ZOJ3551:Bloodsucker(概率DP)

n 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

题意：一共有n个人，每天都会有一个人又一定的几率变成吸血鬼，求所有人变成吸血鬼的期望

思路：概率DP，我们可以逆推，dp[i]代表i个人到所有人变成吸血鬼的概率，往上逆推

则：dp[i]=(dp[i+1]+1)*p1+(dp[i]+1)*p2

移项后化简得： p1*dp[i]=dp[i+1]*p1+1

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

double dp[100005];

int main()
{
    int T,n,i;
    double p;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%lf",&n,&p);
        dp[n] = 0;
        for(i = n-1; i>=1; i--)
        {
            double s1,s2,p1;
            s1 = (double)n*(n-1)/2;//从n个人中选两个人出来的选法，也就是C(n,2)
            s2 = (double)i*(n-i);//i为从i个人中选出一人的方法，n-i为从吸血鬼中选出一个吸血鬼的总数，共有这么多种
            p1 = s2/s1*p;//人与吸血鬼相遇的概率
            dp[i] = (dp[i+1]*p1+1)/p1;
        }
        printf("%.3f
",dp[1]);
    }

    return 0;
}