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: Time Limit Exceeded)

// I - Coins
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
using DBL = double;
int N;
vector<DBL> P(2999); // 0 index
DBL ans = 0;

// n:[N-1,0], p: pre probability, c: front count
void dfs(int n, DBL p, int c){
	if(n < 0){ // finished: out of range
		if((N+2-1)/2 <= c) ans += p; // ceil(N/2)
		return;
	}
	dfs(n-1, p*P[n], c+1); // front: p*pn
	dfs(n-1, p*(1-P[n]), c+0); // back: p*¬pn
}

int main(){
	cin>>N;
	rep(i, N) cin>>P[i];
	dfs(N-1, 1, 0);
	cout<<setprecision(10)<<ans<<endl;
}