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.

#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.

Conclusion

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

Hafeezul Kareem

Published on 28-Jan-2021 06:49:07

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