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<bits/stdc++.h>
#define ll long long
#define dl double
using namespace std;

//***********************************************************************
int MAXNUM = 5e3+5;
vector <int>vis(MAXNUM);
vector<int> pr;
void sieve() {
	pr.push_back(2);
	for(ll i=3;i<MAXNUM;i+=2) if(!vis[i]) {
		pr.push_back(i);
		for(ll j=i*i;j<MAXNUM;j+=2*i) vis[j]=true;
	}
}
//************************************************************************

void solve()
{

	int n;
    cin >> n;
    vector<dl> v(n);
    vector<vector<dl>> dp(n);
    for(int i=0;i<n;i++)
    {
        cin >> v[i];
        dp[i].resize(2*n+1);
    }
    dp[0][n-1]=v[0];
    dp[0][n+1]=1-v[0];
    for(int i=1;i<n;i++)
    {
        for(int j=0;j<2*n;j++)
        {
            dp[i][j]+=v[i]*dp[i-1][j+1];
        }
        for(int j=1;j<=2*n;j++)
        {
            dp[i][j]+=(1-v[i])*dp[i-1][j-1];
        }

    }
    /*for(int i=0;i<n;i++)
    {
        for(int j=0;j<2*n+1;j++)
        {
            cout << dp[i][j] << " ";
        }
        cout << endl;
    }*/
    
    dl ans=0;
    for(int i=0;i<n;i++)
    {
        ans+=dp[n-1][i];
    }
    cout <<setprecision(10) <<fixed<< ans << endl;
}


int main()
{ 
	ios_base::sync_with_stdio(false); 
    cin.tie(NULL);
	ll t;
	//cin >> t;
	t=1;
	while(t--)
	{
		solve();
	}

	return 0;
}