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: Wrong Answer)

/*author: hyperion_1724
  date: 
*/
#include <bits/stdc++.h>
using namespace std;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	// chrono::steady_clock sc;
 //  	auto start = sc.now();
 //  	ifstream cin;
 //  	cin.open("INPUT.txt");

	int N;
	cin>>N;
	cout<<fixed<<showpoint;
	cout<<setprecision(11);
	long double P[N];
	for (int i = 0; i < N; ++i)
	{
		cin>>P[i];
	}
	long double ans=P[0];
	long double h3=0,h1=P[0],t1=1-P[0];
	for (int i = 2; i < N; i+=2)
	{
		long double hh=P[i]*P[i-1];
		long double ht=P[i]*(1-P[i-1]);
		long double th=(1-P[i])*P[i-1];
		long double tt=(1-P[i])*(1-P[i-1]);
		ans=t1*hh+h1*(hh+ht+th)+h3*(tt+hh+ht+th);
		long double h3c=h3,h1c=h1,t1c=t1;
		h3=h3c*(ht+th)+h1c*hh;
		h1=h3c*tt+h1c*(ht+th)+t1c*hh;
		t1=t1c*(ht+th)+h1c*(tt);
	}
	cout<<ans<<endl;
	
	// cin.close();
	// auto end = sc.now();
 //   	auto time_span = static_cast<chrono::duration<double>>(end - start);
 //   	cout<<"Operation took: "<<time_span.count()<<" seconds";

	return 0;
}