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;
using static System.Console;
using static System.Convert;
using static System.Math;

class Program
{
    static void Main(string[] args)
    {
        var num = ToInt32(ReadLine());
        var p = Array.ConvertAll(ReadLine().Split(' '), double.Parse);
        var dp = new double[num+1, num+1];
        dp[1, 0] = 1-p[0];
        dp[1, 1] = p[0];
        for(var i = 2; i <= num; i++)
        {
            for(var j = 0; j <= num; j++)
            {
                dp[i, j] = j == 0 ? dp[i - 1, j] * (1 - p[i - 1]) : dp[i - 1, j] * (1 - p[i - 1]) + dp[i - 1, j - 1] * p[i - 1];
            }
        }
        var sum = 0d;
        for (var i = num / 2 + 1; i <= num; i++)
            sum += dp[num, i];
        WriteLine(sum);
    }
    private const string ALFA = "abcdefghijklmnopqrstuvwxyz";
    private const int MOD = 1000000007;
}