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

#include <bits/stdc++.h>
#include <string>
#include <vector>
using namespace std;
#define l long long
#define pb push_back
#define mem(a, n) memset(a, n, sizeof(a))
#define find(v, n) find(v.begin(), v.end(), n) - v.begin()
#define sort1(v) sort(v.begin(), v.end())
#define lb(v, n) lower_bound(v.begin(), v.end(), n) - v.begin()
#define ub(v, n) upper_bound(v.begin(), v.end(), n) - v.begin()
#define mod 1000000007
#define reverse(s) reverse(s.begin(), s.end())
// #define pi pair<l,l>
const double pi = 3.141592653589793;
bool isprime(l n) {
  for (l i = 2; i * i <= n; i++) {
    if (n % i == 0)
      return false;
  }
  return true;
}
/*bool fact(l n)
{
        vector<l> v;
        for(l i=2;i*i<=n;i++)
        {
                if(n%i==0)
                {
                        if(i%2==1||(n/i)%2==1)
                        {
                                return true;
                        }
                }
        }
        return false;
}*/
l power(l a, l b) {
  l ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a);
    }
    b >>= 1;
    a = (a * a);
  }
  return ans;
} // prefixsums
void frooty() {
  l n;
  cin >> n;
  l sum = 0;
  l div[n + 1];
  mem(div, 0);
  for (l i = 1; i <= n; i++) {
    for (l j = i; j <= n; j += i) {
      div[j]++;
    }
  }
  for (l i = 1; i <= n; i++) {
    sum += (i * div[i]);
  }
  cout << sum << "\\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  //	cin>>q;
  while (q--) {
    frooty();
  }
}


Error: TLE

Fixed submission:

#include <bits/stdc++.h>
#include <string>
#include <vector>
using namespace std;
#define l long long
#define pb push_back
#define mem(a, n) memset(a, n, sizeof(a))
#define find(v, n) find(v.begin(), v.end(), n) - v.begin()
#define sort1(v) sort(v.begin(), v.end())
#define lb(v, n) lower_bound(v.begin(), v.end(), n) - v.begin()
#define ub(v, n) upper_bound(v.begin(), v.end(), n) - v.begin()
#define mod 1000000007
#define reverse(s) reverse(s.begin(), s.end())
// #define pi pair<l,l>
const double pi = 3.141592653589793;
bool isprime(l n) {
  for (l i = 2; i <= n; i++) {
    if (n % i == 0)
      return false;
  }
  return true;
}
/*bool fact(l n)
{
        vector<l> v;
        for(l i=2;i*i<=n;i++)
        {
                if(n%i==0)
                {
                        if(i%2==1||(n/i)%2==1)
                        {
                                return true;
                        }
                }
        }
        return false;
}*/
l power(l a, l b) {
  l ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a);
    }
    b >>= 1;
    a = (a * a);
  }
  return ans;
} // prefixsums
// C++ program to co

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
}
void frooty() {
  l n;
  cin >> n;
  l sum = 0;
  vector<l> v;
  for (l i = 1; i <= n; i++) {
    sum += (countDivisors(i) * i);
    //	cout<<sum<<"\\n";
    v.pb(sum);
  }
  cout << v[n - 1] << "\\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  //	cin>>q;
  while (q--) {
    frooty();
  }
}