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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<cstring>
#include<iomanip>
#define int long long
using namespace std;
double dp[3001][3001];
double func(double a[], int i, int x)
{
	if (x == 0)
	{
		return 1;
	}
	if (i == 0)
	{
		return 0;
	}
	if (dp[i][x] > -0.9)
	{
		return dp[i][x];
	}
	return dp[i][x] = a[i] * func(a, i - 1, x - 1) + (1 - a[i]) * (func(a, i - 1, x));
}
int32_t main()
{
	ios_base:: sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);

	int n;
	cin >> n;
	double a[n + 1];
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	memset(dp, -1, sizeof(dp));
	double ans = func(a, n, (n + 1) / 2);
	cout << fixed << setprecision(12) << ans << "\n";
}