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)

#![allow(unused)]

use std::cmp::{min, max};

macro_rules! dbg {
    ($($a:expr),*) => {
        writeln!(
            std::io::stderr(),
            concat!($(stringify!($a), " = {:?}, "),*),
            $($a),*);
    }
}

//
// Based on EbTech's Scanner https://stackoverflow.com/q/56449027
//
struct Scanner<'a, R: std::io::BufRead> {
    reader: R,
    buf: String,
    iter: std::str::SplitWhitespace<'a>,
}

impl<'a, R: std::io::BufRead> Scanner<'a, R> {
    fn new(reader: R) -> Scanner<'a, R> {
        Scanner { reader: reader, buf: String::new(), iter: "".split_whitespace() }
    }

    fn get_str(&mut self) -> &str {
        if let Some(t) = self.iter.next() {
            return t;
        }
        self.buf.clear();
        self.reader.read_line(&mut self.buf).expect("I/O error");
        self.iter = unsafe { std::mem::transmute(self.buf.split_whitespace()) };
        self.get_str()
    }

    fn get<T: std::str::FromStr>(&mut self) -> T {
        self.get_str().parse().ok().expect("Parse error")
    }
}

fn main() {
    let stdin = std::io::stdin();
    solve(Scanner::new(stdin.lock()));
}

fn solve<R: std::io::BufRead>(mut sc: Scanner<R>) {
    let n: usize = sc.get();
    let ps: Vec<f64> = (0..n).map(|_| sc.get()).collect();

    // dp[i][j] = i 枚中 j 枚が表である確率
    // 最初は 0 枚中 0 枚表の確率を便宜上 1. とする
    let dp = (1..(n + 1)).fold(vec![1.], |dp, i| {
        let p = ps[i - 1];
        let mut dp_ = vec![0.; i + 1];
        dp_[0] = dp[0] * (1. - p);
        for j in 1..i {
            dp_[j] = dp[j] * (1. - p) + dp[j - 1] * p;
        }
        dp_[i] = dp[i - 1] * p;
        dp_
    });

    println!("{}", (((n + 1) / 2)..(n + 1)).map(|i| dp[i]).sum::<f64>());
}