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: Wrong Answer)

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

const int MXN = 3e3 + 20;

int n;
vector<double> a;
double dp[MXN][MXN], ans;

void npt() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		double tmp;
		cin >> tmp;
		a.push_back(tmp);
	}
}

void slv() {
	for (int i = 0; i < MXN; i++)
		for (int j = 0; j < MXN; j++)
			if (i >= j)
				dp[i][j] = 1;
	dp[1][0] = 1 - a[0];
	for (int i = 1; i <= n; i++)
		dp[i][0] = dp[i - 1][0] * (1 - a[i - 1]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			if (i >= j)
				dp[i][j] = dp[i - 1][j] * (1 - a[i - 1]) + 
					dp[i - 1][j - 1] * a[i - 1];
}

void tpt() {
	for (int j = n / 2 + 1; j <= n; j++)
		ans += dp[n][j];
	cout << ans << endl;
}

int main() {
	npt();
	slv();
	tpt();
	return 0;
}