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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

#ifdef LOCAL
#include "prabh.hpp"
#endif

#define int long long
#define pb push_back
#define range(v) v.begin(),v.end()
#define rrange(v) v.rbegin(), v.rend()
#define mod (int)(1e9 + 7)
#define inf (int)(5e18)
#define N (int)(3e3 + 10)

using namespace std;

int n;
long double a[N];
long double dp[N][N];

long double coins(int pos, int bal) {
	if (pos == n) return bal > 0;
	long double &ans = dp[pos][bal];
	if (abs(ans + 1) < 1e9) return ans;

	ans = coins(pos + 1, bal + 1) * a[pos] + coins(pos + 1, bal - 1) * (1 -  a[pos]);

	return ans;
}

void solve() {
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	memset(dp, -1.0, sizeof(dp));
	cout << fixed << setprecision(10) << coins(0, 0) << endl;
}
// Remember to clear the global containers

// Got stuck and want to leave ? look at the bottom once..
int32_t main() {
	ios_base::sync_with_stdio(false), cin.tie(nullptr), cin.tie(nullptr);
	int t_c = 1;
	// cin >> t_c;
	while (t_c--)
		solve();
}
/* Come on Dude .. Dont get up until u get the solution
 * Giving up not gonna help ryt ? So get ur lazy ass to code already
 * You can do it ... Stay Calm and Dont Panic :)
 */