prkagrawal

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prkagrawal

"""Autogenerated input type of AcceptTopicSuggestion"""
input AcceptTopicSuggestionInput {
  """The Node ID of the repository."""
  repositoryId: ID!

  """The name of the suggested topic."""
  name: String!

  """A unique identifier for the client performing the mutation."""
  clientMutationId: String

prkagrawal

t = int(input())
while(t>0):
   t = t - 1
   n = int(input())
   if(n==1) :
      print(1)
      continue
   if(n%2==0):
      print(int(n/2))
      continue

prkagrawal

# This is a "dynamic programming" (aka dp) question and the solution is "recursive"
# Using loop to make the code run until provided non-acceptable input
while True:
    n = str(input())
    # check if given no is zero
    if n == "0":
        break
    dp = []
    # Make an array with a length len(n)+1 with each element as zero
    # We will use this array 

prkagrawal

k,n,w = str(input()).split()
k = int(k)
n = int(n)
w = int(w)
mn = 0
for i in range(w):
    mn = mn + (i+1)*k
if n<mn: 
    print(mn-n)
else:

prkagrawal

n = int(input())
# Part 1
for i in range(n):
    s = []
    for j in range(n):
        if i+j-n >= 0:
            s.append(str(i+j-n))
        else:
            s.append(" ")
    t = s[::-1]

prkagrawal

def stringPatternSearch(text,pattern):
    n = len(text)
    m = len(pattern)
    count = 0
    for i in range(n - m + 1):
        for j in range(m):
            if pattern[j] != text[i+j]:
                break
            if j == m-1 :
                count = count + 1

prkagrawal

import random

t = str(random.randint(1000,9999))

def check():
    k = str(input("Enter a four digit number : "))
    if k == t:
        print("Correct : Game Over")
        return
    elif k != t: 

prkagrawal

n,t = map(int,input().split())
s = str(input())

l = list(s)
z = len(s) - 1
for i in range(t):
    for a in range(z):
        if l[a]=='B' and l[a+1]=='G':
            l[a] = 'G'
            l[a+1] = 'B'

prkagrawal

def trailing_zeroes(n):
    # 0s are produced when 2 and 5 are multiplied
    # so you'll need to count how many 2s and 5s are there
    # 2s are always more than 5s so count just 5s
    res = 0
    k = 5
 
    # find all powers of 5
    # 25 has 2 5s, 125 has 3 5s, etc
    while k <= n: