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

#include <algorithm>
#include <iostream>
using namespace std;

long long n, ans, c[10000001];

int main() {
  cin >> n;

  for (int i = 1; i <= n; ++i) {
    for (int j = 0; (j += i) <= n;) {
      c[j]++;
    }
  }

  for (int i = 1; i <= n; i++) {
    ans += (long long)i * c[i];
  }

  cout << ans << endl;
}

/*
#include <algorithm>
#include <chrono>
#include <iostream>
#include <utility>
#include <vector>
using namespace std;

long long n, ans;

/*
long long div_cnt(long long num)
{
        /*
        vector<int> pf;
        int temp = num, ans = 1;
        for (int i = 2; i <= temp; i++)
        {
                if (!(temp % i))
                {
                        int x = temp / i;
                        pf.push_back(x + 1);
                        temp /= i;
                }
        }

        for (int j = 0; j < pf.size(); j++)
        {
                ans *= pf[j];
        }

        return ans;


        long long div = 1;
        vector<long long> pf;
        for (long long i = 2; i <= num && num > 1; i++)
        {
                if (!(num % i))
                {
                        long long x = 0;
                        while (!(num % i))
                        {
                                num /= i;
                                x++;
                        }

                        pf.push_back(x + 1);
                }
        }

        for (long long j = 0; j < pf.size(); j++)
        {
                div *= pf[j];
        }

        return div;
}


void SieveOfEratosthenes(long long n, bool prime[],
        bool primesquare[], long long 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 (long long 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 (long long i = 0; i <= (n * n + 1); i++)
                primesquare[i] = false;

        // 1 is not a prime number
        prime[1] = false;

        for (long long 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 (long long i = p * 2; i <= n; i += p)
                                prime[i] = false;
                }
        }

        long long j = 0;
        for (long long 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
long long countDivisors(long long 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[10000001], primesquare[100000000000001]; //-- prime[n + 1],
primesquare[n * n + 1]

        long long a[10000000]; // for storing primes upto n-- a[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
        long long ans = 1;

        // Loop for counting factors of n
        for (long long 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.
                long long 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
}


int main()
{
        cin >> n;

        for (long long i = 2; i <= n; i++)
        {
                ans += (i * countDivisors(i));
        }

        cout << ans + 1 << endl;
}
*/

Error: -11

Fixed submission:

#include <algorithm>
#include <chrono>
#include <iostream>
#include <utility>
#include <vector>
using namespace std;

long long n, ans;

/*
long long div_cnt(long long num)
{
        /*
        vector<int> pf;
        int temp = num, ans = 1;
        for (int i = 2; i <= temp; i++)
        {
                if (!(temp % i))
                {
                        int x = temp / i;
                        pf.push_back(x + 1);
                        temp /= i;
                }
        }

        for (int j = 0; j < pf.size(); j++)
        {
                ans *= pf[j];
        }

        return ans;


        long long div = 1;
        vector<long long> pf;
        for (long long i = 2; i <= num && num > 1; i++)
        {
                if (!(num % i))
                {
                        long long x = 0;
                        while (!(num % i))
                        {
                                num /= i;
                                x++;
                        }

                        pf.push_back(x + 1);
                }
        }

        for (long long j = 0; j < pf.size(); j++)
        {
                div *= pf[j];
        }

        return div;
}
*/

void SieveOfEratosthenes(long long n, bool prime[], bool primesquare[],
                         long long 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 (long long 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 (long long i = 0; i <= (n * n + 1); i++)
    primesquare[i] = false;

  // 1 is not a prime number
  prime[1] = false;

  for (long long 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 (long long i = p * 2; i <= n; i += p)
        prime[i] = false;
    }
  }

  long long j = 0;
  for (long long 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
long long countDivisors(long long 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[10000001],
      primesquare[100000000000001]; //-- prime[n + 1], primesquare[n * n + 1]

  long long a[10000000]; // for storing primes upto n-- a[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
  long long ans = 1;

  // Loop for counting factors of n
  for (long long 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.
    long long 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
}

int main() {
  cin >> n;

  for (long long i = 2; i <= n; i++) {
    ans += (i * countDivisors(i));
  }

  cout << ans + 1 << endl;
}