Codechefs starters 106->div 3 : B-> Playing with OR

Statement:  Playing with OR click the link and also click the link what u get after clicking on Playing with OR link.that will direct u to the problem statement.

Explanation:  

look if we have any odd number in any sub-array of length k.then it is obvious that the OR of all elements of this sub-array will also be odd.

Example: 5->101 ,7->111 look 5,7 has 1 as set bit on the

 0'th bit,so if u do OR with any other number like 4->100 or 3->011 the 0'th bit of the resultant number

 will also be 1. because  0|1 and 1|1 is 1 always so,we can eliminate 1 from the 0'th bit.So,when we do OR

 in the sub-array of length K ,then one odd number will be enough to give us odd OR value .

So,we just need to check if there is any odd number in the all sub-array of length K in array a.

Code:

#include<bits/stdc++.h>

using namespace std;

#define ll unsigned long long

#define db double

#define p_b push_back

#define pp endl

int main() {

    ll t; cin>>t;

    while(t--){

        ll n,k,c=0,cnt=0,res=0,d=0,p; cin>>n>>k;

        ll a[n];

        ll i,f1,j;

        for(i=0;i<n;i++) {

            cin>>a[i];

         

           

        }

        for(i=0;i<=n-k;i++){

            for(j=i;j<i+k;j++){

                if(a[j]%2){

                    cnt++;

                    break;

                }

            }

        }

        

        cout<<cnt<<pp;

    }

    

}🍌