ZOJ3551 Bloodsucker(概率dp)

转载请注明出处: http://www.cnblogs.com/fraud/           ——by fraud

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 (TT ≤ 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:已知有n-1个人,1个吸血鬼,每天n个中会随机有两个碰面,若是人和吸血鬼相遇,则人有p的概率变成吸血鬼,问最终全部变吸血鬼的期望天数

很明显,这是个概率dp题,dp[i]表示还剩i个人,那么dp[i] = dp[i+1] + 1 / P[i+1]

其中P[i] = p*C(i,1)*C(n-i ,1) / C(n,2)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author xyiyy @https://github.com/xyiyy
 */

#include <iostream>
#include <fstream>

//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>

using namespace std;
#define dep(X, R, L) for(int X=R;X>=L;X--)

double dp[100010];

class TaskC {
public:
    void solve(std::istream &in, std::ostream &out) {
        int t;
        in >> t;
        while (t--) {
            int n;
            double p;
            in >> n >> p;
            dp[n - 1] = 0;
            double tot = (double) n * (n - 1) / 2 / p;
            dep(i, n - 2, 0) {
                dp[i] = dp[i + 1] + tot / (i + 1) / (n - i - 1);
            }
            out << fixed << setprecision(3) << dp[0] << endl;
        }
    }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    TaskC solver;
    std::istream &in(std::cin);
    std::ostream &out(std::cout);
    solver.solve(in, out);
    return 0;
}

 

posted on 2015-08-16 00:49  xyiyy  阅读(...)  评论(...编辑  收藏

导航

统计