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)

use std::io::Read;

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let mut dp = vec![0f64; n + 1];
    dp[0] = 1.0;
    for s in it {
        let p: f64 = s.parse().unwrap();
        for i in (1..(n + 1)).rev() {
            dp[i] = dp[i - 1] * p + dp[i] * (1.0 - p);
        }
        dp[0] *= 1.0 - p;
    }
    let mut ans = 0.0;
    for i in (n / 2 + 1)..(n + 1) {
        ans += dp[i];
    }
    println!("{:.10}", ans);
}

fn main() {
    run();
}