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> 
#define ll long long
#define M 1000000007
using namespace std;
int main() 
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); 
	int n,m,a,b;
    cin>>n;
    vector<vector<double>>dp(n, vector<double>(n+1)); 
    vector<double>jp(n); 
    for(int i=0;i<n;i++)
    {
         cin>>jp[i];
    }
    dp[0][0]=(1.0000-jp[0]),dp[0][1]=jp[0];
    for(int i=1;i<n;i++)
    {
        dp[i][0]=dp[i-1][0]*(1.0-jp[i]);
    }
    for(int i=1;i<n;i++)
    {  // cout<<"i: ";
        for(int j=1;j<=i+1;j++)
        {
            dp[i][j]=dp[i-1][j-1]*(jp[i])+dp[i-1][j]*(1.00-jp[i]);
            //cout<<dp[i][j]<<" ";
        }
       // cout<<endl;
    }
    double sum=0.00;
    for(int i=n/2+1;i<=n;i++)
    {
        sum+=dp[n-1][i];

    }
    
   
  
    cout<<setprecision(10)<<sum;
	return 0; 
}