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>

#ifndef ONLINE_JUDGE 
const bool DEBUG = true;
#else
const bool DEBUG = false;
#endif

typedef unsigned long long ull;
typedef long long ll;

#define F first
#define S second
#define all(x) begin(x), end(x)

using namespace std;

void exec()
{
	int n;
	cin >> n;

	vector<double> dp(n + 1);
	dp[0] = 1;

	double ans = 0;
	for (int i = 1; i <= n; ++i) {
		double head;
		cin >> head;

		for (int j = i; j > 0; --j) {
			dp[j] = dp[j] * (1 - head) + dp[j - 1] * head;
		}
	}

	/*for (double i : dp)
		cout << i << endl;*/

	int heads = n, tails = 0;
	for (int i = n; i >= 0; --i) {
		if (heads < tails) break;
		ans = dp[i];
		--heads, ++tails;
	}

	cout << ans << endl;
}

















int main()
{
	exec();

	return 0;
}