Problem p02702 (C++) — buggy submission, verdict: Time Limit Exceeded

#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 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 ll mod = 1e9 + 7;
const ld PI = 2 * acos(0.0);
ll m(ll a) { return (a + mod) % mod; }
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;
}
// 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);

  string s;
  cin >> s;
  reverse(all(s));
  map<int, int> freq;
  int rem = 0, p = 1, ans = 0;
  freq[0] = 1;
  for (int i = 0; i < sz(s); i++) {
    int d = s[i] - '0';
    d *= p;
    d %= 2019;
    rem += d;
    rem %= 2019;
    ans += freq[rem];
    freq[rem]++;
    p *= 10;
    p %= 2019;
  }
  cout << ans << endl;
  return 0;
}

Error: TLE

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 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 ll mod = 1e9 + 7;
const ld PI = 2 * acos(0.0);
ll m(ll a) { return (a + mod) % mod; }
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;
}
// 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);
  string s;
  cin >> s;
  int n = sz(s);
  vv dp(n, v(2019, 0));
  int ans = 0;
  dp[0][s[0] - '0'] = 1;
  ans += dp[0][0];
  unordered_map<int, int> mapp;
  mapp[s[0] - '0'] = 1;
  for (int i = 1; i < n; i++) {
    int d = s[i] - '0';
    dp[i][d] = 1;
    queue<int> que;
    for (auto it : mapp) {
      int j = it.fi;
      int rem = (j * 10 + d);
      if (rem >= 2019)
        rem %= 2019;
      dp[i][rem] += dp[i - 1][j];
      if (mapp.count(rem) == 0)
        que.push(rem);
    }
    while (!que.empty()) {
      mapp[que.front()] = 1;
      que.pop();
    }

    mapp[d] = 1;
    ans += dp[i][0];
  }
  cout << ans << endl;

  return 0;
}