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

// Delete the dynamically allocated memory

#include <bits/stdc++.h>
using namespace std;

#define ff first
#define ss second
#define pb push_back
#define pr pair<long long, long long>
#define mp(a, b) make_pair(a, b)
#define vr vector<long long>
#define mod1 1000000007
#define mod2 998244353
#define inf 1e18
#define ps(x, y) fixed << setprecision(y) << x
#define fill(a, b) memset(a, b, sizeof(a))
#define mk(arr, n, type) type *arr = new type[n];
#define ll long long
#define ld long double
#define w(x)                                                                   \\
  int x;                                                                       \\
  cin >> x;                                                                    \\
  while (x--)

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll n;
  cin >> n;
  ll sum = 0;
  for (ll i = 1; i <= n; i++) {
    sum += (n / i);
    if (n % i == 0) {
      sum--;
    }
  }
  cout << sum << "\\n";
  return 0;
}


Error: TLE

Fixed submission:


// Problem : C - A x B + C
// Contest : AtCoder - AtCoder Beginner Contest 179
// URL : https://atcoder.jp/contests/abc179/tasks/abc179_c
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

// Delete the dynamically allocated memory

#include <bits/stdc++.h>
using namespace std;

#define ff first
#define ss second
#define pb push_back
#define pr pair<long long, long long>
#define mp(a, b) make_pair(a, b)
#define vr vector<long long>
#define mod1 1000000007
#define mod2 998244353
#define inf 1e18
#define ps(x, y) fixed << setprecision(y) << x
#define fill(a, b) memset(a, b, sizeof(a))
#define mk(arr, n, type) type *arr = new type[n];
#define ll long long
#define ld long double
#define w(x)                                                                   \\
  int x;                                                                       \\
  cin >> x;                                                                    \\
  while (x--)
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 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(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int sum = 0;
  for (int i = 1; i < n; i++) {
    int v = countDivisors(i);
    sum += v;
  }
  cout << sum << "\\n";
  return 0;
}