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

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000000000000000
#define io                                                                     \\
  ios_base::sync_with_stdio(false);                                            \\
  cin.tie(NULL);
#define pb push_back
#define endl '\\n'
#define pii pair<ll, pair<ll, char>>
const ll mod = 1000000007;
// ll n;
const int MAX = 1000001;

// array to store prime factors
int factor[MAX] = {0};

// function to generate all prime factors
// of numbers from 1 to 10^6
void generatePrimeFactors() {
  factor[1] = 1;

  // Initializes all the positions with their value.
  for (int i = 2; i < MAX; i++)
    factor[i] = i;

  // Initializes all multiples of 2 with 2
  for (int i = 4; i < MAX; i += 2)
    factor[i] = 2;

  // A modified version of Sieve of Eratosthenes to
  // store the smallest prime factor that divides
  // every number.
  for (int i = 3; i * i < MAX; i++) {
    // check if it has no prime factor.
    if (factor[i] == i) {
      // Initializes of j starting from i*i
      for (int j = i * i; j < MAX; j += i) {
        // if it has no prime factor before, then
        // stores the smallest prime divisor
        if (factor[j] == j)
          factor[j] = i;
      }
    }
  }
}

// function to calculate number of factors
int calculateNoOFactors(int n) {
  if (n == 1)
    return 1;

  int ans = 1;

  // stores the smallest prime number
  // that divides n
  int dup = factor[n];

  // stores the count of number of times
  // a prime number divides n.
  int c = 1;

  // reduces to the next number after prime
  // factorization of n
  int j = n / factor[n];

  // false when prime factorization is done
  while (j != 1) {
    // if the same prime number is dividing n,
    // then we increase the count
    if (factor[j] == dup)
      c += 1;

    /* if its a new prime factor that is factorizing n,
       then we again set c=1 and change dup to the new
       prime factor, and apply the formula explained
       above. */
    else {
      dup = factor[j];
      ans = ans * (c + 1);
      c = 1;
    }

    // prime factorizes a number
    j = j / factor[j];
  }

  // for the last prime factor
  ans = ans * (c + 1);

  return ans;
}
ll solve() {
  ll n;
  cin >> n;
  ll ans = 0;
  generatePrimeFactors();
  for (int i = 1; i <= n - 1; i++) {
    ans += calculateNoOFactors(n - i);
  }
  return ans;
}
int main() {
  io ll t = 1;
  // cin>>t;
  while (t--) {
    cout << solve() << endl;
    // solve();
  }
}


Error: TLE

Fixed submission:

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000000000000000
#define io                                                                     \\
  ios_base::sync_with_stdio(false);                                            \\
  cin.tie(NULL);
#define pb push_back
#define endl '\\n'
#define pii pair<ll, pair<ll, char>>
const ll mod = 1000000007;
ll n;
ll countWays(int n1) {
  int count = 0;
  for (int i = 1; i * i <= n1; i++) {
    if (n1 % i == 0 && i * i != n1)
      count += 2;
    else if (n1 % i == 0 && i * i == n1)
      count += 1;
  }

  return count;
}
ll solve() {
  cin >> n;
  ll ans = 0;
  for (int i = 1; i <= n - 1; i++) {
    ans += countWays(n - i);
  }
  return ans;
}
int main() {
  io ll t = 1;
  // cin>>t;
  while (t--) {
    cout << solve() << endl;
    // solve();
  }
}