Divisors of n-square that are not divisors of n in C++ Program

In this tutorial, we are going to write a program that finds the divisors count of n-square and not n.

It's a straightforward problem. Let's see the steps to solve the problem.

• Initialize the number n.

• Initialize a counter for divisors.

• Iterate from 2 to n^2n2?.

  • If the n^2n2? is divisible by the current number and nn? is not divisible by the current number, then increment the count.

• Print the count.

Example

Let's see the code.

 Live Demo

#include <bits/stdc++.h> using namespace std; int getNumberOfDivisors(int n) {    int n_square = n * n;    int divisors_count = 0;    for (int i = 2; i <= n_square; i++) {       if (n_square % i == 0 && n % i != 0) {          divisors_count++;       }    }    return divisors_count; } int main() {    int n = 6;    cout << getNumberOfDivisors(n) << endl;    return 0; }

Output

If you execute the above program, then you will get the following result.

5

Conclusion

If you have any queries in the tutorial, mention them in the comment section.