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

// In the name of God //

#include <bits/stdc++.h>
#define ll long long
#define all(x) (x).begin(), (x).end()
#define pii pair<int, int>
using namespace std;
const int N = 1e6 + 5;
const int MAXN = 1e7 + 50;
const int mod = 1e9 + 7;

// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
vector<int> spf(MAXN);
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < MAXN; i++)

    // marking smallest prime factor for every
    // number to be itself.
    spf[i] = i;

  // separately marking spf for every even
  // number as 2
  for (int i = 4; i < MAXN; i += 2)
    spf[i] = 2;

  for (int i = 3; i * i < MAXN; i++) {
    // checking if i is prime
    if (spf[i] == i) {
      // marking SPF for all numbers divisible by i
      for (int j = i * i; j < MAXN; j += i)

        // marking spf[j] if it is not
        // previously marked
        if (spf[j] == j)
          spf[j] = i;
    }
  }
}

void solve() {
  int n;
  cin >> n;
  sieve();
  vector<int> dp(MAXN);
  long long ans = 1;
  dp[1] = 1;
  for (int i = 2; i <= n; i++) {
    int x = i;
    int p = spf[x];
    long long s = 0;
    s += dp[x / p];
    x = x / p;
    int q = 0;
    while (x % p == 0) {
      q++;
      x /= p;
    }
    q++;
    s += dp[i / p] / q;
    dp[i] = s;
    ans += i * s;
    // cout << s << " ";
  }
  cout << ans;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tt = 1;
  // cin >> tt;
  for (int tc = 1; tc <= tt; ++tc) {
    solve();
  }
}

Error: TLE

Fixed submission:

// In the name of God //

#include <bits/stdc++.h>
#define ll long long
#define all(x) (x).begin(), (x).end()
#define pii pair<int, int>
using namespace std;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;

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
ll 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 solve() {
  int n;
  cin >> n;
  ll ans = 0;
  for (ll i = 1; i <= n; i++) {
    ans += i * countDivisors(i);
  }
  cout << ans;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tt = 1;
  // cin >> tt;
  for (int tc = 1; tc <= tt; ++tc) {
    solve();
  }
}