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--)