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>
//I:Coins

using namespace::std;

const int SIZE = 3e3 + 1;
double DP[SIZE][SIZE];

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

	int N;
	cin >> N;

	vector<double> P(N + 1);
	for(int i = 1;i <= N;i++)
		cin >> P[i];

	DP[1][1] = P[1];
	DP[1][0] = 1.0 - P[1];
	for(int i = 2;i <= N;i++){
		for(int j = 0;j <= i;j++){
			DP[i][j] = DP[i - 1][j - 1] * P[i] + DP[i - 1][j] * (1 - P[i]);
		}
	}

	double sol = 0;
	for(int i = 0;i <= N;i++){
		if(i > N - i)
			sol += DP[N][i];
	}

	cout << setprecision(10) << fixed << sol << '\n';

	return 0;
}