My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       
#include <iostream>
       
#include <vector>
       
 
       
#define ORIGINAL
       
 
       
int main()
       
{
       
  // almost like a reverse prime sieve ...
       
  unsigned int limit = 10000000;
       
#ifdef ORIGINAL
       
  std::cin >> limit;
       
#endif
       
 
       
  // count divisors of the number immediately after "limit", too
       
  limit++;
       
 
       
  // will have the number of divisors for each number
       
  std::vector<unsigned short> divisors(limit, 0);
       
 
       
  // all numbers which can be a divisor ...
       
  for (unsigned int i = 1; i <= limit / 2; i++)
       
    // and all of their multiples
       
    for (unsigned int j = i; j <= limit; j += i)
       
      divisors[j]++;
       
 
       
#ifdef ORIGINAL
       
  unsigned int result = 0;
       
  for (unsigned int i = 2; i < limit; i++)
       
    if (divisors[i] == divisors[i + 1])
       
      result++;
       
  std::cout << result << std::endl;
       
#else
       
  // [index] => [numbers up to index which match the "neighbor" condition]
       
  std::vector<unsigned int> count(limit + 1, 0);
       
  // count numbers whose bigger neighbors has the same number of divisors
       
  for (unsigned int i = 2; i < limit; i++)
       
  {
       
    count[i] = count[i - 1];
       
    if (divisors[i] == divisors[i + 1])
       
      count[i]++;
       
  }
       
 
       
  // simple lookup of results
       
  unsigned int tests = 1;
       
  std::cin >> tests;
       
  while (tests--)
       
  {
       
    unsigned int index;
       
    std::cin >> index;
       
    std::cout << count[index - 1] << std::endl;
       
  }
       
#endif
       
 
       
  return 0;
       
}

This solution contains 8 empty lines, 8 comments and 8 preprocessor commands.