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;
#define ll long long 
 
const ll mod=1e9+7;
 
ll n;long double p[10000000];
long double dp[3001][3001];
map<pair<ll,ll> ,double> m;
long double solve(ll c=0,ll d=0){
    // Base Case
    // dp[i][j] represents the probability of having j heads from first i tossed coins
    dp[1][0]=(1-p[0]);
    dp[1][1]=p[0];
    
    for(ll i=2;i<=n;i++){
        for(ll j=0;j<=n;j++){
            if(j>i)break;
            else{
                if(j==0){
                    dp[i][j]=dp[i-1][j]*(1-p[i-1]);
                }
                else
                    dp[i][j]=dp[i-1][j]*(1-p[i-1])+dp[i-1][j-1]*(p[i-1]);
            }
        }
    }
    long double ans=0;
    for(ll i=1;i<=n;i++){
        if(i>n-i)
            ans+=dp[n][i];
    }
    return ans;
}
 
int main() {
    cin>>n;
    memset(dp,0,sizeof(dp));
    for(ll i=0;i<n;i++){
        cin>>p[i];
    }    
 
    long double ans=solve();
    cout<<fixed<<setprecision(10)<<ans<<endl;
 
}