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>

using namespace std;

typedef long long int ll;
#define pb push_back
#define ALL(a) (a).begin() , a.end()

int mod = 1e9 + 7;
int add(int a, int b){
	if(a + b >= mod)return a+ b - mod;
	return a + b;
}
int mul(int a, int b){
	return 1LL * a * b  % mod;
}
int const N = 100 * 1010;

/********************i**********************************/
void solve(){
	int n, m;
	cin >> n;
	vector<vector<long double>> dp(n+1, vector<long double> (n + 1, 1));
	vector<long double> vec(n);
	for(auto &a : vec)cin >> a;
	dp[1][1] = vec[0];
	dp[1][0] = 1.0 - vec[0];
	for(int x  =1; x<n; x++){
		long double d = 1.0 - vec[x];
		for(int y = 0; y<=x; y++)
		{
			dp[x+1][y] = dp[x][y] * d;
			if(y > 0)
			dp[x+1][y] += dp[x][y-1] * vec[x];
		}
		dp[x+1][x+1] = dp[x][x] * vec[x];
	}
	long double ans = 0;
	for(int x = n/2+1; x<= n; x++)ans += dp[n][x];
	cout << setprecision(15) << fixed << ans;
}


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