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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
using namespace std;
#include<algorithm>
#include<cmath>
#include<iomanip>
#define int long long int
#define mod 1000000007
double function(double arr[],int initial,int final,double memo[3001][3001],int count)
{
	if(count==0)
	  return 1.000000000;
	if(initial>final)
	{
		if(count>0)
		  return 0.00000000000;
		else
		  return 1.00000000000;
	}
	if(memo[initial][count]!=-1)
	  return memo[initial][count];
	else
	{
		double val1=arr[initial]*function(arr,initial+1,final,memo,count-1);
		double val2=(1-arr[initial])*function(arr,initial+1,final,memo,count);
		memo[initial][count]=(val1+val2);
		return memo[initial][count];
	}
}
double memo[3001][3001];
int32_t main()
{
    int n;
    cin>>n;
    double arr[n];
    int j;
    for(j=0;j<n;j++)
      cin>>arr[j];
    int i;
    for(i=0;i<=n;i++)
    {
		for(j=0;j<=n;j++)
		{
			memo[i][j]=-1;
	    }
    }
    double answer=function(arr,0,n-1,memo,n/2+1);
    cout<<setprecision(12)<<answer<<endl;
}