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 <iostream>
#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    int n;
    cin >> n;
    
    double arr[n+1];
    for(int i=1;i<=n;i++)
        cin >> arr[i];
    
    double dp[3000][3000];
    memset(dp,0,sizeof(dp));
    dp[0][0] = 1;
    
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=n;j++)
        {
            if(i==0 && j!=0)
            {
                dp[i][j] = 0;
            }
            if(j==0 && i!=0)
            {
                dp[i][j] = (1-arr[i])*dp[i-1][j];
            }
        }
    }
    
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            dp[i][j] = (arr[i])*dp[i-1][j-1] + (1-arr[i])*dp[i-1][j];
        }
    }
    
    double ans = 0;
    
    int k = (n+1)/2;
    for(int j=n;j>=k;j--)
        ans += dp[n][j];
    
    cout << fixed << setprecision(9) << ans << endl;
    return 0;
}