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.Scanner;

/*
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 * and open the template in the editor.
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/**
 *
 * @author aakas
 */
public class Main{
    static int n;
    static double[] p;
    static boolean v[][];
    static double dp[][];
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        p = new double[n];
        v = new boolean[n+1][n+1];
        dp = new double[n+1][n+1];
        for(int i=0;i<n;i++)
        {
            p[i] = sc.nextDouble();
        }
        double sum = 0;
        for(int i=0;i<Math.ceil((double)n/2.0);i++)
        {
            sum += s(n-i,i);
        }
        System.out.println(sum);
    }
    public static double s(int i,int j)
    {
        if(i==0 && j==1)
            return 1-p[i+j-1];
        if(i==1 && j==0)
            return p[i+j-1];
        if(i==0 && j==0)
            return 0;
        if(i<0 || j<0)
            return 0;
        if(v[i][j])
            return dp[i][j];
        v[i][j] = true;
        return dp[i][j] = p[i+j-1]*s(i-1,j) + (1-p[i+j-1])*(s(i,j-1)); 
    }
}