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>
#define pb emplace_back
#define fi first
#define se second
using namespace std;
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int l) { uniform_int_distribution<int> uid(0, l - 1); return uid(rang); }

long double dp[5000];
int32_t main() {
	ios::sync_with_stdio(false); cin.tie(NULL); cout.precision(10); srand(chrono::high_resolution_clock::now().time_since_epoch().count());
	int n; cin >> n;
	vector<long double> a(n); for (int o = 0; o < n; o++) { cin >> a[o]; }
	dp[0] = 1.0;
	for (int i = 0; i < n; ++i)
		for (int j = 3000; j >= 0; --j)
			dp[j] = (j == 0 ? 0.0l : dp[j - 1] * a[i]) + dp[j] * (1.0l - a[i]);
	long double pp = 0;
	for (int i = (n + 2) / 2; i < 3001; ++i)
		pp += (dp[i]);
	cout << pp;
}