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# 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

#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

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