zoj3551---Bloodsucker

最新推荐文章于 2018-08-22 18:17:48 发布

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最新推荐文章于 2018-08-22 18:17:48 发布

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分类专栏： DP zoj 文章标签： dp zoj

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本文探讨了一个关于吸血鬼转变人群的概率问题，并提供了一种基于动态规划的解决方案。通过数学期望计算，在特定概率下，所有人转变为吸血鬼所需的平均天数。

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Bloodsucker

Time Limit: 2 Seconds Memory Limit: 65536 KB

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.

这题应该有2个思路,我是按吸血鬼数目来dp的,设dp[i] 表示当前有i个吸血鬼,达到目标状态的期望值

则 dp[i] = C(1, i) * C(1, n - i) / C(2, n) * (p * dp[i+1] + (1-p)*dp[i]) + (1 - C(1, i) * C(1, n - i) / C(2, n)) * dp[i] + 1;

化简以后得 dp[i] = dp[i + 1] + (n - 1) * n / (2 * p * i * (n - i))

/*************************************************************************
    > File Name: zoj3551.cpp
    > Author: ALex
    > Mail: 405045132@qq.com 
    > Created Time: 2014年12月22日 星期一 20时44分55秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 100010;
double dp[N];
double p;
int n;

double dfs(int i)
{
	if (i == n)
	{
		return 0;
	}
	if (dp[i] != -1)
	{
		return dp[i];
	}
	double pr = (double)n * (n - 1);
	pr /= (double)2 * p * i * (n - i);
//	printf("%f\n", pr);
	dp[i] = dfs(i + 1) + pr;
	return dp[i];
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%lf", &n, &p);
		for (int i = 0; i<= n; ++i)
		{
			dp[i] = -1;
		}
		dp[n] = 0.0;
		printf("%.3f\n", dfs(1));
	}
	return 0;
}