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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define all(x) x.begin(), x.end()
#define pb push_back
#define mp make_pair
#define re reverse
#define ps push
#define in insert
#define sv(v) shuffle(all(v), default_random_engine())
#define sa(a, n) shuffle(a, a+n, default_random_engine())
#define ca(v) for(auto i:v) cout<<i<<" ";
#define nl cout<<"\n"
#define del(s, x) s.erase(s.find(x))
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) (a*b/gcd(a, b))
#define debug cout<<"DEBUG\n"
const int MOD = 1e9 + 7;
const int MAXN = 3e3+5;

double dp[MAXN][MAXN];
int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    int n; cin>>n;
    double ar[n];
    for(int i=0; i<n; i++) cin>>ar[i];
    dp[0][0] = 1-ar[0];
    dp[0][1] = ar[0];
    for(int i=1; i<n; i++){
        for(int j=n; j>0; j--) {
            dp[i][j]+=dp[i-1][j-1] * ar[i];
            dp[i][j]+=dp[i-1][j] * (1-ar[i]);
        }
        dp[i][0] = dp[i-1][0] * (1-ar[i]);
    }
    double tot = 0;
    for(int i=n/2+1; i<=n; i++){
        tot+=dp[n-1][i];
    }
    cout<<tot;
}