Ilya and Escalator - CodeForces 518 D 概率dp

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：一共有n个人，每个人在每秒的时候进入电梯的概率为p，问第t秒的时候电梯里面人数的期望。

思路：dp[i][j]表示在第i秒，电梯里人数为j的概率，由dp[i-1][j-1]和dp[i-1][j]转移未来。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
int T,t,n,m;
double dp[2010][2010],p,q,ans;
int main()
{
    int i,j,k;
    scanf("%d%lf%d",&n,&p,&t);
    dp[0][0]=1;
    q=1-p;
    for(i=1;i<=t;i++)
    {
        dp[i][0]=dp[i-1][0]*q;
        for(j=1;j<n;j++)
           dp[i][j]=dp[i-1][j-1]*p+dp[i-1][j]*q;
        dp[i]
=dp[i-1][n-1]*p+dp[i-1]
;
    }
    for(i=1;i<=n;i++)
       ans+=i*dp[t][i];
    printf("%.7f\n",ans);
}