Bloodsucker ZOJ

概率论
题目：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
题意：n-1人，1吸血鬼，每天有两个人遇见，只有人和吸血鬼遇见会触发概率事件，同化吸血鬼，全部同化的期望。

#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define ll long long
using namespace std;
double f[100010],g[100010],a[100010];
int main()
{int n,t;double s1,s2,s3,d;scanf("%d",&n);while(n--){scanf("%d%lf",&t,&d);f[t]=0;for(int i=t-1;i>=1;i--){s1=(double)t*(t-1)/2;s2=(double)i*(t-i);s3=s2/s1*d;f[i]=(f[i+1]*s3+1)/s3;}printf("%.3lf\n",f[1]);}return 0;
}

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