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.util.*;

public class Main
{
	public static void main(String [] Args)
	{
		Scanner in = new Scanner(System.in);
		int n=in.nextInt();
		double [][] dp=new double [n+1][n];
		double [] p=new double [n];
		for(int i=0;i<n;i++)
		{
			p[i]=in.nextDouble();
		}
		dp[0][0]=1.0d-p[0];
		dp[1][0]=p[0];
		for(int i=1;i<n;i++)
		{
			dp[0][i]=dp[0][i-1]*(1.0000000000d-p[i]);
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<n;j++)
			{
				dp[i][j]=dp[i-1][j-1]*p[j]+dp[i][j-1]*(1.0000000000d-p[j]);
			}
		}
		Double ans=0.0;
		for(int i=n/2+1;i<=n;i++)
		{
			ans=ans+dp[i][n-1];
		}
		System.out.println(ans);
	}
}