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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String arg[]) {

		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		double[] p = new double[n + 1];

		for (int i = 1; i <= n; i++) {
			p[i] = scan.nextDouble();
		}

		double dp[][] = new double[n + 1][n + 1];
		
		double ph = p[1];
		for(int i =2;i<=n;i++) {
			ph *= p[i];
		}
		
		System.out.println(String.format("%.9f",probability(n, (n +1)/2, p, dp)));
	}

	public static double probability(int n, int x, double[] p, double[][] dp) {
		//System.out.println(n + "  " + x);
		if(x == 0)return 1;
		if (n == 0)
			return 0;
		
		if (dp[n][x] != 0) {
			return dp[n][x];
		}
		double withHead = p[n] * probability(n - 1, x - 1, p, dp);
		double withTail = (1 - p[n]) * probability(n - 1, x, p, dp);

		dp[n][x] = withHead + withTail;
		return dp[n][x];

	}
}