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 <iostream>
#include <iomanip>

using namespace std;

int n;
double p[3005];

double dp[3005][3005];

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		cin >> p[i];
	}
	dp[0][0] = 1.0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= i; ++j) {
			dp[i][j] = dp[i-1][j]*(1.0-p[i]);
			if (j != 0) {
				dp[i][j] += dp[i-1][j-1]*p[i];
			}
		}
	}
	double res = 0.0;
	for (int j = n; j > n/2; --j) {
		res += dp[n][j];
	}
	cout << fixed << setprecision(15) << res << "\n";
	return 0;
}

/*
	dp[i][j] = probability of getting j heads with coins 1..i

	dp[i][j] = dp[i-1][j] * P(i_tail) + dp[i-1][j-1] * P(i_head)
*/