D. Ilya and Escalator

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

题目抽象：n个人排队上电梯，排头每秒上去的概率为p，一共t秒，求t秒都电梯内人数的期望。

思路：简单的概率dp，dp[i][j]表示第i秒电梯上有j个人的概率，最后累计一下期望

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <algorithm>
 7 #include <string>
 8 #include <vector>
 9 #include <map>
10 #include <set>
11 #include <stack>
12 #include <queue>
13 #include <iomanip>
14 using namespace std;
15 const int INF=0x7fffffff;
16 const double EXP=1e-8;
17 const int MS=2005;
18 const int mod=100000007;
19 typedef long long LL;
20 double dp[MS][MS];   
21 //  dp[i][j]   在i second 的时间内 进入 了j个人的概率，  
22 
23 ///                dp[i-1][n-1]*p  
24 //  dp[i][n]    
25 //                dp[i-1][n]*1; 
26 
27 int main()
28 {
29     
30     int n,t,i,j;
31     double p,ans=0;
32     memset(dp,0,sizeof(dp));
33     dp[0][0]=1.0;
34     cin>>n>>p>>t;
35     for(i=0;i<t;i++)
36     {
37         for(j=0;j<n;j++)
38         {
39             dp[i+1][j+1]+=dp[i][j]*p;
40             dp[i+1][j]+=dp[i][j]*(1-p); 
41         }
42         dp[i+1][n]+=dp[i][n];    
43         //特别注意这里。 因为n特殊一点
44         //                        dp[i][n-1]*p
45         //       dp[i+1][n] 
46         //                        dp[i][n]*p; 
47     }
48     for(i=1;i<=n;i++)
49         ans+=i*dp[t][i];
50     cout<<setiosflags(ios::fixed)<<setprecision(6)<<ans<<endl;
51     return 0;
52 }