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 Java7 (OpenJDK 1.7.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        double[] p = new double[n];

        for(int i = 0; i < n; i++){
            p[i] = sc.nextDouble();
        }

        int T = 0;
        int H = 1;
        double[][][] a = new double[n][n+1][2];

        a[0][0][T] = 1 - p[0];
        a[0][1][H] = p[0];

        for(int i = 1; i < n; i++){
            for(int j = 0; j <= i+1; j++){
                if(j != 0)
                a[i][j][H] = p[i]*(a[i-1][j-1][H]+a[i-1][j-1][T]);
                a[i][j][T] = (1-p[i])*(a[i-1][j][H]+a[i-1][j][T]);
            }
        }

        double ans = 0;

        for(int i = (int) Math.ceil(n/2.0); i <= n; i++){
            ans += a[n-1][i][H]+a[n-1][i][T];
        }

        System.out.println(ans);
    }
}