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)

//-----by LY02------------------------------------
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define M 1000000007
//------------------------------------------------

double a[3000],b[3000];
double dp[3000][3000];

//------------------------------------------------
int main() {
//------------------------------------------------
//	freopen("In.txt","r",stdin);
//	freopen("Out.txt","w",stdout);
//------------------------------------------------
	ios::sync_with_stdio(0);
	cin.tie(0);
//------------------------------------------------

int n; cin>>n;
for(int i=0;i<n;i++)
{
	cin>>a[i];
	b[i]=1-a[i];
}
dp[0][0]=b[0];
dp[0][1]=a[0];
for(int i=1;i<n;i++)
{
	for(int j=0;j<=i;j++)
	{
		dp[i][j]+=dp[i-1][j]*b[i];
		dp[i][j+1]+=dp[i-1][j]*a[i];
	}
}
double ans=0;
for(int i=n/2+1;i<=n;i++)
{
	ans+=dp[n-1][i];
}
cout<<fixed<<setprecision(10)<<ans<<endl;


//------------------------------------------------
	return 0;
}





//-----by LY02------------------------------------