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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int n;
double dp[3000][3000] ;

double a[3000] = {0.0};
double f(int h,  int i) {

	if (h == 0 ) return 1;
	if (i == n + 1) return 0;
	if (dp[i][h] > -0.9)
		return dp[i][h];

	return dp[i][h] = a[i] * f(h - 1,  i + 1) + (1 - a[i]) * f(h,  i + 1);
}

int main()
{	
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

	memset(dp, -1, sizeof dp);
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	int k = ( n + 1) / 2 ;
	double ans = 0;
	// for (int i = k; i <= n; i++) {
	// 	ans +=
	// }
	//cout <<
	cout << fixed << setprecision(10) << f(k,  1);
	return 0;
}