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: Time Limit Exceeded)

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

public class Main {

	public static void main(String[] args) {
		Main ans = new Main();
		System.out.printf("%.10f\n",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[][] dp = new double[N+1][N+1];

		dp[0][0]=1.0;
		for(int i=1;i<=N;i++){
			Arrays.fill(dp[i], -1);
		}

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

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

		if (x==0) return 1.0;
		if (N==0) return 0;

		if(dp[N][x]>-0.9) return dp[N][x];
		for(int i=N;i>=1;i--){
			dp[i][x] = ((1-array[i])* solve(array, dp, i-1, x)) + (array[i]* solve(array, dp, i-1, x-1)); 
			//System.out.printf("dp[%d][%d]: %d\n", i, x, dp[i][x]);
		}
		return dp[N][x];
	}

}