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 (2.4.0) (judge verdict: Runtime Error)

import numpy as np
n=int(input())
l=list(map(float,input().split()))
t=(n+1)//2
dp=[[-1 for i in range(n+1)]for i in range(n+1)]
def recur(i,k,ans):
    if i==n:
        # global ans
        if k>=t:
            return 1
            
        return 0
    # a=dp[i][k]
    # dp[i][k]=recur(i+1,k+1,p*l[i],ans)+recur(i+1,k,p*(1-l[i]),ans)
    # b=dp[i][k]
    # if a!=0 and a!=b:
    #     print(a,b)
    # return dp[i][k]
    if dp[i][k]!=-1:
        return dp[i][k]
    
    dp[i][k]=l[i]*recur(i+1,k+1,ans)+recur(i+1,k,ans)*(1-l[i])
    # print(dp[i][k])
    # if dp[i+1][k+1] == 0:
    #     dp[i+1][k+1]=recur(i+1,k+1,p*l[i])
    # if dp[i+1][k]==0:
    #     dp[i+1][k]=recur(i+1,k,p*(1-l[i]))
    return dp[i][k]
# ans=0
# print(dp)
print("%0.10f"%recur(0,0,0))
# print(dp)
# print(ans)