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 lli long long int
#define ld long double
#define rep(i,n) for(int i = 0; i < n; i++)
#define reps(i,a,n) for(int i = a; i < n; i++)
#define s(n) cin >> n
#define csl ios_base::sync_with_stdio(false); cin.tie(NULL)
#define rsz(x,n) x.clear(); x.resize(n)
#define rrsz(x,a,n) x.clear(); x.resize(n,a)
#define wt while(true)
#define all(x) x.begin(),x.end()
#define pb push_back
#define ls(x) ((2*(x)) + 1)
#define rs(x) ((ls(x)) + 1)
#define md(a,b) (((a) + (b))/2)
#define INF (1e9 + 7)

using namespace std;

int main()
{
    lli n;s(n);
    vector<ld>a(n+1);
    rep(i,n)
    {
        ld b;s(b);
        a[i+1]=b;
    }
    vector<vector<ld>>vec(n+1,vector<ld>(n+1,0));
    vec[0][0]=1;
    reps(i,1,n+1)
    vec[i][0]=vec[i-1][0]*(1-a[i]);
    vec[1][1]=a[1];
    reps(i,2,n+1)
    {
        reps(j,1,i+1)
        {
            vec[i][j]+=((vec[i-1][j-1])*(a[i])+(vec[i-1][j])*(1-a[i]));
        }
    }
     /*reps(i,1,n+1)
    cout<<vec[n][i]<<" ";*/
    
    ld sum=0;
    reps(i,(n/2)+1,n+1)
    sum+=vec[n][i];
    cout<<fixed<<setprecision(10)<<sum; 
    return 0;
}