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 Rust (1.15.1) (judge verdict: Accepted)

// Practice Educational DP Contest
// author: Leonardone @ NEETSDKASU

fn main() {
	let mut stdin = String::new();
	std::io::Read::read_to_string(
		&mut std::io::stdin(),
		&mut stdin).unwrap();
	let mut stdin = stdin.split_whitespace();
	let mut get = || stdin.next().unwrap();
	macro_rules! get {
		($t:ty) => (get().parse::<$t>().unwrap());
		() => (get!(usize));
	}
	let n = get!();
	let mut ps = vec![];
	for _ in 0..n {
		ps.push(get!(f64));
	}
	
	let mut dp = vec![0.0; (n+1)/2+1];
	dp[0] = 1.0;
	
	for i in 0..n {
		for j in 0..dp.len()-1 {
			let j = dp.len()-1-j-1;
			dp[j+1] += dp[j] * ps[i];
			dp[j] *= 1.0 - ps[i];
		}
	}
	
	println!("{}", dp[(n+1)/2]);
}