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>

using namespace std;

typedef vector<int> vi;
typedef pair<int, int> ii;
typedef vector<ii> vii;

#define endl '\n'

#define pb push_back
#define mp make_pair

#define fi first
#define se second

#define r(a) cin>>a
#define r2(a,b) cin>>a>>b
#define r3(a,b,c) cin>>a>>b>>c
#define r4(a,b,c,d) cin>>a>>b>>c>>d

#define p(a) cout<<a<<' '
#define pn(a) cout<<a<<endl
#define p2n(a,b) cout<<a<<' '<<b<<endl
#define p3n(a,b,c) cout<<a<<' '<<b<<' '<<c<<endl
#define p4n(a,b,c,d) cout<<a<<' '<<b<<' '<<c<<' '<<d<<endl
#define psn cout<<endl

#define repeat(i,n) for(int i=0;i<n;++i)
#define repeat2(i,a,b) for(int i=a;i<b;++i)
#define rfor(i,n) for(int i=n-1;i>=0;--i)
#define range(i,a) for(auto &i:a)

#define readall(t) range(i,t)r(i)
#define printall(t) range(i,t)p(i);psn

#define chmin(a,b) a=min(a,b)
#define chmax(a,b) a=max(a,b)

#define sz(a) (int)a.size()
#define len(a) (int)a.length()
#define all(a) a.begin(),a.end()

const int MAXN = 3000;

double dp[MAXN][MAXN];
double pr[MAXN];

int32_t main()
{
    ios::sync_with_stdio(false);
    int n ;
    r(n);
    repeat(i, n)
        r(pr[i]);
    
    dp[0][0] = 1 - pr[0];
    dp[0][1] = pr[0];

    repeat2(i, 1, n)
    {
        repeat(j, n + 1)
        {
            dp[i][j] = dp[i - 1][j] * (1 - pr[i]);
            if (j > 0)
                dp[i][j] += dp[i - 1][j - 1] * pr[i];
        }
    }

    double ans = 0;

    repeat2(i, n / 2 + 1, n + 1)
        ans += dp[n - 1][i];

    cout << setprecision(10) << ans << endl;
    return 0;
}