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: Runtime Error)

#include<bits/stdc++.h>

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
using namespace std;
#define ll long long
#define endl "\n"
#define ar array
#define pb push_back
#define sz(X) ((int)(X).size())
#define ordered_set tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>
#define sort_unique(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define get_pos(c, x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define all(X) (X).begin(), (X).end()
#define  No cout<<"NO"<<endl;
#define  Yes cout<<"YES"<<endl;
#define  nl cout<<endl;

template<typename T>
ostream &operator+(ostream &out, const vector<T> &vec) {
    for (const auto &x : vec) {
        out << x << " ";
    }
    out << "\n";
    return out;
}

template<typename T>
ostream &operator*(ostream &out, const vector<T> &vec) {
    for (const auto &x : vec) {
        out + x;
    }
    return out;
}

template<typename T>
istream &operator>>(istream &in, vector<T> &vec) {
    for (auto &x : vec) {
        in >> x;
    }
    return in;
}

// DEBUG
void DBG() { cerr << "]" << endl; }

template<class H, class... T>
void DBG(H h, T... t) {
    cerr << ts(h);
    if (sizeof...(t)) cerr << ", ";
    DBG(t...);
}

#ifdef _DEBUG // compile with -DLOCAL
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

template<class T>
bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }

template<class T>
bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }


const int mod = 1e9 + 7;
const int mod2 = 998244353;
const ll INF = 1e9 + 6;

ll add(ll a, ll b) {
    a += b;
    return a % mod;
}

ll mul(ll a, ll b) {
    a *= b;
    return a % mod;
}

inline ll power(ll x, ll y, ll p) {
    ll res = 1;
    x = x % p;
    while (y > 0) {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}

inline ll modInverse(ll x, ll p) {
    return power(x, p - 2, p);
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

const int MXN = 1e3+1;
double p[MXN];
double cache[MXN][MXN];
int n;
double dp(int idx,int heads){
    if(idx>n){
        if(heads>n/2)return 1;
        else return 0;
    }
    double &ans=cache[idx][heads];
    if(ans!=-1){
        return ans;
    }

    ans=dp(idx+1,heads+1)*p[idx]+dp(idx+1,heads)*(1-p[idx]);
    return ans;
}
void solve() {

    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>p[i];
    }
    for(int i=0;i<MXN;i++){
        for (int j = 0; j <MXN ; ++j) {
            cache[i][j]=-1;
        }
    }
    cout<<fixed<<setprecision(10)<<dp(1,0);
}

int main() {

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
#endif
    int t = 1;
    //cin >> t;
    while (t--)solve();
    // cerr <<endl<< "Time elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '\n';
    return 0;
}
//look if it requires ll