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 PyPy3 (7.3.0) (judge verdict: Accepted)

#!/usr/bin/env python3

# from pprint import pprint

def recur(parr, curp, curi, heads):
    # if dp[curi][heads] != -1:
    # return dp[curi][heads]
    if curi >= len(parr):
        if heads >= len(parr)//2 + 1:
            return curp
        else:
            return 0

    return (recur(parr, curp*parr[curi], curi+1, heads+1)
            + recur(parr, curp*(1-parr[curi]), curi+1, heads))

n = int(input())

parr = [float(i) for i in input().split(' ')]

# print(recur(parr, 1, 0, 0))

dp = [0]*(n+1)
dp[0] = 1

for i in range(n):
    for j in range(i+1, -1, -1):
        # heads
        heads = dp[j-1] if j > 0 else 0
        # tails
        tails = dp[j]
        dp[j] = heads*parr[i-1] + tails*(1-parr[i-1])

print(sum(dp[n//2+1:]))