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: Runtime Error)

#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;

#define MAXN ((int)1e3+5)
#define MOD ((int)1e9 + 7) 
#define ll long long
#define _ << " " <<
#define TRACE(x) cout << #x << " = " << x << endl;
#define pb push_back
#define MP make_pair
#define pi pair<int,int>
#define vi vector<int>
#define vii vector<pi>
#define all(x) x.begin(), x.end()

int n,m,t,k;
double p[MAXN];
double dp[MAXN][MAXN];

int main()
{
	ios_base::sync_with_stdio(false);cin.tie(0);

	cin >> n;
	for (int i = 0; i < n; ++i)
		cin >> p[i];

	// for (int i = 0; i < n; ++i)
	// {
	// 	cout << 1-p[i] << endl;
	// }

	dp[0][0] = 1.0f-p[0];
	dp[0][1] = p[0];

	for (int i = 1; i < n; ++i)
	{
		for (int j = 0; j <= i+1; ++j)
		{
			if(j)
				dp[i][j] = dp[i-1][j] * (1.0f-p[i])  + dp[i-1][j-1] * (p[i]);
			else
				dp[i][j] = dp[i-1][j] * (1.0f-p[i]);
		}
	}

	double ans = 0;
	for (int i = n/2+1; i <= n; ++i)
	{
		// cout << i _ dp[n-1][i] << endl;
		ans += dp[n-1][i];
	}

	cout << std::setprecision(12) << ans;
	// for (int i = 0; i < n; ++i)
	// {
	// 	cout << i << " : ";
	// 	for (int j = 0; j <= i+1; ++j)
	// 	{
	// 		cout << dp[i][j] << " ";
	// 	}
	// 	cout << endl;
	// }
	return 0;
}