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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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