HDU 2601 An easy problem

(i+1)*(j+1)=n+1

转换成上面这个式子，也就是问n+1的因子有几个

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        long long n,y,i;
        scanf("%lld",&n);
        n++;
        int anss=0;
        y=sqrt(1.0*n);
        for(i=2;i<=y;i++)
            if(n%i==0)
                anss++;
        printf("%d
",anss);
    }
    return 0;
}