Codeforces 518D Ilya and Escalator【概率dp】

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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.

Examples

Input

1 0.50 1

Output

0.5

Input

1 0.50 4

Output

0.9375

Input

4 0.20 2

Output

0.4

题目大意：

一共有N个人站成一队，一共有t秒，每秒第一个人有p概率上电梯，有（1-p）概率不上电梯。

问t秒结束时，上电梯的期望人数。

思路：

1、很明显，概率dp.设定dp【i】【j】【2】:

①dp【i】【j】【0】表示到时间i，电梯上已经有了j个人，且这一秒没有人上电梯的概率。

②dp【i】【j】【1】表示到时间i，电梯上已经有了j个人，且这一秒有人上电梯的概率。

2、那么不难推出其状态转移方程：

①dp【i】【j】【0】=（dp【i-1】【j】【0】+dp【i-1】【j】【1】）*（1-p）

②dp【i】【j】【1】=（dp【i-1】【j-1】【0】+dp【i-1】【j-1】【1】）*p

3、那么最终统计ans即可。（注意该结束的时候必须结束）；

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
double dp[2005][2005][2];
int main()
{
    int n,t;
    double p;
    while(~scanf("%d%lf%d",&n,&p,&t))
    {
        memset(dp,0,sizeof(dp));
        dp[0][0][0]=1;
        for(int i=1;i<=t;i++)
        {
            for(int j=0;j<=n;j++)
            {
                dp[i][j][0]=(dp[i-1][j][0]+dp[i-1][j][1])*(1-p);
                dp[i][j][1]=(dp[i-1][j-1][0]+dp[i-1][j-1][1])*p;
            }
        }
        double ans=0;
        for(int i=1;i<=t;i++)ans+=dp[i][n][1]*n;
        for(int i=1;i<n;i++)
        {
            ans+=dp[t][i][0]*i;
            ans+=dp[t][i][1]*i;
        }
        printf("%lf\n",ans);
    }
}