int(task[1]) - q)
        data.append(task)

Fixed, accepted submission:
import collections
import sys

n,q = map(int,input().split())
data = [[i for i in input().split()]for i in range(n)]
time = 0
while data:
    task = data[0]
    del data[0]
    if int(task[1]) <= q:
        time += int(task[1])
        print(task[0],time)
      
    else:
        time += q
        task[1] = str(int(task[1]) - q)
        data.append(task)
Problem:
肥満は多くの成人病の原因として挙げられています。過去においては、一部の例外を除けば、高校生には無縁なものでした。しかし、過度の受験勉強等のために運動不足となり、あるいはストレスによる過食症となることが、非現実的なこととはいえなくなっています。高校生にとっても十分関心を持たねばならない問題になるかもしれません。  そこで、あなたは、保健室の先生の助手となって、生徒のデータから肥満の疑いのある生徒を探し出すプログラムを作成することになりました。  方法は BMI (Body Mass Index) という数値を算出する方法です。BMIは次の式で与えられます。      BMI = 22 が標準的で、25 以上だと肥満の疑いがあります。  各々の生徒の体重、身長の情報から BMI を算出し、25 以上の生徒の学籍番号を出力するプログラムを作成してください。

Incorrect submission (verdict: Wrong Answer):
def bmi(w,h):
    return w / (h ** 2)

s,w,h = [],[],[]
while True:
    try:
        tmp = input().split(',')
        s.append(float(tmp[0]))
        w.append(float(tmp[1]))
        h.append(float(tmp[2]))
    except EOFError:
        break
f = []
for i in range(len(s)):
    if bmi(w[i],h[i]) >= 25:
      f.append(s[i])
for i in range(len(f)-1):
    print(f[i])
if len(f) != 1 or len(f) != 0:
    print(f[-1])

Fixed, accepted submission:
def bmi(w,h):
    return w / (h ** 2)

s,w,h = [],[],[]
while True:
    try:
        tmp = input().split(',')
        s.append(int(tmp[0]))
        w.append(float(tmp[1]))
        h.append(float(tmp[2]))
    except EOFError:
        break
f = []
for i in range(len(s)):
    if bmi(w[i],h[i]) >= 25:
      f.append(s[i])
if len(f) != 1 and len(f) != 0:
    for i in range(len(f)-1):
        print(f[i])
if len(f) != 1 and len(f) != 0:
    print(f[-1])
elif len(f) == 1:
    print(f[0])
Problem:
We will call a string x _good_ if it satisfies the following condition:    * Condition: x can be represented as a concatenation of two copies of another string y of length at least 1.  For example, `aa` and `bubobubo` are good; an empty string, `a`, `abcabcabc` and `abba` are not good.  Eagle and Owl created a puzzle on good strings. Find one string s that satisfies the following conditions. It can be proved that such a string always exists under the constraints in this problem.    * 1 ≤ |s| ≤ 200   * Each character of s is one of the 100 characters represented by the integers 1 through 100.   * Among the 2^{|s|} subsequences of s, exactly N are good strings.

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

#

"""
agc012 C
"""


n = int(input())
m = len(bin(n+1))-2
n = n-(2**(m-1)-1)


fact = []
tmp = 1
for i in range(1, m+2):
    fact.append(tmp)
    tmp = tmp*i

binom = [[0]*(m+1) for i in range(m+1)]
for i in range(m+1):
    for j in range(i+1):
        binom[i][j] = (fact[i]//(fact[j]*fact[i-j]))

binomsums = []
for j in range(m//2+1):
    tmp = 1
    for i in range(1, j+1):
        tmp += binom[j][i]*binom[m-j][i]
    binomsums.append(tmp)

a = [0]*(m//2+1)
ind = m//2
while n != 0:
    b = binomsums.pop()
    c, n = divmod(n, b)
    a[ind] = c
    ind -= 1

ans = []
i = 2
for j in range(m//2+1):
    for k in range(a[j]):
        ans.append(i)
        i += 1
    ans.append(1)
ans = ans + [1 for j in range(m-m//2-1)]
i -= 1
while i >= 2:
    ans.append(i)
    i -= 1
print(" ".join(map(str, ans)))


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

#

"""
agc012 C
"""


n = int(input())
m = len(bin(n+1))-2
n = n-(2**(m-1)-1)


fact = []
tmp = 1
for i in range(1, m+2):
    fact.append(tmp)
    tmp = tmp*i

binom = [[0]*(m+1) for i in range(m+1)]
for i in range(m+1):
    for j in range(i+1):
        binom[i][j] = (fact[i]//(fact[j]*fact[i-j]))

binomsums = []
for j in range(m//2+1):
    tmp = 1
    for i in range(1, j+1):
        tmp += binom[j][i]*binom[m-j][i]
    binomsums.append(tmp)

a = [0]*(m//2+1)
ind = m//2
while n != 0:
    b = binomsums.pop()
    c, n = divmod(n, b)
    a[ind] = c
    ind -= 1

ans = []
i = 2
for j in range(m//2+1):
    for k in range(a[j]):
        ans.append(i)
        i += 1
    ans.append(1)
ans = ans + [1 for j in range(m-m//2-1)]
i -= 1
while i >= 2:
    ans.append(i)
    i -= 1
print(len(ans))
print(" ".join(map(str, ans)))
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):
x,a,b = map(int, input().split())
print(min(abs(x-a), abs(x-b)))

Fixed, accepted submission:
x,a,b = map(int, input().split())
if abs(x-a) > abs(x-b):
    print("B")
else:
    print("A")
Problem:
There is a directed graph G with N vertices and M edges. The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i. G **does not contain directed cycles**.  Find the length of the longest directed path in G. Here, the length of a directed path is the number of edges in it.

Incorrect submission (verdict: Wrong Answer):
# -*- coding: utf-8 -*-
"""
Created on Fri Jan 24 14:37:00 2020

@author: matsui
"""

N,M=map(int,input().split())

g=dict()
h=[0]*(N+1)
for i in range(1,N+1):
    g[i]=[]
  
for i in range(M):
    x,y=map(int,input().split())
    g[x].append(y)
    h[y]+=1
     
st=[]

for i in range(1,N+1):
    if h[i]==0: 
        st.append(i)
    
ans=[]
dp=[0]*(N+1)

while(len(st)):
    i=st[0]
    st.remove(st[0])
    
    ans.append(i)
    
    for j in g[i]:
        h[j]-=1
        if h[j]==0: 
            st.append(j)
            if dp[i]+1>dp[j]: 
                dp[j]=dp[i]+1
        
print(ans)
print(max(dp))

Fixed, accepted submission:
# -*- coding: utf-8 -*-
"""
Created on Fri Jan 24 14:37:00 2020

@author: matsui
"""

N,M=map(int,input().split())

g=dict()
h=[0]*(N+1)
for i in range(1,N+1):
    g[i]=[]
  
for i in range(M):
    x,y=map(int,input().split())
    g[x].append(y)
    h[y]+=1
     
st=[]

for i in range(1,N+1):
    if h[i]==0: 
        st.append(i)
    
ans=[]
dp=[0]*(N+1)

while(len(st)):
    i=st[0]
    st.remove(st[0])
    
    ans.append(i)
    
    for j in g[i]:
        h[j]-=1
        if h[j]==0: 
            st.append(j)
            if dp[i]+1>dp[j]: 
                dp[j]=dp[i]+1
        
#print(ans)
print(max(dp))
Problem:
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i.  How many squares i satisfy both of the following conditions?    * The assigned number, i, is odd.   * The written integer is odd.

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

n = int(input())
for i in range(n):
  high_lim = int(math.sqrt(i+1)//1) + 1

  l1 = [i for i in range(1, high_lim)]
  l2 = [i for i in range(1, high_lim)]
  l3 = [i for i in range(1, high_lim)]

  p = itertools.product(l1, l2, l3)

  result = 0
  for pair in set([tuple(sorted(pair)) for pair in p]):
    if pair[0]**2 + pair[1]**2 + pair[2]**2 + pair[0]*pair[1] + pair[1]*pair[2] + pair[0]*pair[2] == i + 1:
      result += 1
  print(result)

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

result = 0
for i, num in enumerate(num_lst):
  if i % 2 == 0:
    if num % 2 == 1:
      result += 1

print(result)
Problem:
You are given three strings A, B and C. Check whether they form a _word chain_.  More formally, determine whether both of the following are true:    * The last character in A and the initial character in B are the same.   * The last character in B and the initial character in C are the same.  If both are true, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
s=input().split()
if s[0][-1] == s[1][0] and s[1][-1] == s[2][0]:
  print("Yes")
else:
  print("No")

Fixed, accepted submission:
s=input().split()
if s[0][-1] == s[1][0] and s[1][-1] == s[2][0]:
  print("YES")
else:
  print("NO")
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):
def main():
    s = input()
    ans = []
    for op in s:
        if op == '0':
            ans.append(op)
        elif op == '1':
            ans.append(op)
        else:
            if len(ans) > 0:
                ans.pop()
    print(*ans)
if __name__ == '__main__':
    main()


Fixed, accepted submission:
def main():
    s = input()
    ans = []
    for op in s:
        if op == '0':
            ans.append(op)
        elif op == '1':
            ans.append(op)
        else:
            if len(ans) > 0:
                ans.pop()
    print(*ans, sep='')
if __name__ == '__main__':
    main()
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())
road = [0]*N
for i in range(M):
    a, b = map(int,input().split())
    road[a-1] += 1
    road[b-1] += 1
for i in road:
    print(road)

Fixed, accepted submission:
N, M = map(int, input().split())
road = [0]*N
for i in range(M):
    a, b = map(int,input().split())
    road[a-1] += 1
    road[b-1] += 1
for i in road:
    print(i)
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())

if a>b:
  print("GRATER")
elif a<b:
  print("LESS")
else:
  print("EQUAL")


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

if a<b:
  print("LESS")
elif a>b:
  print("GREATER")
else:
  print("EQUAL")
Problem:
Write a program which manipulates a sequence A = {a0, a1, . . . , an-1} with the following operations:    * find(s, t): report the minimum element in as, as+1, . . . ,at.   * update(i, x): change ai to x.  Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.

Incorrect submission (verdict: Wrong Answer):

Fixed, accepted submission:
# Range Minimum Query (RMQ)
INF = 2**31 - 1
D = [INF]
[n, q] = list(map(int, input().split()))

def initRMQ(n_):
    global D, n
    size = 1
    while size < n_:
        size *= 2
    i = 1
    while i <= 2*size - 1 - 1:
        D.append(INF)
        i += 1
    n = size

def update(k, a):
    global n
    k += n - 1
    D[k] = a
    while k > 0:
        k = int((k - 1)/2)
        D[k] = min(D[k*2 + 1], D[k*2 + 2])

def query(a, b, k, l, r):
    global INF, D
    if r <= a or b <= l:
        return INF
    if a <= l and r <= b:
        return D[k]

    vl = query(a, b, k*2 + 1, l, int((l + r)/2))
    vr = query(a, b, k*2 + 2, int((l + r)/2), r)
    return min(vl, vr)

def findMin(a, b):
    global n
    return query(a, b + 1, 0, 0, n)

initRMQ(n)

for i in range(q):
    data = list(map(int, input().split()))
    if data[0] == 0:
        update(data[1], data[2])
    if data[0] == 1:
        print(findMin(data[1], data[2]))
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):
x=input().split()
a=int(x[0])
op=x[1]
b=int(x[2])
if op=="+":
  print(a+b)
elif op=="-":
   print(a-b)
elif op=="*":
   print(a*b)
elif op=="/":
   print(a/b)
else:
    print()


Fixed, accepted submission:
while True:
    x=input().split(" ")
    a=int(x[0])
    op=x[1]
    b=int(x[2])
    if op == "+":
        print(a + b)
    elif op == "-":
        print(a - b)
    elif op == "*":
        print(a * b)
    elif op == "/":
        print(a // b)
    elif op == "?":
        break
Problem:
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.

Incorrect submission (verdict: Wrong Answer):
n,k=map(int,input().split())
print("Yes" if n//2+n%2>=k else "No")

Fixed, accepted submission:
n,k=map(int,input().split())
print("YES" if n//2+n%2>=k else "NO")
Problem:
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it.  The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).  Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer.  If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

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

x=N*100/108
print((N*100)%108)
if (N*100)%108==0:
  print('{:.0f}'.format(x))
else:
  print(":(")



Fixed, accepted submission:
import math

N = int(input())*100

x=N//108

if (math.floor(x*108/100)==N/100):
  print('{:.0f}'.format(x))
elif (math.floor((x+1)*108/100)==N/100):
  print('{:.0f}'.format(x+1))
else:
  print(":(")
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):
from collections import Counter

A, B, C = map(int, input().split())

c = Counter([A, B, C])
if c[5] == 2 and c[7] == 1:
    print('Yes')
else:
    print('No')

Fixed, accepted submission:
from collections import Counter

A, B, C = map(int, input().split())

c = Counter([A, B, C])
if c[5] == 2 and c[7] == 1:
    print('YES')
else:
    print('NO')
Problem:
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up.   Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1.   Here, both the x\- and y-coordinates before and after each movement must be integers.   He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy).   Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty).   Under this condition, find a shortest path for him.

Incorrect submission (verdict: Wrong Answer):
sx,sy,tx,ty = map(int, input().split())
ans = ""
if sx > tx: ans += "L" * (sx-tx)
else: ans += "R" * (tx-sx)
if sy > ty: ans += "D" * (sy-ty+1)
else: ans += "U" * (ty-sy+1)

if sx > tx: ans += "R" * (sx-tx+1)
else: ans += "L" * (tx-sx+1)
if sy > ty: ans += "U" * (sy-ty+1)
else: ans += "D" * (ty-sy+1)

if sx > tx: ans += "L"
else: ans += "R" 

print(ans)

Fixed, accepted submission:
sx, sy, tx, ty = map(int, input().split())

ans = []
ans += (ty-sy) * "U" + (tx-sx) * "R" + (ty-sy) * "D" + (tx-sx) * "L"

ans += "L" + (ty-sy+1) * "U" + (tx-sx+1) * "R" + "D"
ans += "R" + (ty-sy+1) * "D" + (tx-sx+1) * "L" + "U"

print("".join(ans))
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("odd")
else:
    print("even")

Fixed, accepted submission:
a, b = map(int, input().split())
if (a*b) % 2 != 0:
    print("Odd")
else:
    print("Even")
Problem:
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode:                 for i = 1 to A.length-1     key = A[i]     /* insert A[i] into the sorted sequence A[0,...,j-1] */     j = i - 1     while j >= 0 and A[j] > key         A[j+1] = A[j]         j--     A[j+1] = key   Note that, indices for array elements are based on 0-origin.  To illustrate the algorithms, your program should trace intermediate result for each step.

Incorrect submission (verdict: Wrong Answer):

n=int(input("n:"))
A=[0 for i in range(n)]

A=input().split()
for i in range(n):
    A[i]=int(A[i])

print(A)

for i in range(n):
    key=A[i]
    j=i-1
    while j>=0 and A[j]>key:
        A[j+1]=A[j]
        j=j-1
    A[j+1]=key
    print(A)



Fixed, accepted submission:
n=int(input())
A=[0 for i in range(n)]

A=input().split()

for i in range(n):
    A[i]=(int)(A[i])

for i in range(n):
    key=A[i]
    j=i-1
    while j>=0 and A[j]>key:
        A[j+1]=A[j]
        j=j-1
        A[j+1]=key

    for i in range(n):
        if i!=n-1:
            print(A[i],end=" ")
        if i==n-1:
            print(A[i])