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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Main {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());

		double[] a = new double[n];
		
		String[] input = br.readLine().split(" ");
		
		for (int i = 0; i < n; i++) {
			a[i] = Double.parseDouble(input[i]);
		}
		
		double ans = solve(a);
		
		System.out.println(ans);
	}

	private static double solve(double[] a) {
		
		double[][] dp = new double[a.length+1][a.length+1];
		
		dp[0][0]=1;
		
		for (int i = 1; i < dp.length; i++) {
			for (int j = 0; j < dp.length; j++) {
				if(j==0) {
					dp[i][j] = (1-a[i-1])*dp[i-1][j];
				}else {
					dp[i][j] = (1-a[i-1])*dp[i-1][j]+a[i-1]*dp[i-1][j-1];
				}
			}
		}
		
		double res = 0;
		
		for(int i=((a.length/2)+1); i<=a.length; i++) {
			res += dp[a.length][i];
		}
		
		return res;
	}
}