zoj 3551 Bloodsucker 概率DP

最新推荐文章于 2015-11-20 17:42:08 发布

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本文探讨了一个涉及人与吸血鬼互动的模型，通过概率转换机制，分析了从人类群体到完全吸血鬼群体的演变过程。重点在于计算特定条件下，人类群体完全转变为吸血鬼群体的数学期望天数。

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 ofD.

Input

The number of test cases (T, T ≤ 100) is given in the first line of the input. Each case consists of an integern and a float numberp (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

/////

#include<iostream>
#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;
int n;
double p;
int main()
{
    int ci;scanf("%d",&ci);
    while(ci--)
    {
        scanf("%d%lf",&n,&p);
        if(n==1)
        {
            printf("0.000\n");continue;
        }
        double pre;
        double total=(double)n*(n-1)/2;
        pre=(1.0)/(1-2.0/n*(1-p)-1+2.0/n);
        for(int i=2;i<=n-1;i++)
        {
            double tmp=(double)i*(n-i);
            pre=(1+tmp/total*p*pre)/(1-tmp/total*(1-p)-1+tmp/total);
        }
        printf("%.3lf\n",pre);
    }
    return 0;
}