codeforces--518D--Ilya and Escalator--概率DP

D. Ilya and Escalator

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.

Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.

Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.

Your task is to help him solve this complicated task.

Input

The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.

Output

Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.

Sample test(s)

input

1 0.50 1

output

0.5

input

1 0.50 4

output

0.9375

input

4 0.20 2

output

0.4

题意：输入n,p,t表示n个人排成一队上电梯，每个时间单位只能上去队首的那一个人，而且这个人上电梯成功的概率是p，电梯无限载重，可以容下所有人，问你在t单位时间这个时间点电梯中人数的期望值

题解：dp[x][y]表示y时刻电梯中有x个人的概率，考虑上去成功、上去失败两种情况、没有人排队这三种情况

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <cmath>
#include <algorithm>
#define Max(a,b) (a>b?a:b)
#define Min(a,b) (a<b?a:b)
#define mod 1000000007
#define LL __int64
#define M 2010
using namespace std;
double dp[M][M];
int main()
{
    int n,t,i,j,k,l;
    double p;
    while(cin>>n>>p>>t)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(i=1;i<=t;i++)dp[0][i]=dp[0][i-1]*(1-p);
        for(i=1;i<=n;i++)//现求出概率
        for(j=i;j<=t;j++)
        {
            //dp[i][j]=p;
            dp[i][j]+=dp[i-1][j-1]*p;//加上 前一个状态*成功上电梯的概率
            if(i==n)dp[i][j]+=dp[i][j-1];//如果没有人可以上电梯了，直接传递概率值
            else dp[i][j]+=dp[i][j-1]*(1.0-p);//加上 前一个状态*未成功上电梯的概率
        }
        double sum=0;
        for(i=1;i<=n;i++)//然后求期望
        sum+=dp[i][t]*i;
        printf("%.7lf\n",sum);
    }
	return 0;
}