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: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int N = input.nextInt();
		double[][] dp = new double[N+1][N+1];
		double[] p = new double[N+1];
		double answer = 0.0;
		for (int i = 1; i <= N; i++) {
			p[i] = input.nextDouble();
		}
		dp[1][1] = p[1];
		for (int i = 2; i <= N; i++) { //Filling diagonals of DP table
			dp[i][i] = dp[i-1][i-1]*p[i];
		}
		for (int j = 2; j <= N; j++) { //Filling 1st row of DP table
			dp[1][j] = dp[1][j-1]*(1.0-p[j]);
			if (j==2) dp[1][j]+=p[j]*(1.0-dp[1][j-1]);
			else dp[1][j]+=p[j]*(1.0-dp[j-1][j-1]-dp[1][j-1]);
		}
		for (int i = 2; i < N; i++) { //dp[N][N] is already filled, last row not necessary
			for (int j = i+1; j <= N; j++) {
				dp[i][j]=dp[i][j-1]*(1.0-p[j])+dp[i-1][j-1]*p[j]; 
			}
		}
		/*for (int i = 1; i <= N; i++) {
			for (int j = 1; j <= N; j++) {
				System.out.print(dp[i][j]+" ");
			}
			System.out.print("\n");
		}*/
		for (int i = N/2+1; i <= N; i++) {
			answer+=dp[i][N];
		}
		System.out.println(answer);
	}
}