题解 | #约数的个数#

#include <stdio.h>
#include <math.h>
int ApproxiNum(long long x) {
    if (x == 1)
        return 1;
    else {
        long long y = sqrt(x);
        int count = 0;
        for (long i = 1; i <= y; i++) {
            if (x % i == 0 && i * i != x)
                count += 2;
            else if (x % i == 0 && i * i == x)
                count++;
        }
        return count;
    }
}
int main() {
    int n;
    scanf("%d", &n);
    int Arr[n];
    int i = 0;
    while (n--) {
        long long m = 0;
        scanf("%lld", &m);
        Arr[i++] = ApproxiNum(m);
    }
    for (int k = 0; k < i; k++)
        printf("%d\n", Arr[k]);
    return 0;
}