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;

#define int long long
#define pb push_back
#define mp make_pair
#define f(i,a,n) for(int i=a ; i<n ; i++)
#define all(c) (c).begin(),(c).end()
#define F first
#define S second
#define fast ios:: sync_with_stdio(false),cin.tie(nullptr);
typedef long double ld;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef pair<pii,int> ppi;
typedef vector<pii> vpi;

const int inf = 1e9;
const int inf64 = 1e18;
const int MOD = inf + 7;
const ld PI = acos(-1.0);

const int N = 3000;
int n;
ld a[N] , dp[N][N];
ld solve(int id , int tot)
{
	if(id == 0)
	{
		if(tot == 1)
			return a[0];
		if(tot == 0)
			return (1 - a[0]);
		return 0;
	}

	ld res = dp[id][tot];
	if(res != -1)   return res;
	
	res = 0;
	res += a[id]*solve(id - 1, tot - 1);
	res += (1.00 - a[id])*(solve(id - 1, tot));
	
	return dp[id][tot] = res;
}
int32_t main(){
	fast;

	int n;
	cin >> n;
	
	f(i,0,n)    cin >> a[i];
	
	f(i,0,N)    f(j,0,N)    dp[i][j] = -1;
	
	ld ans = 0;
	
	for(int i=n/2 + 1 ; i<=n ; i++)
	{
		ans += solve(n-1,i);
	}
	cout << fixed << setprecision(10) << ans;
}