zoj 3551 Bloodsucker

最新推荐文章于 2020-02-16 18:32:04 发布

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最新推荐文章于 2020-02-16 18:32:04 发布

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分类专栏： acm bloodsucker 文章标签： output each float input

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探讨了在一个特定模型下所有人被转化为血族所需的天数期望值计算问题。该模型规定每天只有两个人会相遇，若一人为普通人而另一人为血族，则普通人有一定概率转化为血族。通过数学推导给出了具体的计算方法及实现代码。

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 numberp (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

2 1

Sample Output

1.000

一道概率题，可以这样考虑：

根据题意，每天只能增加一个Bloodsucker,也就是说n-1个正常人是一个接着一个变成Bloodsucker的，那么所有正常人变成Bloodsucker的天数期望等于所有正常人变成Bloodsucker的天数期望之和。考虑n-1个正常人中第i个变成Bloodsucker的正常人的转变概率为：


 
  P[i] = i*(n-i)/(n*(n-1)/2)


  
   则天数期望为（直接从概率的定义考虑）：


   
    D[i] = 1/P[i]

   
  
 
于是：


 
  D = sum(D[i])

代码如下

#include <stdio.h>



int main() {

    int T;

    int n;

    double p;

    scanf("%d", &T);

    while(T--) {

    scanf("%d %lf", &n, &p);

    

    double res = 0;

    double tmp = n*0.5*(n-1)/p;

    int i;

    for( i=1; i<=(n-1)/2; i++) {

     res += tmp/i/(n-i);

    }

    res *= 2;

    res += (n%2==0)?(4*tmp/n/n):0;

    

    printf("%.3lf\n", res);



    }

    return 0;

}

原文链接 http://blog.chinaunix.net/space.php?uid=26372629&do=blog&id=2996052