hdu4586期望公式

1. if (sum == 0) ans = 0

2. if (n == m) ans = inf , everyE =sum / n

3. stE = P, ndE = (m / n) * p, rdE = (m / n)^2 * p, ... so sumAll = P * (1 - (m / n)^N) / (1 - q = (m / n)),N@limit= p / q 

#include <cstdio>int main() {    int n, m, x, i, sum;    while (~scanf("%d", &n)) {        for (sum = 0, i = 0; i < n; ++i) scanf ("%d", &x), sum += x;        for (scanf ("%d", &m), i = 0; i < m; ++i) scanf ("%d", &x);        if (!sum) printf ("0.00\n");        else if (n == m) printf ("inf\n");        else {            double p = sum * 1.0 / n, q = m * 1.0 / n;            printf ("%.2f\n", p / (1.0 - q));        }    }}