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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import Data.Array
import qualified Data.Vector.Unboxed as VU
import qualified Data.ByteString.Char8 as BS
import Control.Monad
import Data.List

solve :: VU.Vector Double -> Int -> Array (Int,Int) Double
solve ps n = dp
    where
        dp :: Array (Int,Int) Double
        dp = listArray ((0,0),(n,n)) [ f i j | i <- [0..n], j <- [0..n]]
        
        f :: Int -> Int -> Double
        f 0 0 = 1
        f _ 0 = 0
        f 0 j = dp ! (0,j-1) * (1 - p j)
        f i j
            | i == j = (dp ! (i-1,j-1)) * p j
            | otherwise = (dp ! (i-1,j-1) * p j) + (dp ! (i,j-1) * (1 - p j)) 
        p n = ps VU.! (n-1)

main = do
    n <- readLn :: IO Int
    ps <- VU.fromListN n . map (read . BS.unpack) . BS.words <$> BS.getLine :: IO (VU.Vector Double)
    let dp = solve ps n
    print $ foldl' (\acc i -> dp ! (i,n) + acc) 0.0 [n `rem` 2 + 1..n]