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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define  forloop for(int i=0;i<n;i++)
#define pb push_back 
#define ull unsigned long long
#define ll long long
#define mod 1000000007
#define MOD 998244353;
#define unmap unordered_map
#define pi 3.1415926536
#define testcases ll t;cin>>t;while(t--)
#define  mp make_pair
#define  vcp vector<pair<ll,ll>>
#define preci(x,y)  fixed<<setprecision(y)<<x
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define fast ios::sync_with_stdio(false);   cin.tie(0); cout.tie(0);
#define preinc priority_queue <ll, vector<ll>, greater<ll>>
#define predec priority_queue<ll> 
#define vll vector<ll>
#define f(a,b) for(ll i=a;i<=b;i++)
double coins( double a[],int n,vector<vector<double>>&v)
{
    v[0][0]=1.0;
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<=i;j++)
        {
            if(j==0)
                v[i][j]=(1-a[i-1])*v[i-1][j];
            else
                v[i][j]=(a[i-1])*v[i-1][j-1]+(1-a[i-1])*v[i-1][j];
        }
    }
    double ans=0.0;
    for(int i=n/2+1;i<=n;i++)
    {
        ans+=v[n][i];
    }
    return ans;

}
int main() {
    #ifndef ONLINE_JUDGE
   freopen("inp.txt", "r", stdin);
   freopen("output.txt", "w", stdout);
   #endif
   fast
   int n;
   cin>>n;
   double a[n];
   forloop
   cin>>a[i];
   vector<vector<double>>v(n+1,vector<double>(n+1,0.0));
   cout<<preci(coins(a,n,v),10);
    // your code goes here
    return 0;
}