dp（Codeforces Round #293 (Div. 2)D. Ilya and Escalator）

D. 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 probabilityp, 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 from1 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 aftert 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 two digits after the decimal point.

Output

Print a single real number — the expected number of people who will be standing on the escalator aftert seconds. The absolute or relative error mustn't exceed10^- 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：n个人上电梯，每秒钟有一个人能上电梯，概率是p，一共t秒，问电梯上人数的期望

思路：dp[i][j]表示第i秒电梯上有j个人的概率

#include <cstdio>#include <cstring>double dp[2005][2005];int main(){    int n, t;    double p, ans = 0;    memset(dp, 0, sizeof(dp));    dp[0][0] = 1;    scanf("%d %lf %d", &n, &p, &t);    for(int i = 1; i <= t; i++)    {        for(int j = n; j >= 0; j--)        {            if(j == n)                dp[i][j] = dp[i - 1][j - 1] * p + dp[i - 1][j];            else if(j != 0)                dp[i][j] = dp[i - 1][j - 1] * p + dp[i - 1][j] * (1 - p);            else                dp[i][j] = dp[i - 1][j] * (1 - p);        }    }    for(int i = 1; i <= t; i++)        ans += (dp[t][i] * i);    printf("%.7f\n", ans);}