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++14 (GCC 5.4.1) (judge verdict: Accepted)

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

int main(){

	int n;
	scanf("%d", &n);

	double p[n];
	for (int i = 0; i < n; ++i) {
		scanf("%lf", &p[i]);
	}

	double dp[n+1][n+1];
	memset(dp, 0.0, sizeof dp);

	dp[0][0] = 0.0;
	dp[1][0] = p[0];
	dp[0][1] = 1.0 - p[0];

	for(int sum = 2; sum <= n; sum++) {
		for (int i = 0; i <= sum; ++i) {
			if(i == 0) {
				//cout << i << " " << sum-i << ": " << dp[i][sum-i-1] << "*" << (1 - p[sum-1]) << endl;
				dp[i][sum-i] = dp[i][sum-i-1]*(1 - p[sum-1]);
			}
			else if(i == sum) {
				//cout << i << " " << sum-i << ": " << dp[i-1][sum-i] << "*" << (p[sum-1]) << endl;
				dp[i][sum-i] = dp[i-1][sum-i]*(p[sum-1]);
			}
			else {
				//cout << i << " " << sum-i << ": " << dp[i][sum-i-1] << "*" << (1 - p[sum-1]) << " + " << dp[i-1][sum-i] << "*" << (p[sum-1]) << endl;
				dp[i][sum-i] = dp[i][sum-i-1]*(1 - p[sum-1]) + dp[i-1][sum-i]*(p[sum-1]);
			}
		}
	}
	
	double ans = 0.0;
	for (int i = n; i >= 0; --i) {
		if(i > n-i)ans += dp[i][n-i];
		else break;
	}

	cout << setprecision(10) << fixed << ans << endl;
	
	return 0;
}