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;


// type definitions
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector< vi > vvi;
typedef pair<int, int> pii;
typedef vector<pii> vpii;
typedef complex<int> ci;
typedef complex<double> cd;
typedef complex<long double> cld;

// macros
#define rep(i, a, b) for(int i=a; i<b; i++)
#define TIME clock() * 1.0 / CLOCKS_PER_SEC
#define all(c) c.begin(), c.end()
#define tr(c, it) for(auto it=(c).begin(); it!=c.end(); it++)
#define pb push_back
#define mp make_pair
#define bitcount __builtin_popcount
#define ispresent(c, x) ((c).find(x) != (c).end())
#define watch(x) cout<< (#x) << " is " << (x) <<"\n";
#define sz(x) ((int)((x).size()))
#define UNIQUE(c) (c).resize(unique(all(c)) - (c).begin())
#define pii2ll(p) ((ll)(p).first<<32 | (p).second)

// template functions
template <typename T>
inline void printV(vector<T>& a) { rep(i, 0, sz(a)) { cout<<a[i]<<" "; } cout<<"\n"; }
template <typename T>
inline void printV(vector<T>& a, int n) { rep(i, 0, n) { cout<<a[i]<<" "; } cout<<"\n"; }
template <typename T>
inline void printVV(vector<vector<T>>& a)
{
	rep(i, 0, sz(a)) {
		printV(a[i]);
	}
	cout<<"\n";
}

template <typename T>
inline void clearVV(vector<vector<T>>& a)
{
	for(auto& x:a) {
		x.clear();
	}
	a.clear();
}
template <typename T>
inline void assignVV(vector<vector<T>> &a, T val)
{
	for(int i=0; i<sz(a); i++) {
		for(int j=0; j<sz(a[i]); j++) {
			a[i][j]=val;
		}
	}
}


// constants
const long double eps = 1e-9;
const int inf = 5e8;
const int modn = 1e9+7;
const int MAX = 1e5;

typedef long double ld;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    // Your code here
    cout<<fixed<<setprecision(10);
	int n;
	cin>>n;
	vector<ld> p(n);
	rep(i,0, n) cin>>p[i];
	vector<vector<ld>> dp(n, vector<ld>(n));
	assignVV(dp, (ld)-1);
	
	function<ld(int, int)> solve=[&](int i, int no)
	{
		if(i==n-1) {
			if(no==0) return 1-p[i];
			else if(no<0) return (ld)0;
			else return (ld)1;
		}
		if(no<0) return (ld)0;
		//~ if(no>=n-i+1) return (ld)1;
		
		if(fabs(dp[i][no]+1)>eps) return dp[i][no];
		ld &memo=dp[i][no];
		memo=p[i]*solve(i+1, no-1);
		memo+=(1-p[i])*solve(i+1, no);
		return memo;
	};
	cout<<1-solve(0, n/2)<<"\n";
	//~ printVV(dp);
    return 0;
}