這題目是算概率並不難 想了一會就有結果了， 程式碼是對的，可是因為用cout 格式要自己控制的 ，在WA了之後沒有想到，也是在比賽最後一點時間了，比賽結束時候就無語了 不過也好給我個教訓吧

題目連線 ：http://codeforces.com/contest/518/problem/D

題意：有一個電梯 ， 門口有N個人排隊依次進入 ， 每個人進入的概率為P ，在每一秒排在隊伍第一位的人可以進入或者不進入，不進入後這一秒就過去了，而且排在隊伍後面的人必須等自己前面的人都進入電梯後才能進入 ，問我們在t秒內進去電梯人的期望。

思路： 開一個數組pp[i]  表示當前 t 秒時的前一秒進入電梯總人數（i）的概率，然後從第一秒開始 DP每一秒進入電梯人數的狀態，

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 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

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<set>
using namespace std;


int main()
{
     double pp[2005] = {0},  p , ans = 0;
     int n , t;
     cin >> n >> p >> t;
     pp[0] = 1;
     if(t <= n)
     {
          for(int i = 1 ; i <= t ; i++)
          {
               for(int j = i - 1 ; j > -1 ; j--)
               {
                    pp[j+1] += pp[j] * p;
                    ans += pp[j] * p;
                    pp[j] -= pp[j] * p;
               }
          }
     }
     else
     {
          for(int i = 1 ; i <= t ; i++)
          {
               for(int j = min(i - 1 , n-1) ; j > -1 ; j--)
               {
                    pp[j+1] += pp[j] * p;
                    ans += pp[j] * p;
                    pp[j] -= pp[j] * p;
               }
          }
     }
     printf("%lf\n", ans);
     return 0;
}