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

#include <bits/stdc++.h>

#define int long long
#define endl '\\n'
#define mod 1000000007
#define pb push_back
#define fi first
#define se second
#define ii pair<int, int>
#define vi vector<int>
#define vii vector<ii>
#define fastIO                                                                 \\
  ios_base::sync_with_stdio(false);                                            \\
  cin.tie(NULL);                                                               \\
  cout.tie(NULL);
#define rep(i, l, r) for (int i = (int)(l); i <= (int)(r); i++)
#define repi(i, l, r) for (int i = (int)(l); i >= (int)(r); i--)
const int inf = 1e18;

using namespace std;

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;
}

void solve() {
  int n;
  cin >> n;

  int cnt = 0;
  generatePrimeFactors();
  rep(i, 1, n - 1) {
    int x = calculateNoOFactors(n - i);
    cnt += x;
  }

  cout << cnt << endl;
}

int32_t main() {
  fastIO int t = 1;
  //    cin >> t;
  while (t--) {
    solve();
  }
}


Error: TLE

Fixed submission:

#include <bits/stdc++.h>

#define int long long
#define endl '\\n'
#define mod 1000000007
#define pb push_back
#define fi first
#define se second
#define ii pair<int, int>
#define vi vector<int>
#define vii vector<ii>
#define fastIO                                                                 \\
  ios_base::sync_with_stdio(false);                                            \\
  cin.tie(NULL);                                                               \\
  cout.tie(NULL);
#define rep(i, l, r) for (int i = (int)(l); i <= (int)(r); i++)
#define repi(i, l, r) for (int i = (int)(l); i >= (int)(r); i--)
const int inf = 1e18;

using namespace std;

int countDivisors(int n) {
  int cnt = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      // If divisors are equal,
      // count only one
      if (n / i == i)
        cnt++;

      else // Otherwise count both
        cnt = cnt + 2;
    }
  }
  return cnt;
}

void solve() {
  int n;
  cin >> n;

  int cnt = 0;
  rep(i, 1, n - 1) {
    int x = countDivisors(n - i);
    cnt += x;
  }

  cout << cnt << endl;
}

int32_t main() {
  fastIO int t = 1;
  //    cin >> t;
  while (t--) {
    solve();
  }
}