[UVA11605] Lights inside a 3d Grid && 数学期望

来源：互联网发布：网络大电影的发展趋势编辑：程序博客网时间：2024/06/05 17:13

先求每盏灯被操作的概率p 那么就等于（x坐标被操作的概率）px * （y坐标概率）py * （z坐标概率）pz

对于每一盏灯设K次亮着概率d[i] 不亮概率f[i] 则 1 = d[i] + f[i]

则d[i] = d[i-1] * (1-p) + f[i-1] * p = d[i-1] * (1-p) + (1-d[i-1]) * p = d[i-1] * (1 - 2*p) + p

构造新数列 d[i] - 0.5 = (d[i-1] - 0.5) * (1 - 2 * p)

得到 d[i] = (d[1] - 0.5) * (1-2p) ^ (i-1) + 0.5

d[i] = 0.5 - (p - 0.5) * (1 - 2p) ^ (i-1) = 0.5 + (1-2p) ^ i;

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define SF scanf#define PF printfusing namespace std;double cnt(int x, int n){    return ((n - x + 1) * x * 2 - 1) * 1.0 / n / n;}double cal(double p, int k){    return (1 - pow(1 - 2 * p, k)) / 2;}int main(){    int n, m, h, K;    int T, kase = 0; SF("%d", &T); while(T--) {        SF("%d%d%d%d", &n, &m, &h, &K);        double ans = 0;        for(int i = 1; i <= n; i++)        {            double Px = cnt(i, n);            for(int j = 1; j <= m; j++)            {                double Py = cnt(j, m);                for(int k = 1; k <= h; k++)                {                    double Pz = cnt(k, h);                    ans += cal(Px * Py * Pz, K);                }            }        }        PF("Case %d: %.10lf\n", ++kase, ans);    }}

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