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: Time Limit Exceeded)

import java.util.*;
import java.lang.*;
import java.io.*;

class Main
{
	static int MAX;
	static Double dp[][];

	static double maxProbility(double ar[], int n, int k)
	{
		if(n==0)
			return (k==0)?1.0: 0;
		if(k==0)
		{
			double prod=1;
			for(int i=0;i<n;i++)
				prod*=(1-ar[i]);
			return prod;
		}

		if(dp[n][k]!=null)
			return dp[n][k];
		
		return dp[n][k]=((maxProbility(ar, n-1, k-1)*ar[n-1]) + (maxProbility(ar, n-1, k)*(1-ar[n-1])));
	}

	public static void main(String[] args)
	{
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		double ar[]=new double[n];
		for(int i=0;i<n;i++)
		{
			ar[i]=scan.nextDouble();
		}

		double ans=0;
		dp=new Double[n+1][n+1];

		for(int k=(n/2)+1;k<=n;k++)
		{
			ans += maxProbility(ar, n, k);
		}

		System.out.println(ans);
	}
}