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::collections::*;
// use std::io::prelude;
use std::io;
use std::io::Read;
use std::cmp::*;

fn main(){
    let t_cases = 1;
    for _ in 0..t_cases{
        solve();
    }
}

fn solve(){
    // let MOD = 10i32.pow(9)+7;
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    let n = buf.trim().to_owned().parse::<usize>().unwrap();
    buf.clear();
    io::stdin().read_line(&mut buf).unwrap();
    let a = buf
                .trim()
                .split_whitespace()
                .map(|x| x.parse::<f64>().unwrap())
                .collect::<Vec<_>>();
    let mut dp = vec![vec![0f64;n+1];n+1];
    dp[0][0] = 1.0;
    for i in 1..(n+1){
        dp[i][0] = dp[i-1][0] * (1.0-a[i-1]);
        for j in 1..(i+1){
            dp[i][j]+=dp[i-1][j]*(1.0-a[i-1])+dp[i-1][j-1]*a[i-1];
        }
    }
    let mut res = 0.0f64;
    for i in (n/2+1)..(n+1){
        res+=dp[n][i];
    }
    println!("{}",res);
}