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

这题目是算概率并不难 想了一会就有结果了， 代码是对的，可是因为用cout 格式要自己控制的 ，在WA了之后没有想到，也是在比赛最后一点时间了，比赛结束时候就无语了 不过也好给我个教训吧

题目连接 ：http://codeforces.com/contest/518/problem/D

题意：有一个电梯 ， 门口有N个人排队依次进入 ， 每个人进入的概率为P ，在每一秒排在队伍第一位的人可以进入或者不进入，不进入后这一秒就过去了，而且排在队伍后面的人必须等自己前面的人都进入电梯后才能进入 ，问我们在t秒内进去电梯人的期望。

思路： 开一个数组pp[i]  表示当前 t 秒时的前一秒进入电梯总人数（i）的概率，然后从第一秒开始 DP每一秒进入电梯人数的状态，

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;
}