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# 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 ``#include ` `// 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 ``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!

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