Program to Find All Factors of a Natural Number

A Naive Solution would be to iterate all the numbers from 1 to n, checking if that number divides n and printing it. Below is a program for the same:
// C implementation of Naive
// method to print all divisors
#include<stdio.h>

// Function to print the divisors
void printDivisors(int n)
{
	for (int i = 1; i <= n; i++)
		if (n % i == 0)
			printf("%d ", i);
}

// Driver code
int main()
{
	printf("The divisors of 100 are: ");
	printDivisors(100);
	return 0;
}

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// A Better (than Naive) Solution
// to find all divisors
#include <stdio.h>
#include <math.h>

// Function to print the divisors
void printDivisors(int n)
{
	// Note that this loop runs till
	// square root
	for (int i = 1; i <= sqrt(n); i++)
	{
		if (n % i == 0)
		{
			// If divisors are equal,
			// print only one
			if (n / i == i)
				printf("%d ", i);

			// Otherwise print both
			else
				printf("%d %d ", i, n / i);
		}
	}
}

// Driver code
int main()
{
	printf("The divisors of 100 are: ");
	printDivisors(100);
	return 0;
}