Count numbers have all 1s together in binary representation in C++

We are given a positive integer N. The goal is to count the numbers less than or equal to N that have all 1’s in their binary representation. For example 1 is 1, 3 is 11, 7 is 111, 15 is 1111... so on.

If we see the numbers then all of them are 2i-1. Where i start from 1. To check such numbers less than n. We’ll compare if 2i-1<=n. Then increment count.

Let’s understand with examples.

Input − N=15

Output − Number having all 1's in binary : 4

Explanation − Numbers as sum of same primes − The numbers will be 1 3 7 15

Input − N=50

Output − Number having all 1's in binary : 5

Explanation − Numbers as sum of same primes −

Approach used in the below program is as follows

We take a positive integer N..
Function allOnes(int n) takes n as input and returns the numbers that have all 1’s in binary representation.
Take the initial variable count as 0 for such numbers..
Traverse from i=1 to i<=n using for loop.
For each i, if pow(2,i)-1 is less than or equal to n. Increment count.
Return the count as a result at the end of the for loop.

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int allOnes(int n){
   int count = 0;
   for(int i=1;i<=n;i++){
      if(n>=pow(2,i)-1){
         count++;
         //cout<<" "<<pow(2,i)-1;
      }
   }
   return count;
}
int main(){
   int N=23;
   cout <<endl<< "Number having all 1's in binary : "<<allOnes(N);
   return 0;
}