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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import sys
sys.setrecursionlimit(3001)

N = int(input())
p_f = [0]
p_b = [0]
for p in list(map(float,input().split())):
    p_f.append(p)
    p_b.append(1.0-p)

p_majority = 0.0
n_majority = N // 2 + 1

def cointoss(n_toss,n_front,n_back,p):
    global p_majority
    n_toss += 1
    #print(n_toss,n_front,n_back,p,p_majority,n_majority)
    if n_front+1 == n_majority:
        p_majority = p_majority + p * p_f[n_toss]
    else:
        cointoss(n_toss, n_front+1, n_back, p*p_f[n_toss])
    if n_back+1 == n_majority:
        return
    cointoss(n_toss, n_front, n_back+1, p*p_b[n_toss])

cointoss(0,0,0,1.0)
print(p_majority)