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.*;
import java.io.*;

class Main{

     /*
     * https://atcoder.jp/contests/dp/tasks/dp_i
     */ 
 
    public static void main(String[] args){
        
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();

        double[] dp = new double[n + 1];
        dp[0] = 1.0;
        for(int i = 0; i < n; i++){ // read coin probabilities
            double probabilityHead = in.nextDouble();
            for(int j = i + 1; j >= 0; j--){
                /**
                 * the probability taht we have j heads is the probaility that 
                 * we got a head * the probability of getting j-1 heads or the
                 * probability of not getting a head (1 - prob_head), which means
                 * we still have j heads to pick, hence dp[j].
                 */
                dp[j] = (j == 0 ? 0: dp[j - 1] * probabilityHead) + 
                dp[j] * (1 - probabilityHead);
            }

        }
        double ret = 0;
        for(int index = (n / 2) + 1; index <= n; index++) ret += dp[index];
        System.out.println(ret);
    }
}