Count numbers whose XOR with N is equal to OR with N in C++

We are a number N. The goal is to find numbers between 0 and N whose OR with N is equal to XOR with N.

We will do this by traversing no. from i=0 to i<=N and for each i, if (N^i==i | N) then increment count.

Let us understand with examples.

Input − X=6

Output − Count of numbers whose OR with N == XOR with N: 2

Explanation − Numbers are 0 1.

Input − X=20

Output − Count of numbers whose OR with N == XOR with N: 8

Explanation − Numbers are 0 1 2 3 8 9 10 11

Approach used in the below program is as follows

• We take integer N.

• Function orisXOR(int n) takes n and returns a count of numbers whose OR with n is equal to XOR with n.

• Take the initial count as 0.

• Traverse from i=0 to i<=n.

• If i|n==i^n. Increment count

• At the end of for loop count will have the desired result.

• Return count and print.

Example

 Live Demo

#include <bits/stdc++.h>
#include <math.h>
using namespace std;
int orisXOR(int n){
   int count = 0;
   for (int i = 0; i <= n; i++){
      if((n|i)==(i^n))
         { count++; }
   }
   return count;
}
int main(){
   int N = 15;
   int nums=orisXOR(N);
   cout <<endl<<"Count of numbers whose OR with N == XOR with N: "<<nums;
   return 0;
}

Output

If we run the above code it will generate the following output −

Count of numbers whose OR with N == XOR with N: 1