2-4子序列的和(subsequence)

#include "stdafx.h"
#include <math.h>

int main()
{
    int   m, n;
    int q=0;
    while (scanf_s("%d %d", &n, &m) == 2 && m&&n)
    {
        double p = 0,a;
        for (; n <= m; n++)
        {
            a = (1. / (long long)(pow(n,2)));
            p += a;
        }
        q++;
        printf("Case %d: %.5f\n", q, p);
    }
}