Problem p02624 (C++)

Buggy submission (status: Time Limit Exceeded):
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define ff first
#define ss second
#define file_se_input                                                          \
  freopen("input.txt", "r", stdin);                                            \
  freopen("output.txt", "w", stdout);
#define pp long long int
#define setbits(x) __builtin_popcountll(x)
#define zrobits(x) __builtin_ctzll(x)
#define pb push_back
#define mp make_pair
#define pi 3.141592653589793238
#define eb emplace_back
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define zip(i, a, b) for (int i = a; i < b; i++)
#define rzip(i, a, b) for (int i = a; i >= b; i--)
#define ll unsigned long long int
#define test                                                                   \
  int t;                                                                       \
  cin >> t;                                                                    \
  while (t--)
#define um unordered_map
#define en '\n'
#define us unordered_set
typedef pair<int, int> pii;
typedef pair<char, int> pci;
typedef pair<char, char> pcc;
typedef vector<pii> vii;
typedef long double ld;
#define all(v) v.begin(), v.end()
#define INF (1e18 + 5)
#define inf (1e9 + 5)
#define mod 1000000007

bool check_prime(long long n) {
  int flag = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      flag = 1;
      break;
    }
  }
  if (n == 1)
    return false;
  else if (flag == 0 || n == 2 || n == 3) {
    return true;
  } else {
    return false;
  }
}

int BE(int x, int n, int m) // function to calculate x raise to power n modulo m
{
  int result = 1;
  while (n > 0) {
    if (n % 2 == 1)
      result = result * x % m;
    x = x * x % m;
    n = n / 2;
  }
  return result;
}

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
}

int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
#ifndef ONLINE_JUDGE
  if (fopen("input.txt", "r")) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
#endif
  //_______________________________-code
  //starts-_______________________________________________
  int n;
  cin >> n;
  int ans = 0;
  rep(i, 1, n) { ans += i * (countDivisors(i)); }
  cout << ans << en;

  return 0;
}


Fixed submission:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define ff first
#define ss second
#define file_se_input                                                          \
  freopen("input.txt", "r", stdin);                                            \
  freopen("output.txt", "w", stdout);
#define pp long long int
#define setbits(x) __builtin_popcountll(x)
#define zrobits(x) __builtin_ctzll(x)
#define pb push_back
#define mp make_pair
#define pi 3.141592653589793238
#define eb emplace_back
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define zip(i, a, b) for (int i = a; i < b; i++)
#define rzip(i, a, b) for (int i = a; i >= b; i--)
#define ll unsigned long long int
#define test                                                                   \
  int t;                                                                       \
  cin >> t;                                                                    \
  while (t--)
#define um unordered_map
#define en '\n'
#define us unordered_set
typedef pair<int, int> pii;
typedef pair<char, int> pci;
typedef pair<char, char> pcc;
typedef vector<pii> vii;
typedef long double ld;
#define all(v) v.begin(), v.end()
#define INF (1e18 + 5)
#define inf (1e9 + 5)
#define mod 1000000007

bool check_prime(long long n) {
  int flag = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      flag = 1;
      break;
    }
  }
  if (n == 1)
    return false;
  else if (flag == 0 || n == 2 || n == 3) {
    return true;
  } else {
    return false;
  }
}

int BE(int x, int n, int m) // function to calculate x raise to power n modulo m
{
  int result = 1;
  while (n > 0) {
    if (n % 2 == 1)
      result = result * x % m;
    x = x * x % m;
    n = n / 2;
  }
  return result;
}

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
}

int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
#ifndef ONLINE_JUDGE
  if (fopen("input.txt", "r")) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
#endif
  //_______________________________-code
  //starts-_______________________________________________
  int n;
  cin >> n;
  int ans = 0;
  int j = 1;
  rep(i, 1, n) {
    ans += (i * ((n / j + 1) * (n / j)) / 2);
    j++;
  }
  cout << ans << en;

  return 0;
}