Problem:
Score: 400 points

Problem Statement

To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives.

He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows:

A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day.

In the i-th day, M-kun can make the following trade any number of times (possibly zero), within the amount of money and stocks that he has at the time.

Buy stock: Pay A_i yen and receive one stock.

Sell stock: Sell one stock for A_i yen.

What is the maximum possible amount of money that M-kun can have in the end by trading optimally?

Constraints

2 \leq N \leq 80

100 \leq A_i \leq 200

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N

Output

Print the maximum possible amount of money that M-kun can have in the end, as an integer.

Sample Input 1

7
100 130 130 130 115 115 150

Sample Output 1

1685

In this sample input, M-kun has seven days of trading. One way to have 1685 yen in the end is as follows:

Initially, he has 1000 yen and no stocks.

Day 1: Buy 10 stocks for 1000 yen. Now he has 0 yen.

Day 2: Sell 7 stocks for 910 yen. Now he has 910 yen.

Day 3: Sell 3 stocks for 390 yen. Now he has 1300 yen.

Day 4: Do nothing.

Day 5: Buy 1 stock for 115 yen. Now he has 1185 yen.

Day 6: Buy 10 stocks for 1150 yen. Now he has 35 yen.

Day 7: Sell 11 stocks for 1650 yen. Now he has 1685 yen.

There is no way to have 1686 yen or more in the end, so the answer is 1685.

Sample Input 2

6
200 180 160 140 120 100

Sample Output 2

1000

In this sample input, it is optimal to do nothing throughout the six days, after which we will have 1000 yen.

Sample Input 3

2
157 193

Sample Output 3

1216

In this sample input, it is optimal to buy 6 stocks in Day 1 and sell them in Day 2, after which we will have 1216 yen.

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#ifdef LOCAL
//#define _GLIBCXX_DEBUG
#endif
//#pragma GCC target("avx512f,avx512dq,avx512cd,avx512bw,avx512vl")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<int, int> Pi;
typedef vector<ll> Vec;
typedef vector<int> Vi;
typedef vector<string> Vs;
typedef vector<P> VP;
typedef vector<vector<ll>> VV;
typedef vector<vector<int>> VVi;
typedef vector<vector<vector<ll>>> VVV;
typedef vector<vector<vector<vector<ll>>>> VVVV;

#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define PER(i, a, b) for(ll i=(a); i>=(b); i--)
#define rep(i, n) REP(i, 0, n)
#define per(i, n) PER(i, n, 0)
const ll INF=1e18+18;
const ll MOD=1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl;
#define ALL(v) v.begin(), v.end()
#define rALL(v) v.rbegin(), v.rend()
#define pb(x) push_back(x)
#define mp(a, b) make_pair(a,b)
#define Each(a,b) for(auto &a :b)
#define rEach(i, mp) for (auto i = mp.rbegin(); i != mp.rend(); ++i)
#ifdef LOCAL
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbgmap(mp) cerr << #mp << ":"<<endl; for (auto i = mp.begin(); i != mp.end(); ++i) { cerr << i->first <<":"<<i->second << endl;}
#define dbgarr(n,m,arr) rep(i,n){rep(j,m){cerr<<arr[i][j]<<" ";}cerr<<endl;}
#define dbgdp(n,arr) rep(i,n){cerr<<arr[i]<<" ";}cerr<<endl;
#define dbgmint(n,arr) rep(i,n){cerr<<arr[i].x<<" ";}cerr<<endl;
#define dbgarrmint(n,m,arr) rep(i,n){rep(j,m){cerr<<arr[i][j].x<<" ";}cerr<<endl;}
#else
#define dbg(...)
#define dbgmap(...)
#define dbgarr(...)
#define dbgdp(...)
#define dbgmint(...)
#define dbgarrmint(...)
#endif
#define out(a) cout<<a<<endl
#define out2(a,b) cout<<a<<" "<<b<<endl
#define vout(v) rep(i,v.size()){cout<<v[i]<<" ";}cout<<endl
#define Uniq(v) v.erase(unique(v.begin(), v.end()), v.end())
#define fi first
#define se second

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }

template<typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; }

template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,v.size())i>>v[j];return i;}

// vector
template<typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
    int len=v.size();
    for(int i=0; i<len; ++i) {
        s<<v[i];
        if(i<len-1) s<<" ";
    }
    return s;
}

// 2 dimentional vector
template<typename T>
ostream &operator<<(ostream &s, const vector<vector<T> > &vv) {
    int len=vv.size();
    for(int i=0; i<len; ++i) {
        s<<vv[i]<<endl;
    }
    return s;
}

//mint
struct mint {
    ll x; // typedef long long ll;
    mint(ll x=0):x((x%MOD+MOD)%MOD){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += MOD-a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= MOD; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime MOD
    mint inv() const { return pow(MOD-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
typedef vector<mint> Vmint;

struct combination {
    vector<mint> fact, ifact;
    combination(int n):fact(n+1),ifact(n+1) {
        assert(n < MOD);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n]*ifact[k]*ifact[n-k];
    }
};

ll h,w,k;
vector<int> buf;
vector<vector<int>> bs;

void dfs(int i, const int size, const int range_start, const int range_end)
{
    //dbg(buf);
    if (i == size) {
        string tmp = "";
        bs.pb(buf);
    }
    else{
        for(int j = range_start; j <= range_end; ++j){
            buf[i] = j;
            dfs(i + 1, size, range_start, range_end);
        }
    }
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<std::setprecision(10);

    ll n;
    cin>>n;
    vector<ll> a(n);
    for(ll i=0; i<n; i++) {
    	cin>>a[i];
    }

    ll ans = 0;

    //全列挙
    int range_start = 0;
    int range_end = 1;
    buf.resize(n);
    dfs(0, n, range_start, range_end);

    for (ll i = 0; i < bs.size(); i++) {
        ll mon=1000;
        ll kabu=0;
        for (ll j = 0; j < n; j++) {
            if(bs[i][j] == 1){
                kabu += mon / a[j];
                mon = mon % a[j];
            } else {
                mon += kabu * a[j];
                kabu = 0;
            }
//            dbg(mon);
        }
        chmax(ans,mon);
    }
//    dbg(bs);

    cout<<ans<<endl;
    return  0;
}