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 pb push_back
#define mp make_pair
#define vll vector<long long>
#define ll long long
#define pii pair<int,int>
const int mod=1e9+7;
int n;
long double solve(int i,int j,long double* p,vector<vector<long double>> &dp,vector<vector<bool>> &vis){
    //cout<<i<<" "<<j<<"\n";
    if(i>=n) return 1;
    if(j<=0) return 1;
    if(j>n-i) return 0;
    if(vis[i][j]) return dp[i][j];
    vis[i][j]=1;
    if(n-i==j){
        //cout<<i<<"\n";
        dp[i][j]=p[i]*solve(i+1,j-1,p,dp,vis);
    }
    else{
        dp[i][j]=p[i]*solve(i+1,j-1,p,dp,vis)+((1-p[i])*(solve(i+1,j,p,dp,vis)));
    }
    //cout<<dp[i][j]<<"\n";
    return dp[i][j];
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("inputf.txt","r",stdin);
    freopen("outputf.txt","w",stdout);
#endif
    cin>>n;
    long double p[n];
    for(int i=0;i<n;i++){
        cin>>p[i];
        //cout<<p[i]<<" ";
    }
    //cout<<"\n";
    cout<<fixed<<setprecision(10);

    //memset(dp,-1,sizeof(dp));
    vector<vector<long double>> dp((n+1),vector<long double>((n+3)/2,1));
    vector<vector<bool>> vis((n+1),vector<bool>((n+3)/2,0));
    cout<<solve(0,(n+1)/2,p,dp,vis);
    return 0;
}