Problem:
Score : 100 points

Problem Statement

Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins.
Find the probability of having more heads than tails.

Constraints

N is an odd number.

1 \leq N \leq 2999

p_i is a real number and has two decimal places.

0 < p_i < 1

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 \ldots p_N

Output

Print the probability of having more heads than tails.
The output is considered correct when the absolute error is not greater than 10^{-9}.

Sample Input 1

3
0.30 0.60 0.80

Sample Output 1

0.612

The probability of each case where we have more heads than tails is as follows:

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 × 0.6 × 0.8 = 0.144;

The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 × 0.6 × 0.8 = 0.336;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 × 0.4 × 0.8 = 0.096;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 × 0.6 × 0.2 = 0.036.

Thus, the probability of having more heads than tails is 0.144 + 0.336 + 0.096 + 0.036 = 0.612.

Sample Input 2

1
0.50

Sample Output 2

0.5

Outputs such as 0.500, 0.500000001 and 0.499999999 are also considered correct.

Sample Input 3

5
0.42 0.01 0.42 0.99 0.42

Sample Output 3

0.3821815872

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

public class S
{

    private const long DP_INIT_MINUS1 = -1;
	private const long MOD = 1000000007;

    private static double[,] dp;


    public static void Main()
	{
		/*
        var input = Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse);
        var N = input[0];
        var M = input[1];
        */

        solver(int.Parse(Console.ReadLine()), Array.ConvertAll(Console.ReadLine().Split(' '), double.Parse));
    }

    public static void solver(int N, double[] coins){

        /*
            たぶんdp[N][N]
            [枚数][ここまでの表枚数]

            ある時点でのコイン枚数xについて、
            ここまでの表の枚数がy,x枚目のコインの表率をpとしたとき
            dp[x][y]は
            ・dp[x-1][y]*(1-p)、裏で変動しないことによるもの
            ・dp[x-1][y-1]*p、表による変動、1枚前の表枚数-1から遷移する
            この二つの合計値になる
            
        
         */

        dp = new double[N+1,N+1];
        
        // i: コイン枚数
        for(int i = 0; i <=N; i++){
            
            // k: 表枚数
            for(int k = 0; k <=N; k++){
                if(i == 0){
                    dp[i,k] = k == 0 ? 1 : 0;
                    continue;
                }

                // すべて裏
                if(k == 0){
                    dp[i,k] = dp[i-1,k] * (1-coins[i-1]);
                    continue;
                }

                dp[i,k] = (dp[i-1,k-1] * coins[i-1]) + (dp[i-1,k] * (1-coins[i-1]));
            }

        }



        
        double sum = 0.0;

        for(int k = (N/2) + 1; k <= N; k++){
            sum += dp[N,k];
        }

        Console.WriteLine(sum);


    }

}