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 ll long long int 

using namespace std;

int main(){
    int n;
    cin >> n;
    double arr[n+1];
    for(int i=0;i<n;i++) cin >> arr[i];
    double ans[n+1][n+1][2];
    for(int i=0;i<=n;i++){
        ans[0][i][0]=(double)1.000000000;
        ans[0][i][1]=(double)1.000000000;
    }
    for(int i=1;i<=n;i++){
        for(int j=0;j<=i;j++){
            if(j==0){
                ans[i][j][0]=(double)(ans[i-1][j][0])*(double)((double)1-arr[i-1]);
                ans[i][j][1]=(double)0;
            }else if(j==i){
                ans[i][j][0]=(double)0.0000000000;
                ans[i][j][1]=(double) (ans[i-1][j-1][1])*(double)arr[i-1];
            }
            else{
                ans[i][j][0]=(double)(ans[i-1][j][0])*((double)1-arr[i-1])+(double)(ans[i-1][j][1])*((double)1-arr[i-1]);
                ans[i][j][1]=(double)(ans[i-1][j-1][0])*(arr[i-1])+(double)(ans[i-1][j-1][1])*(double)(arr[i-1]);
            }
        }
    }
    double res=0;
    for(int i=1;i<=n;i++){
        if(i>=(n+1)/2){
            //cout <<ans[n][i][0]<<endl;
            //cout <<ans[n][i][1]<<endl;
            if(ans[n][i][0]!=0){
            res=(double)res+(double)ans[n][i][0];}
            if(ans[n][i][1]!=0){
            res=(double)res+(double)ans[n][i][1];}
        }
    }
    cout <<fixed<<setprecision(10)<<res<<endl;
    //cout << ans[n][n][0]<<endl;
    //cout << ans[n][n][1]<<endl;
}