Problem p02624 (C++) — buggy submission, verdict: Time Limit Exceeded

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const int mod = 1000000007;
const double PI = 3.14159265359;
#define mxe(v) *max_element(v.begin(), v.end())
#define mne(v) *min_element(v.begin(), v.end())
#define INF 1000000000000
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define FOR(i, n) for (int i = 0; i < (n); i++)
#define FORA(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
#define FAST                                                                   \\
  ios_base::sync_with_stdio(false);                                            \\
  cin.tie(NULL);
#define count_1(n) __builtin_popcountll(n)
#define parity(n) __builtin_parityll(n) // Even parity 0 as output
void show(int a[], int arraysize) {
  for (int i = 0; i < arraysize; ++i) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int gcd(int a, int b) {
  if (b == 0)
    return a;
  return gcd(b, a % b);
}
void sieve(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));

  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p)
        prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p])
      cout << p << " ";
}
int main() {
  FAST ll n;
  cin >> n;
  ll count = 0;
  for (ll i = 1; i <= n; i++) {
    count += i * (((n / i) * (n / i + 1)) / 2);
  }
  cout << count << endl;
  return 0;
}


Error: TLE

Fixed submission:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const int mod = 1000000007;
const double PI = 3.14159265359;
#define mxe(v) *max_element(v.begin(), v.end())
#define mne(v) *min_element(v.begin(), v.end())
#define INF 1000000000000
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define FOR(i, n) for (int i = 0; i < (n); i++)
#define FORA(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
#define FAST                                                                   \\
  ios_base::sync_with_stdio(false);                                            \\
  cin.tie(NULL);
#define count_1(n) __builtin_popcountll(n)
#define parity(n) __builtin_parityll(n) // Even parity 0 as output
void show(int a[], int arraysize) {
  for (int i = 0; i < arraysize; ++i) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int gcd(int a, int b) {
  if (b == 0)
    return a;
  return gcd(b, a % b);
}
void sieve(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));

  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p)
        prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p])
      cout << p << " ";
}
void SieveOfEratosthenes(int n, bool prime[], bool primesquare[], int 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 (int 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 (int i = 0; i <= (n * n + 1); i++)
    primesquare[i] = false;

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

  for (int 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 (int i = p * 2; i <= n; i += p)
        prime[i] = false;
    }
  }

  int j = 0;
  for (int 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
int countDivisors(int 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];

  int 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
  int ans = 1;

  // Loop for counting factors of n
  for (int 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 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 casse
  else if (n != 1)
    ans = ans * 4;

  return ans; // Total divisors
}
int main() {
  FAST int n;
  cin >> n;
  ll ans = 0;
  FOR(i, n) { ans += ((i + 1) * countDivisors(i + 1)); }
  cout << ans << endl;
  return 0;
}