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>
typedef long long int ll;
#define vaibhavv06               \
	ios::sync_with_stdio(false); \
	cin.tie(NULL);               \
	cout.tie(0);
#define endl '\n'
#define int long long
#define MM(a,b) memset(a,b,sizeof(a))
const ll MOD = 1000000007;
using namespace std;


int32_t main(){
    vaibhavv06;
	int n; cin>>n;
    double arr[n];
    //for(int i=0;i<n;i++) cin>>arr[i];
    cout<<fixed<<setprecision(10);
    double dp[n+1]={0};
    dp[0]=1;
    for(int j=0;j<n;j++){
        double p; cin>>p;
        for(int i=j+1;i>=0;--i){
            dp[i]= (i==0 ? 0: dp[i-1]*p) + dp[i]*(1-p);
        }
        // for(int i=0;i<=n;i++) cout<<dp[i]<<" ";
        // cout<<endl;
    }
    double ans=0;
    for(int i=0;i<=n;i++){
        int j=n-i;
        if(i>j) ans+=dp[i];
    }
    cout<<ans<<endl;
}