ZOJ 3551 Bloodsucker（概率dp啊 ）

标签：dp   数学   hdu   

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4530

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：
开始有 n-1个人，1个吸血鬼，以后每天这 n 个中其中有两个会相遇，如果一个是吸血鬼，一个是人，那这个人有一定的概率 p 变成吸血鬼。

求这 n 个最后都变成吸血鬼所需天数的期望值。

PS:

用dp[i]来表示有 i 个吸血鬼时的期望值,dp[n]=0;

dp[1]即为所求。

移项得：dp[i] = (dp[i+1]*p3+1)/p3;

代码如下：

#include <cstdio>
#include <cstring>
double dp[100017];
int main()
{
    int t, n;
    double p;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%lf",&n,&p);
        dp[n] = 0;
        double p1 = (double)n*(n-1)/2;//c n 2 n个人中选2人
        for(int i = n-1; i >= 1; i--)
        {
            double p2 = (double)i*(n-i);
            double p3 = p2/p1*p;//相遇并变成吸血鬼的概率
            dp[i] = (dp[i+1]*p3+1)/p3;
        }
        printf("%.3lf\n",dp[1]);
    }
    return 0;
}

ZOJ 3551 Bloodsucker（概率dp啊 ）

标签：dp   数学   hdu   

原文地址：http://blog.csdn.net/u012860063/article/details/43700367