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>
using namespace std;
using ll=long long;
using ull=unsigned long long;
using pii=pair<ll,ll>;
#define For(i,a,b)  for(int i=a;i<=b;i++)
#define Forr(i,a,b) for(int i=a;i>=b;i--)
#define F first
#define S second
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define sz(x) ((int)x.size())
#define bpc __builtin_popcount
#define MOD (ll)(1e9+7)
#define INF (2e9)
#define EPS (1e-6)
#define NATHANLEE_IS_EVIL_AND_ELECTRIC true
#define int ll
#ifdef LOCAL_TEST
#define debug(...) do{\
	cerr<<__PRETTY_FUNCTION__<<" - "<<__LINE__<<\
	" : ("#__VA_ARGS__<<") = ";\
	_OUT(__VA_ARGS__);\
}while(0)
template<typename T>void _OUT(T x){cerr<<x<<"\n";}
template<typename T,typename...I>void _OUT(T x,I ...tail){cerr<<x<<", ";_OUT(tail...);}
#else
#define debug(...)
#endif
void INIT(){
#ifdef LOCAL_TEST
	freopen("D:\\input.in","r",stdin);
	freopen("D:\\output.out","w",stdout);
	freopen("D:\\error.out","w",stderr);
#endif
	ios::sync_with_stdio(!NATHANLEE_IS_EVIL_AND_ELECTRIC);
	cin.tie(0);
}

ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); }
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll fpow(ll b,ll p){
	b%=MOD;
	ll ans=1;
	while(p>0){
		if(p&1) ans=(ans*b)%MOD;
		p/=2;
		b=(b*b)%MOD;
	}
	return ans;
}

double pos[3010][3010]; //first i with j head

int32_t main() {
	INIT();
	//code...
	int n; cin>>n;
	memset(pos,0,sizeof(pos));
	pos[0][0]=1;
	For(i,1,n){
		double p; cin>>p;
		pos[i][0]=pos[i-1][0]*(1-p);
		For(j,1,n) pos[i][j]=pos[i-1][j]*(1-p)+pos[i-1][j-1]*p;
	}
	double ans=0;
	For(i,n/2+1,n) ans+=pos[n][i];
	cout<<fixed<<setprecision(15)<<ans<<"\n";
	return 0;
}