ZOJ 3551 Bloodsucker

A - Bloodsucker

Time Limit:2000MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3551

Description

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 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

概率DP

设dp[i]表示从第i个人到所有人变成吸血鬼的期望（平均值）。

则dp[n]=0。（n个人全都是吸血鬼了，那么平均变0次）

开始逆推。

dp[i]=(dp[i+1]+1)*p1+(dp[i]+1)*(1-p1)
第i个人到所有人变成吸血鬼的期望等于两个期望的和。
当第i个人与吸血鬼相遇时。
有p1的概率变成吸血鬼。p1的值可通过排列组合得到。从n个人里选两个人的方法数有n*(n-1)/2种，其中如果是人和吸血鬼相遇的方法种数有i*(n-i)种。
那么n个人中人和吸血鬼相遇的概率为 ((double)i*(n-i))/((double)n*(n-1)*0.5)
进一步得到n个人中，人和吸血鬼相遇并且人变成吸血鬼的概率p1为
double p1=((double)i*(n-i))/((double)n*(n-1)*0.5)*p;
接着
dp[i]可分为两部分（全期望公式），如果第i个人变成了吸血鬼，那么期望便是，从第i+1人到所有人的期望+1（第i个人自己）再乘以变吸血鬼的概率p1.如果第i个人没变成吸血鬼，期望是从第i个人到所有人变吸血鬼期望+1,再乘上不变吸血鬼的概率(1-p1).
最终dp[1]便是答案。

#include <stdio.h>
#define N 100005
double dp[N];
int main()
{
        int t;
        scanf("%d",&t);
        while(t--)
        {
                int n;
                double p;
                scanf("%d%lf",&n,&p);
                dp[n]=0;
                for(int i=n-1;i>=1;i--)
                {
                        double p1=((double)i*(n-i))/((double)n*(n-1)*0.5)*p;
                        dp[i] = (dp[i+1]*p1+1)/p1;
                }
                printf("%.3lf\n",dp[1]);
        }
        return 0;
}