ZOJ 3551 Bloodsucker(概率DP)

思路：d[i]表示从i个人已变成吸血鬼到所以都变成吸血鬼天数的期望，那么则有d[i]=(d[i+1]+1)p1*p+(d[i]+1)*(1-p1)*p.   p1=(i(n-i))/C(n,2)

#include<iostream>#include<cstdio>using namespace std;const int maxn=100008;int main(){    int T,n;    double p;    double dp[maxn];    cin>>T;    while(T--)    {        cin>>n>>p;        dp[n]=0;        for(int i=n-1;i>0;i--)        {            double y=(double)n*(n-1)/2;            double x=(double)i*(n-i);            double p1=(x/y)*p;            dp[i]=dp[i+1]+1/p1;        }        printf("%.3f\n",dp[1]);    }    return 0;}