ZOJ - 3551 Bloodsucker（期望dp）

In 0th day, there are n-1 people and 1 bloodsucker. Every day, two and only two of them meet. Nothing will happen if they are of the same species, that is, a people meets a people or a bloodsucker meets a bloodsucker. Otherwise, people may be transformed into bloodsucker with probability p. Sooner or later(D days), all people will be turned into bloodsucker. Calculate the mathematical expectation of D.

Input

The number of test cases (T, T ≤ 100) is given in the first line of the input. Each case consists of an integer n and a float number p (1 ≤ n < 100000, 0 < p ≤ 1, accurate to 3 digits after decimal point), separated by spaces.

Output

For each case, you should output the expectation(3 digits after the decimal point) in a single line.

Sample Input

1
2 1

Sample Output

1.000

题目大概：

给出n个人，有一个吸血鬼，n-1个普通人，每天会有两个人会相遇，如果一个是吸血鬼，一个是普通人，有m概率普通人会变成吸血鬼，问所有人都变成吸血鬼的期望天数。

思路：

dp[i]表示已经有i个吸血鬼，距离所有人变成吸血鬼还差的期望。
第i个人变成吸血鬼的概率可以计算出来 为p=2*m*（n-i）*i/（n-1）/n。
可以得到一般方程dp[i]=(1-p)*dp[i]+p*dp[i+1]+1。
化简一下为dp[i]=dp[i+1]+1/p。

代码：

#include <bits/stdc++.h>

using namespace std;
const int maxn=1e5+10;

double dp[maxn];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        int n;
        double m;
        scanf("%d%lf",&n,&m);
        dp[n]=0;
        for(int i=n-1;i>=1;i--)
        {
            double p=2.0*(double)i*(double)(n-i)/(double)n/(double)(n-1)*m;
            dp[i]=dp[i+1]+1/p;
        }
        printf("%.3lf\n",dp[1]);
    }
    return 0;
}