Problem p02624 (C++) — buggy submission, verdict: Runtime Error

// Created by Kshitij Anand NSIT
#include <bits/stdc++.h>
// #include <ext/numeric>
// using namespace __gnu_cxx;

using namespace std;

#define int long long
#define pb push_back
#define P pair<int, int>
#define F first
#define S second
#define vi vector<int>
#define vc vector<char>
#define vb vector<bool>
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define mp(a, b) make_pair(a, b)
#define min3(a, b, c) min(min(a, b), c)
#define min4(a, b, c, d) min(min(a, b), min(c, d))
#define max3(a, b, c) max(max(a, b), c)
#define max4(a, b, c, d) max(max(a, b), max(c, d))
#define fill(arr, val) memset(arr, val, sizeof(arr))
#define inf 1e18
#define ps(x, y) fixed << setprecision(y) << x
#define db(x) cout << #x << " : " << x << endl
// #define M 10000000

const int N = 1000000007;

const int MAX = 10000001;

// array to store prime factors
int factor[MAX] = {0};

// function to generate all prime factors
// of numbers from 1 to 10^6
void generatePrimeFactors() {
  factor[1] = 1;

  // Initializes all the positions with their value.
  for (int i = 2; i < MAX; i++)
    factor[i] = i;

  // Initializes all multiples of 2 with 2
  for (int i = 4; i < MAX; i += 2)
    factor[i] = 2;

  // A modified version of Sieve of Eratosthenes to
  // store the smallest prime factor that divides
  // every number.
  for (int i = 3; i * i < MAX; i++) {
    // check if it has no prime factor.
    if (factor[i] == i) {
      // Initializes of j starting from i*i
      for (int j = i * i; j < MAX; j += i) {
        // if it has no prime factor before, then
        // stores the smallest prime divisor
        if (factor[j] == j)
          factor[j] = i;
      }
    }
  }
}

// function to calculate number of factors
int calculateNoOFactors(int n) {
  if (n == 1)
    return 1;

  int ans = 1;

  // stores the smallest prime number
  // that divides n
  int dup = factor[n];

  // stores the count of number of times
  // a prime number divides n.
  int c = 1;

  // reduces to the next number after prime
  // factorization of n
  int j = n / factor[n];

  // false when prime factorization is done
  while (j != 1) {
    // if the same prime number is dividing n,
    // then we increase the count
    if (factor[j] == dup)
      c += 1;

    /* if its a new prime factor that is factorizing n,
       then we again set c=1 and change dup to the new
       prime factor, and apply the formula explained
       above. */
    else {
      dup = factor[j];
      ans = ans * (c + 1);
      c = 1;
    }

    // prime factorizes a number
    j = j / factor[j];
  }

  // for the last prime factor
  ans = ans * (c + 1);

  return ans;
}

void solve() {

  int n, a, b, c, ans = 0, count = 0, sum = 1;
  cin >> n;

  // seive();
  generatePrimeFactors();

  for (int i = 1; i <= n; i++) {

    ans += i * calculateNoOFactors(i);
  }

  cout << ans << endl;
  // cout<<primes[n-1]<<endl;
}

int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  //  int t; cin>>t; while(t--)
  { solve(); }

  return 0;
}


Error: 0

Fixed submission:

// Created by Kshitij Anand NSIT
#include <bits/stdc++.h>
// #include <ext/numeric>
// using namespace __gnu_cxx;

using namespace std;

#define int long long
#define pb push_back
#define P pair<int, int>
#define F first
#define S second
#define vi vector<int>
#define vc vector<char>
#define vb vector<bool>
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define mp(a, b) make_pair(a, b)
#define min3(a, b, c) min(min(a, b), c)
#define min4(a, b, c, d) min(min(a, b), min(c, d))
#define max3(a, b, c) max(max(a, b), c)
#define max4(a, b, c, d) max(max(a, b), max(c, d))
#define fill(arr, val) memset(arr, val, sizeof(arr))
#define inf 1e18
#define ps(x, y) fixed << setprecision(y) << x
#define db(x) cout << #x << " : " << x << endl
// #define M 10000000

const int N = 1000000007;

const int MAX = 1000001;

// array to store prime factors
int factor[MAX] = {0};

// function to generate all prime factors
// of numbers from 1 to 10^6
void generatePrimeFactors() {
  factor[1] = 1;

  // Initializes all the positions with their value.
  for (int i = 2; i < MAX; i++)
    factor[i] = i;

  // Initializes all multiples of 2 with 2
  for (int i = 4; i < MAX; i += 2)
    factor[i] = 2;

  // A modified version of Sieve of Eratosthenes to
  // store the smallest prime factor that divides
  // every number.
  for (int i = 3; i * i < MAX; i++) {
    // check if it has no prime factor.
    if (factor[i] == i) {
      // Initializes of j starting from i*i
      for (int j = i * i; j < MAX; j += i) {
        // if it has no prime factor before, then
        // stores the smallest prime divisor
        if (factor[j] == j)
          factor[j] = i;
      }
    }
  }
}

// function to calculate number of factors
int calculateNoOFactors(int n) {
  if (n == 1)
    return 1;

  int ans = 1;

  // stores the smallest prime number
  // that divides n
  int dup = factor[n];

  // stores the count of number of times
  // a prime number divides n.
  int c = 1;

  // reduces to the next number after prime
  // factorization of n
  int j = n / factor[n];

  // false when prime factorization is done
  while (j != 1) {
    // if the same prime number is dividing n,
    // then we increase the count
    if (factor[j] == dup)
      c += 1;

    /* if its a new prime factor that is factorizing n,
       then we again set c=1 and change dup to the new
       prime factor, and apply the formula explained
       above. */
    else {
      dup = factor[j];
      ans = ans * (c + 1);
      c = 1;
    }

    // prime factorizes a number
    j = j / factor[j];
  }

  // for the last prime factor
  ans = ans * (c + 1);

  return ans;
}

void solve() {

  int n, a, b, c, ans = 0, count = 0, sum = 1;
  cin >> n;

  // seive();
  generatePrimeFactors();

  for (int i = 1; i <= n; i++) {

    ans += i * calculateNoOFactors(i);
  }

  cout << ans << endl;
  // cout<<primes[n-1]<<endl;
}

int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  //  int t; cin>>t; while(t--)
  { solve(); }

  return 0;
}