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: Wrong Answer)

// Author - Saikat_7
#include <bits/stdc++.h> 
using namespace std;
#define fl(i,a,b) for(int i=a;i<b;i++)
#define trav(i,cont) for(const auto& i:cont)
#define cs(x) cout<<x<<" "
#define cn(x) cout<<x<<" "<<endl
#define ll long long int
#define endl '\n'
#define pb push_back
#define ff first
#define ss second
#define mod 1000000007 //998244353 
#define inf 1e17
const int mx = 2999; // 2* (10 to power 5);
double dp[3000][3000];
double coins(double arr[],int n, int h)
{	if(dp[n][h]==-1)
	{
	if(h==0|| n==0)
		dp[n][h]= 1;
	else if(h<n)
		dp[n][h]=arr[n-1]*coins(arr,n-1,h-1) + (1-arr[n-1])*coins(arr,n-1,h);
	else if(h==n)
		dp[n][h]=arr[n-1]*coins(arr,n-1,h-1);
	else
		dp[n][h]=0;
		return dp[n][h];;
	}
	else
		return dp[n][h];

}
int main()
{
    ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
    int tc=1;
    // cin>>tc;
    while(tc-- > 0)
    {
   		int n;
   		cin>>n;
   		double arr[n];
   		int h=n/2+1;
   		for(int i=0;i<=n;i++)
   		{
   			for (int j=0;j<=h;j++)
   				dp[i][j]=-1;
   		}
   		for(int i=0;i<n;i++)
   		{
   			cin>>arr[i];
   		}     
   		double ans=coins(arr,n,(n/2)+1);
   		cout<<ans;
    }
    return 0;
}