CF 518 D. Ilya and Escalator

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

Let's assume that n people stand in the queue for the escalator. At all second one of the following possibilities takes place:either the first person in the queue enters the Escalat or with probability p, or the first person in the queue doesn ' t move with probability (1- P), p Aralyzed 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, which is he'll be standing on the escalator at a NY 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, numberp is real, given with exactly TS after the decimal point.

Output

Print a single real number-the expected number of people who'll be standing on the escalator after T s Econds. The absolute or relative error mustn ' t exceed 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


Simple DP
DP (I,J) indicates the probability that a J person would go in the first minute.
Expected =∑J*DP (T,J)

Note: There are 2 things to do when you push a recursive step:
The queue is still manned, the queue has no one

#include <cstdio>#include<cstring>#include<vector>#include<iostream>#include<algorithm>#include<stack>#include<queue>#defineLL Long Long#defineULL unsigned long Longusing namespacestd;Const intmaxn=2005;DoubleDP[MAXN][MAXN];voidSolveint,Double,int );intMain () {//Loop:    intn,t; DoublePro; scanf ("%d%lf%d",&n,&pro,&t);    Solve (n,pro,t); //Goto Loop;    return 0;}voidSolveintNDoubleProintt) {     for(intI=0; i<maxn;i++)         for(intj=0; j<maxn;j++) Dp[i][j]=0.0; dp[0][0]=1.0;  for(intI=1; i<=t;i++) {dp[i][0]=dp[i-1][0]*(1.0-Pro);  for(intj=1; j<=i;j++){            if(j<N) {Dp[i][j]=dp[i-1][j-1]*pro+dp[i-1][j]* (1.0-Pro); }            Else if(j==N) dp[i][j]=dp[i-1][j-1]*pro+dp[i-1][j]; ElseDp[i][j]=0.0; }    }    Doubleret=0.0;  for(intj=0; j<=t;j++) {ret+=dp[t][j]*J; } printf ("%.10f\n", ret); return ;}

CF 518 D. Ilya and Escalator