Problem p02603 (C++) — buggy submission, verdict: Runtime Error

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ordered_set                                                            \\
  tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
#define int long long
typedef unsigned long long lu;
typedef vector<ll> v;
typedef vector<vector<ll>> vv;
typedef vector<string> vs;
typedef vector<pair<ll, ll>> vpr;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef long double ld;
#define f(i, n) for (ll i = 0; i < n; i++)
#define ff(i, n) for (ll i = 1; i <= n; i++)
#define pb push_back
#define mp make_pair
#define endl "\\n"
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define amax(x, y)                                                             \\
  if (y > x)                                                                   \\
    x = y;
#define amin(x, y)                                                             \\
  if (y < x)                                                                   \\
    x = y;
#define bg(x) x.begin()
#define sz(x) (ll) x.size()
#define in(x, n)                                                               \\
  for (ll i = 0; i < n; i++)                                                   \\
  cin >> x[i]
#define out(x, n)                                                              \\
  for (ll i = 0; i < n; i++)                                                   \\
  cout << x[i] << " "
#define mxt(a) *(max_element(a.begin(), a.end()))
#define mnt(a) *(min_element(a.begin(),a.end())
#define tc                                                                     \\
  ll t;                                                                        \\
  cin >> t;                                                                    \\
  while (t--)
typedef pair<ll, ll> pi;
#define yes cout << "YES\\n";
#define no cout << "NO\\n";
#define yesno(f)                                                               \\
  if (f)                                                                       \\
  yes else no
const v dx = {1, -1, 0, 0};
const v dy = {0, 0, 1, -1};
const ld PI = 2 * acos(0.0);
ll cel(ll x1, ll y1) {
  if ((x1 % y1) == 0)
    return x1 / y1;
  else
    return x1 / y1 + 1;
}
ll power(ll a, ll b, ll m) {
  if (b == 0)
    return 1;
  ll d = power(a, b / 2, m);
  d = (d * d) % m;
  if (b & 1)
    d = (d * a) % m;
  return d;
}
const ll mod = 1e9 + 7;
int MOD(int a) {
  if (a < 0)
    a += mod;
  if (a >= mod)
    a %= mod;
  return a;
}
// set_name.find_by_order(k) It returns to an iterator to the kth element
// (counting from zero) in the set in O(logn) time set_name.order_of_key(k) It
// returns to the number of items that are strictly smaller than our item k in
// O(logn) time.
/*string operations :
str.substr (x,y) : returns a substring str[x],str[x+1],...str[x+y-1]
str.substr (x) : returns a substring str[x],... end of string
str.find(qtr) : returns the first occurenece of qtr in str */
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  // freopen("input.txt", "r", stdin);
  // freopen("output.txt", "w", stdout);
  int n;
  cin >> n;
  vector<int> a(n);
  in(a, n);
  int money = 1000, stock = 0;
  for (int i = 0; i < n; i++) {
    if (i != (n - 1) && a[i + 1] > a[i]) {

      stock += money / a[i];
      money %= a[i];
    } else {
      money += stock * a[i];
      stock = 0;
    }
  }
  cout << money << endl;
  return 0;
}

Error: -11

Fixed submission:

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ordered_set                                                            \\
  tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
#define int long long
typedef unsigned long long lu;
typedef vector<ll> v;
typedef vector<vector<ll>> vv;
typedef vector<string> vs;
typedef vector<pair<ll, ll>> vpr;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef long double ld;
#define f(i, n) for (ll i = 0; i < n; i++)
#define ff(i, n) for (ll i = 1; i <= n; i++)
#define pb push_back
#define mp make_pair
#define endl "\\n"
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define amax(x, y)                                                             \\
  if (y > x)                                                                   \\
    x = y;
#define amin(x, y)                                                             \\
  if (y < x)                                                                   \\
    x = y;
#define bg(x) x.begin()
#define sz(x) (ll) x.size()
#define in(x, n)                                                               \\
  for (ll i = 0; i < n; i++)                                                   \\
  cin >> x[i]
#define out(x, n)                                                              \\
  for (ll i = 0; i < n; i++)                                                   \\
  cout << x[i] << " "
#define mxt(a) *(max_element(a.begin(), a.end()))
#define mnt(a) *(min_element(a.begin(),a.end())
#define tc                                                                     \\
  ll t;                                                                        \\
  cin >> t;                                                                    \\
  while (t--)
typedef pair<ll, ll> pi;
#define yes cout << "YES\\n";
#define no cout << "NO\\n";
#define yesno(f)                                                               \\
  if (f)                                                                       \\
  yes else no
const v dx = {1, -1, 0, 0};
const v dy = {0, 0, 1, -1};
const ld PI = 2 * acos(0.0);
ll cel(ll x1, ll y1) {
  if ((x1 % y1) == 0)
    return x1 / y1;
  else
    return x1 / y1 + 1;
}
ll power(ll a, ll b, ll m) {
  if (b == 0)
    return 1;
  ll d = power(a, b / 2, m);
  d = (d * d) % m;
  if (b & 1)
    d = (d * a) % m;
  return d;
}
const ll mod = 1e9 + 7;
int MOD(int a) {
  if (a < 0)
    a += mod;
  if (a >= mod)
    a %= mod;
  return a;
}
// set_name.find_by_order(k) It returns to an iterator to the kth element
// (counting from zero) in the set in O(logn) time set_name.order_of_key(k) It
// returns to the number of items that are strictly smaller than our item k in
// O(logn) time.
/*string operations :
str.substr (x,y) : returns a substring str[x],str[x+1],...str[x+y-1]
str.substr (x) : returns a substring str[x],... end of string
str.find(qtr) : returns the first occurenece of qtr in str */
const int jmax = 100 * 80 + 1000 + 1;
int dp[82][jmax][82];
vector<vector<vb>> vst(82, vector<vb>(jmax, vb(82, false)));
int n;
vector<int> a;
int fun(int i, int j, int k) {
  if (i == n)
    return j;
  if (vst[i][j][k])
    return dp[i][j][k];

  dp[i][j][k] = fun(i + 1, j, k);
  if (i == (n - 1)) {
    if (k != 0) {
      int val = fun(i, j + a[i], k - 1);
      dp[i][j][k] = max(dp[i][j][k], val);
    }
  } else {
    if (a[i] >= a[i + 1]) {
      if (k != 0) {
        int val = fun(i, j + a[i], k - 1);
        dp[i][j][k] = max(dp[i][j][k], val);
      }
    } else {
      if (a[i] <= j) {
        int val = fun(i, j - a[i], k + 1);
        dp[i][j][k] = max(dp[i][j][k], val);
      }
    }
  }
  vst[i][j][k] = true;
  return dp[i][j][k];
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  // freopen("input.txt", "r", stdin);
  // freopen("output.txt", "w", stdout);
  cin >> n;
  a.resize(n);
  in(a, n);
  cout << fun(0, 1000, 0) << endl;
  return 0;
}