C Program to Check if count of divisors is even or odd

Given a number “n”, find its total number of divisors is even or odd.

Examples :

Input  : n = 10      
Output : Even

Input:  n = 100
Output: Odd

Input:  n = 125
Output: Even

A naive approach would be to find all the divisors and then see if the total number of divisors is even or odd.

C

// Naive Solution to
// find if count of
// divisors is even
// or odd
#include <math.h>
#include <stdio.h>

// Function to count
// the divisors
void countDivisors(int n)
{
    // Initialize count
    // of divisors
    int count = 0;

    // Note that this
    // loop runs till
    // square root
    for (int i = 1; i <= sqrt(n) + 1; i++) {
        if (n % i == 0)

            // If divisors are
            // equal increment
            // count by one
            // Otherwise increment
            // count by 2
            count += (n / i == i) ? 1 : 2;
    }

    if (count % 2 == 0)
        printf("Even\n");

    else
        printf("Odd\n");
}

/* Driver program to test above function */
int main()
{
    printf("The count of divisor: ");
    countDivisors(10);
    return 0;
}

Output:

The count of divisor: Even

Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)

Efficient Solution: We can observe that the number of divisors is odd only in case of perfect squares. Hence the best solution would be to check if the given number is perfect square or not. If it’s a perfect square, then the number of divisors would be odd, else it’d be even. 

CPP

// C++ program for
// Efficient Solution to find
// if count of divisors is
// even or odd
#include <bits/stdc++.h>
using namespace std;

// Function to find if count
// of divisors is even or
// odd
void countDivisors(int n)
{
    int root_n = sqrt(n);

    // If n is a perfect square,
    // then it has odd divisors
    if (root_n * root_n == n)
        printf("Odd\n");
    else
        printf("Even\n");
}

/* Driver program to test above function */
int main()
{
    cout << "The count of divisors of 10 is: \n";

    countDivisors(10);
    return 0;
}

Output:

The count of divisors of 10 is: 
Even

Time Complexity: O(log(n))
Auxiliary Space: O(1). If recursive call stack will be considered then it would be log(n).

Please refer complete article on Check if count of divisors is even or odd for more details!