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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;

public static class Solution
{
  public static void Main()
  {
    Console.ReadLine();
    double[] p = Array.ConvertAll(Console.ReadLine().Split(' '), double.Parse);
    double total = 0;
    for (int i = 0; i < (1 << p.Length); i++)
    {
      total += GetProbability(i, p); // Counts only if heads > tails
    }
    Console.WriteLine(total);
  }

  private static double GetProbability(int n, double[] p)
  {
    //Assume zero-bit means head -> p[i]
    int headsCount = 0;
    double probability = 1;
    for (int i = 0; i < p.Length; i++)
    {
      if (((n >> i) & 1) == 0)
      {
        headsCount++;
        probability *= p[i];
      }
      else
      {
        probability *= 1 - p[i];
      }
    }
    return (headsCount > p.Length / 2) ? probability : 0;
  }
}