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>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
#define fastio() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define pb push_back
#define F first
#define S second
const string nl="\n";
const ll MOD=1e9+7;
const ll ARR_MAX = 1e5+1;
const ll INF=1e14;

void solve(){
    int n;cin>>n;
    ld p[n];
    for(int i=0;i<n;i++){
    	cin>>p[i];
    }
    ld dp[n+1][n+1]={}; //dp[i][j]= probability of having j heads after (i-1)th toss.
    dp[0][0]=1;
    for(int i=0;i<n;i++){
    	for(int j=0;j<=i+1;j++){
    		if(j!=0){
    			dp[i+1][j]=dp[i][j-1]*p[i];
    		}
    		dp[i+1][j]+=dp[i][j]*(1-p[i]);
    	}
    }
    // for(int i=0;i<=n;i++){
    // 	for(int j=0;j<=n;j++){
    // 		cout<<dp[i][j]<<" ";
    // 	}
    // 	cout<<nl;
    // }
    ld prob=0;
    for(int i=n;2*i>n;i--){
    	prob+=dp[n][i];
    }
	cout<<fixed<<setprecision(11);
    cout<<prob<<nl;
}

int main(){
    int t=1;
    // cin>>t;
    while(t--){
        solve();
    }
    return 0;
}