CodeForces 518 D. Ilya and Escalator（概率DP）

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

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n个人都在电梯内的期望。

dp[i][j]表示第i秒电梯上有j个人的概率

//虽然说简单但是自己还是没有分析完整，加油

<pre name="code" class="cpp">#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
double dp[2010][2010];
int main()
{
int n,t;
double p,ans;
while (scanf("%d %lf %d",&n,&p,&t)!=EOF)
{
ans=0;
memset(dp, 0, sizeof(dp));
dp[0][0]=1;  //第0秒里面肯定没有人
for (int i=1; i<=t; i++)
{
//第i秒电梯里没人的概率为第i－1秒电梯里没人的概率乘第i秒时排头不进电梯的概率（1-p）
dp[i][0]=dp[i-1][0]*(1-p);
for (int j=1; j<n; j++)
{

//这里第i秒有n个人的概率等于第i－1秒有j－1个人的概率乘第i秒第j个人进电梯的概率加上第i - 1秒排头不进电梯的概率
dp[i][j]=dp[i-1][j-1]*p+dp[i-1][j]*(1-p);

}
//第i秒有n个人的概率等于第i－1秒有j－1个人的概率乘第i秒第j个人进电梯的概率加上第i - 1秒电梯就已经有n个人的概率
dp[i]
=dp[i-1][n-1]*p+dp[i-1]
;

}
for(int i = 1; i <= t; i++)
ans += (dp[t][i] * i);
printf("%.7f\n", ans);

}
return 0;
}