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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

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

public class Main{

	public static void main(String[] args) {
		Main ans = new Main();
		System.out.printf("%.10f",ans.solve());
	}

	private double solve() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		double [] array = new double[N+1];
		for(int i=1;i<=N;i++){
			array[i] = sc.nextDouble();
		}

		double ans = 0.0;
		double [][] dp = new double[3000][3000];
		for(int i=0;i<3000;i++) {
			Arrays.fill(dp[i], -1);
		}


		return solve(dp, array, N, (N+1)/2);

	}

	private double solve(double [][] dp, double []array, int i, int x) {

		if (x==0) return 1;
		if (i==0) return 0;
		if (dp[i][x]>-0.9) return dp[i][x];
		dp[i][x] = array[i]* solve(dp, array, i-1, x-1) + (1-array[i]) * solve(dp, array, i-1, x);
		return dp[i][x];

	}

}