CodeForces 518D

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120930#problem/D

D - R2D2 and Droid Army

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 518D

Description

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 tseconds.

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 Input

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:电梯前有一队人,每秒只能有一个人进入电梯,每个人进入电梯的概率为p,不进入电梯的概率为(1-p);给出队中的人数,每个人进入电梯的概率,时间t .求t秒时电梯中的人数;

应用的知识;概率dp

思路 :二维数组dp[ i ][ j ]表示第 i 秒电梯里有 j 个人的概率.

 dp[0]0]==1; dp[0][j](j不等于0)=0;进行递推

 分三种情况讨论,

1. dp[i][0]=dp[i-1][0]*(1-p);

          2. dp[i][n]=dp[i-1][n] + dp[i-1][n-1]*p;

3. dp[i][ j ]=dp[i-1][j]*(1-p) + dp[i-1][j-1]*p;

#include<iostream>#include<iomanip>#include<cstdio>#include<cstring>#include<cmath>using namespace std;const int maxn=2000+5;double d[maxn][maxn];int main(){    int n,t;    double p;    d[0][0]=1.0;    while(~scanf("%d%lf%d",&n,&p,&t)){        for(int j=1;j<=n;j++) d[0][j]=0;        for(int i=1;i<=t;i++){            for(int j=0;j<=n;j++){                if(j==0)d[i][j]=d[i-1][j]*(1-p);                else if(j==n)d[i][j]=d[i-1][j]+d[i-1][j-1]*p;                else d[i][j]=d[i-1][j]*(1-p)+d[i-1][j-1]*p;            }        }        double ans=0.0;        for(int j=1;j<=n;j++)            ans+=d[t][j]*j;        printf("%lf\n",ans);    }    return 0;}