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)

// KOI NI YAAR SAB SAHI HOEGA  BAS  CHORRO MAT
// ENJOR KARO YAAR
// HAAR MAT MAANO
// // APNA TIME AAYEGA
// BEILIEVE IN GOD
#include<bits/stdc++.h>
using namespace std;
#define lli int
#define loop(i,a,b) for(lli i=a;i<b;i++)
#define loopb(i,a,b) for (lli i=a;i>=b;i--)
#define pb push_back
#define mod 1000000007
#define fast() ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define mem(a,b) memset(a,b,sizeof(a))
#define gcd(a,b) (__gcd((a),(b)))
#define lcm(a,b)(((a)*(b))/gcd((a),(b)))
#define ma 1e18
#define bc __builtin_popcountll
#define vi vector<lli>
#define pr pair<lli,lli>
#define vp vector<pr>
#define F  first
#define S   second
#define lld long double
#define PI  3.14159265358979
#define pri  priority_queue<ll>
#define prim  priority_queue<ll,vector<ll>,greater<ll>>
#define flush  cout<<flush

const lli  N=3e3+10;
 
vector<vector<lld>> dp(N,vector<lld>(N,-1)); 
lli n;
vector<lld>a;
lli ans;



lld solve(lli index , lli tails)
{if(index == n)
{if(tails<=n/2)
{return 1;
}
else
{return 0;
}
}
if(dp[index][tails]!=-1)
{return dp[index][tails];
}
return dp[index][tails] =  ((a[index] * 1 * solve(index+1,tails)) + (1-a[index]) * 1 *  solve(index+1,tails+1));
}

int main()
{fast();
cin>>n;

a.resize(n);
loop(i,0,n)
{cin>>a[i];
}

cout<<fixed<<setprecision(10)<<solve(0,0)<<"\n";
}