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 F first
#define S second
#define T second.first
#define C second.second
#define FIO ios_base::sync_with_stdio(0);cin.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int,pair<int,int>> piii;

const int N= 1e2+5, L = 14, OO = 0x3f3f3f3f;
const int M= 1e9+7;
const double ESP = 1e-8;

int n;
double A[3000];
double mem[3000][3000];
double solve(int idx, int tot){
    if(idx == n){
        if(tot>n/2)
            return 1;
        return 0;
    }

    double& ret = mem[idx][tot];
    if(ret != -1)
        return ret;

    return ret = (solve(idx+1,tot+1)*A[idx]) + (solve(idx+1, tot)* (1-A[idx]));
}

int main(){
    FIO
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    cin>>n;
    for(int i=0; i<3000; i++)fill(mem[i], mem[i]+3000, -1);
    for(int i=0; i<n; i++){
        cin>>A[i];
    }
    cout<<fixed<<setprecision(10)<<solve(0,0);

}