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

C++Server Side ProgrammingProgramming

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
raja
Published on 31-Oct-2020 05:28:48
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