find_numbers_with_same_amount_of_divisors.cpp

//The integers 14 and 15, are contiguous (1 the difference between them, obvious) and have the same number of divisors.

//14 ----> 1, 2, 7, 14 (4 divisors)
//15 ----> 1, 3, 5, 15 (4 divisors)

//The next pair of contiguous integers with this property is 21 and 22.

//21 -----> 1, 3, 7, 21 (4 divisors)
//22 -----> 1, 2, 11, 22 (4 divisors)

//We have 8 pairs of integers below 50 having this property, they are:

//[[2, 3], [14, 15], [21, 22], [26, 27], [33, 34], [34, 35], [38, 39], [44, 45]]

//Let's see now the integers that have a difference of 3 between them. There are seven pairs below 100:

//[[2, 5], [35, 38], [55, 58], [62, 65], [74, 77], [82, 85], [91, 94]]

//Let's name, diff, the difference between two integers, next and prev, (diff = next - prev) and nMax, an upper bound of the range.

//We need a special function, count_pairsInt(), that receives two arguments, diff and nMax and outputs the amount of pairs of integers that fulfill this property, all of them being smaller (not smaller or equal) than nMax.

//Let's see it more clearly with examples.

//count_pairsInt(1, 50) -----> 8 (See case above)
//count_pairsInt(3, 100) -----> 7 (See case above)

//Happy coding!!!

#include <cmath>

unsigned int getDivisors(const unsigned int num)
{
  int result = 0;
  
  for (int i = 1, root = std::sqrt(num); i <= root; ++i)
  {
    if (!(num % i)) {
      if ((num / i) == i) {
          result++;
      } else {
        result += 2;
      }
    }
  }
  
  return result;
}

unsigned int countDivisors(const unsigned int diff, const unsigned int nMax) 
{
  int divisors = 0;
  
  for (int prev = 0, next = prev + diff; next < nMax; ++prev, ++next)
  {
    if (getDivisors(prev) == getDivisors(next))
      divisors++;
  }
  
  return divisors;
}