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 (3.8.2) (judge verdict: Runtime Error)

import sys
readline = sys.stdin.readline
#import math
#import numpy as np
N = int(input())
#= input()
#= map(int, input().split())
p = list(map(float, readline().split()))
#= [input(), input()]
#= [list(map(int, input().split())) for _ in range(N)]
#= {i:[] for i in range(N)}

dp = [[0.0] * (N + 1) for _ in range(N + 1)]

dp[0][0] = 1

for i, p in enumerate(p, 1):
    pp = 1 - p
    dpij = dp[i][j]
    dpi = dp[i - 1]
    for j in range(i + 1):
        if j == 0:
            dpij = dpi[j] * pp
        elif j == i:
            dpij = dpi[j - 1] * p
        else:
            dpij = dpi[j - 1] * p + dpi[j] * pp

res = 0.0
n = N // 2 + 1

for cnt in dp[-1][n:]:
    res += cnt

print(res)