题解 | #约数的个数#

#include <iostream>
#include <cmath>
using namespace std;

#define MAX 1000

int fnum(int a){
    int sum=0;
    for(int j=1;j<=sqrt(a);j++){
        if(a%j==0){
            sum+=2;
            if(j*j==a){
                sum--;
            }
        }
    }
    return sum;
}

int main() {
    int N;
    cin >> N;
    int num[MAX][2]; //第一行为整数，第二行为对应的约数的个数
    for (int i = 0; i < N; i++) {
        cin >> num[i][0];
        num[i][1] = fnum(num[i][0]);
    }
    for (int i = 0; i < N; i++) {
        cout << num[i][1] << endl;
    }
    return 0;
}