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

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

題意：

開始有一個吸血鬼，n-1個平民百姓。每天一個百姓被感染的概率可求，問每個人都變成吸血鬼的天數期望。

思路：

一般期望題逆推，設dp[i]是目前已經有i個吸血鬼，所有人變成吸血鬼的期望。則dp[n]=0;答案是dp[1]。每一個dp[i]的感染概率可求是p[]=2.0*(n-i)*i/(n-1)/n*p;

則可得遞推公式： dp[i] = (dp[i+1]*p[]+1)/p[];

#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;
double dp[100010],p,tmp;
int main()
{
    int T,n,i;
    scanf("%d",&T);
    while(T--){
        scanf("%d%lf",&n,&p);
        dp[n]=0;
        for(i=n-1;i>=1;i--) {
            tmp=2.0*(n-i)*i/(n-1)/n*p; 
            dp[i] = (dp[i+1]*tmp+1)/tmp;  
        }
        printf("%.3lf\n",dp[1]);
    }
    return 0;
}

ZOJ3551Bloodsucker （數學期望）