求取约数的个数的算法

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/*https://www.nowcoder.com/practice/04c8a5ea209d41798d23b59f053fa4d6?tpId=40&tqId=21334&tPage=1&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking*/

#include <bits/stdc++.h>
using namespace std;

void printnum(int num){
	int n = sqrt(num);
	int count = 0;
	if(n * n == num){
		count++;
		n--;
	}
	for (int i = 1; i <= n; ++i){
		if(num % i == 0)
			count += 2;
	}
	printf("%d\n", count );
}

int main(int argc, char const *argv[]){
	int n,num;
	scanf("%d",&n);
	for (int i = 0; i < n; ++i){
		scanf("%d",&num);
		printnum(num);
	}
	return 0;
}