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 ll long long
#define mod 1e9+7
#define pll pair<long long,long long>
#define pdd pair<long double,long double>
#define vll vector<ll>
#define rep(i,j,n) for(int i=j;i<n;i++)
#define mp make_pair
#define pb push_back
#define pf push_front
#define inf 1e17
priority_queue<pll,vector<pll>,greater<pll> >pq;
priority_queue<ll,vector<ll>,greater<ll> >pq1;

ll n;
vector<double>arr(3000);
vector<vector<double> >dp(3000,vector<double>(3000,-1));

double solve(ll i,ll j){
	if(i==0 && j==0)
		return 1-arr[0];
	if(i==0 && j==1)
		return arr[0];
	if(i==0)
		return 0;
	if(j<0)
		return 0;
	
	if(dp[i][j] != -1)
		return dp[i][j];
	dp[i][j] = arr[i] * solve(i-1,j-1) + (1-arr[i]) * solve(i-1,j) ;
	return dp[i][j];
}

int main(){
	//freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll n;
    cin>>n;
    for(int i=0;i<n;i++){
    	cin>>arr[i];
    }
    if(n==1){
    	cout<<arr[0];
    	return 0;
    }
    dp[0][0]=1;
    double ans=0;
    for(ll i=n/2+1;i<=n;i++){
    	ans=ans+solve(n-1,i);
    }
    
    cout<<setprecision(27)<<ans;
    return 0;
}