Problem p02548 (C++) — buggy submission, verdict: Time Limit Exceeded

// nb_9960
#include <bits/stdc++.h>
using namespace std;
#define fi(i, a, n) for (ll i = a; i < n; i++)
#define ll long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
typedef pair<ll, ll> pl;
typedef vector<ll> vl;
typedef vector<pl> vpii;
#define sz(a) int((a).size())
#define tr(c, i) for (typeof(c).begin() i = c.begin(); i != c.end(); i++)
#define present(c, x) (c.find(x) != c.end())
#define cpresent(c, x) (find(all(c), x) != c.end())
#define MOD 1000000007
#define PI 3.1415926536

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);

  ll t = 1;
  // cin >> t;

  while (t--) {
    ll n;
    cin >> n;
    ll cnt = 0;
    for (ll a = 1; a < n; a++) {
      for (ll b = 1; b < n; b++) {
        if (a * b >= n) {
          break;
        } else
          cnt++;
      }
    }
    cout << cnt << endl;
  }
  return 0;
}


Error: TLE

Fixed submission:

// nb_9960
#include <bits/stdc++.h>
using namespace std;
#define fi(i, a, n) for (ll i = a; i < n; i++)
#define ll long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
typedef pair<ll, ll> pl;
typedef vector<ll> vl;
typedef vector<pl> vpii;
#define sz(a) int((a).size())
#define tr(c, i) for (typeof(c).begin() i = c.begin(); i != c.end(); i++)
#define present(c, x) (c.find(x) != c.end())
#define cpresent(c, x) (find(all(c), x) != c.end())
#define MOD 1000000007
#define PI 3.1415926536

void SieveOfEratosthenes(ll n, bool prime[], bool primesquare[], ll a[]) {
  // Create a boolean array "prime[0..n]" and
  // initialize all entries it as true. A value
  // in prime[i] will finally be false if i is
  // Not a prime, else true.
  for (ll i = 2; i <= n; i++)
    prime[i] = true;

  // Create a boolean array "primesquare[0..n*n+1]"
  // and initialize all entries it as false. A value
  // in squareprime[i] will finally be true if i is
  // square of prime, else false.
  for (ll i = 0; i <= (n * n + 1); i++)
    primesquare[i] = false;

  // 1 is not a prime number
  prime[1] = false;

  for (ll p = 2; p * p <= n; p++) {
    // If prime[p] is not changed, then
    // it is a prime
    if (prime[p] == true) {
      // Update all multiples of p
      for (ll i = p * 2; i <= n; i += p)
        prime[i] = false;
    }
  }

  ll j = 0;
  for (ll p = 2; p <= n; p++) {
    if (prime[p]) {
      // Storing primes in an array
      a[j] = p;

      // Update value in primesquare[p*p],
      // if p is prime.
      primesquare[p * p] = true;
      j++;
    }
  }
}

// Function to count divisors
ll countDivisors(ll n) {
  // If number is 1, then it will have only 1
  // as a factor. So, total factors will be 1.
  if (n == 1)
    return 1;

  bool prime[n + 1], primesquare[n * n + 1];

  ll a[n]; // for storing primes upto n

  // Calling SieveOfEratosthenes to store prime
  // factors of n and to store square of prime
  // factors of n
  SieveOfEratosthenes(n, prime, primesquare, a);

  // ans will contain total number of distinct
  // divisors
  ll ans = 1;

  // Loop for counting factors of n
  for (ll i = 0;; i++) {
    // a[i] is not less than cube root n
    if (a[i] * a[i] * a[i] > n)
      break;

    // Calculating power of a[i] in n.
    int cnt = 1;          // cnt is power of prime a[i] in n.
    while (n % a[i] == 0) // if a[i] is a factor of n
    {
      n = n / a[i];
      cnt = cnt + 1; // incrementing power
    }

    // Calculating the number of divisors
    // If n = a^p * b^q then total divisors of n
    // are (p+1)*(q+1)
    ans = ans * cnt;
  }

  // if a[i] is greater than cube root of n

  // First case
  if (prime[n])
    ans = ans * 2;

  // Second case
  else if (primesquare[n])
    ans = ans * 3;

  // Third case
  else if (n != 1)
    ans = ans * 4;

  return ans; // Total divisors
}

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);

  ll t = 1;
  // cin >> t;

  while (t--) {
    ll n;
    cin >> n;
    ll c = n - 1, cnt = 0, m = n - c;
    while (m <= (n - 1)) {
      cnt += countDivisors(m);
      m++;
    }
    cout << cnt << endl;
  }
  return 0;
}