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 fo(i,s,e_ex) for(i=s;i<e_ex;i++)
#define Fo(i,k,n) for(i=k;k<n?i<=n:i>=n;k<n?i+=1:i-=1)
#define endl '\n'
#define MOD 1000000007//998244353
#define pbb push_back
#define mpp make_pair
#define ff first
#define ss second
#define all(x) x.begin(),x.end()
#define mset(arr,val) memset(arr,val,sizeof(arr))
double dp[3005][3005];
double findAns(double* arr,ll n,ll idx,ll hsf){
	if(idx==n){
		if(hsf>n-hsf) return 1;
		return 0;
	}
	if(dp[idx][hsf]>-1) return dp[idx][hsf];
	double myans=0;
	myans+=arr[idx]*findAns(arr,n,idx+1,hsf+1);//head
	myans+=(1-arr[idx])*findAns(arr,n,idx+1,hsf);//tail
	return dp[idx][hsf]=myans;
}
void solve(){
	ll i,j,n;
	cin>>n;
	double* arr = new double[n];
	fo(i,0,n){
		cin>>arr[i];
	}
	mset(dp,-1);
	double ans = findAns(arr,n,0,0);
	cout<<fixed<<setprecision(12)<<ans<<endl;
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	ll t;
	//cin>>t;
	t=1;
	while(t--){
		solve();
	}
	return 0;
}