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<iostream>
#include<bits/stdc++.h>
#include <stdio.h>
#include<unordered_map>
#define Amr_Abdelazim ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long
#define ld long double
#define pll pair<ll,ll>
#define vll vector<ll>
#define vvll vector<vector<ll> >
#define vpll vector<pair<ll,ll> >
#define P complex<ld>
#define X real()
#define Y imag()
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define rep(i,n)   for(int i=0;i<n;i++)
#define repp(i,s,n)   for(int i=s;i<=n;i++)
#define print(v) {for(auto i:v)cout<<i<<" ";cout<<"\n";}
#define mkp make_pair
#define clr(ar,v) memset(ar,v,sizeof(ar));
#define TC ll amr_abdelazim;cin>>amr_abdelazim;while(amr_abdelazim--)
#define popcnt(x) __builtin_popcount(x)
#define EPS 1e-9
#define INF 1000000000000000009
using namespace std;
//freopen("output.txt", "w", stdout);
//freopen("input.txt", "r", stdin);
const int mod = 1000000007;
const int N = 2 * 100000 + 9;
const double pi = acos(-1);
class cmp{
public:
	bool operator() (ll &A, ll &B){
		return A > B;
	}
};
bool by_sec(const pll &a, const pll &b){
	if ((abs(a.first) == abs(b.first)))
		return (abs(a.second) < abs(b.second));
	return (abs(a.first) < abs(b.first));
}
bool byth(const tuple<ll, ll, ld>& a, const tuple<ll, ll, ld>& b){
	return (get<2>(a) < get<2>(b));
}
ld dist(P a, P b){
	return (ld)abs(a - b);
}
ll fact[1000009], inv[1000009];
ll fp(ll a, ll b){
	if (b == 0)
		return 1;
	ll ans = fp(a, b / 2);
	ans = (ans * ans) % mod;
	if (b % 2 != 0)
		ans = (ans * (a % mod)) % mod;
	return ans;
}
void FacInv(ll n){
	fact[0] = inv[0] = 1;
	for (ll i = 1; i <= n; i++)
	{
		fact[i] = ((i%mod) * fact[i - 1] % mod) % mod;
		inv[i] = fp(fact[i], mod - 2);
	}
}
ll ncr(ll n, ll r){
	return (((fact[n] * inv[r]) % mod) * inv[n - r]) % mod;
}
ll gcd(ll a, ll b){
	if (b == 0)return a;
	return gcd(b, a%b);
}
ll max(ll a, ll b){
	if (a >= b)return a;
	else return b;
}
ll min(ll a, ll b){
	if (a <= b)return a;
	else return b;
}
ld dp[3009][3009];
ll n;
vector<ld>v;
int main()
{
	Amr_Abdelazim	
	cin >> n;
	v = vector<ld>(n);
	rep(i, n)cin >> v[i];
	ld ans = 0;
	dp[0][0] = 1;
	repp(i,1, n)
		rep(j, n + 1){
			if (j == 0)dp[i][j] = (1-v[i - 1]) * dp[i - 1][j];
			else dp[i][j] = v[i - 1] * dp[i - 1][j - 1] + (1 - v[i - 1])*dp[i - 1][j];
		}
	rep(i, n + 1)if (i > n - i)ans += dp[n][i];
	cout << fixed << setprecision(9) << ans << "\n";
	return 0;
}