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.*;

public class Main{
	
	public static double find(double arr[],double dp[][],int tail,int i,int n)
	{
		if(i>=n)
			return 1;
		
		if(dp[tail][i]==-1)
		{
			double op1=(arr[i])*find(arr,dp,tail,i+1,n);
			double op2=0;
			if(tail<n/2)
				op2=(1-arr[i])*find(arr,dp,tail+1,i+1,n);
			
				dp[tail][i]=op2+op1;
				
		}
		
			return dp[tail][i];
	}
    public static void main(String[] args) 
    {
       Scanner s=new Scanner(System.in);
       int n=s.nextInt();
       double arr[]=new double[n];
       for(int i=0;i<n;i++)
    	   arr[i]=s.nextDouble();
       
       double dp[][]=new double[n/2+1][n];
       for(int i=0;i<=n/2;i++)
       {
    	   for(int j=0;j<n;j++)
    		   dp[i][j]=-1;
       }
       System.out.print(find(arr,dp,0,0,n));
    }
}