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;
use std::str::FromStr;

fn input<T:FromStr>() -> T {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.trim().parse().ok().unwrap()
}

fn input_vec() -> Vec<f64> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.split_whitespace().map(|s| s.parse().unwrap()).collect()
}

fn main() {
    let n: usize = input();
    let p_vec = input_vec();
    let mut dp = vec![vec![0f64; (n+1)]; (n+1)];
    dp[0][0] = 1f64;
    for i in 1..(n+1) {
        for j in 0..(n+1) {
            if j == 0 {
                dp[i][j] = dp[i-1][j] * (1f64 - p_vec[i - 1]);
                continue;
            }
            dp[i][j] = dp[i-1][j] * (1f64 - p_vec[i - 1]) + dp[i-1][j-1] * p_vec[i - 1];
        }
    }
    let mut sum = 0f64;
    for j in (n/2 + 1)..(n+1) {
        sum += dp[n][j];
    }
    println!("{}", sum);
}