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.

       
#include <map>
       
#include <set>
       
#include <vector>
       
#include <iostream>
       
 
       
// count how often each digit occurs,
       
// store zero at lowest position, then ones, etc.
       
// e.g. 5063 means 3 zeros, 6 ones, no twos, 5 threes and nothing else
       
// note: can't handle input values with 2^6 or more identical digits

       
unsigned long long fingerprint(unsigned long long x)
       
{
       
  unsigned long long result = 0;
       
  while (x > 0)
       
  {
       
    // extract lowest digit
       
    auto digit = x % 10;
       
    x /= 10;
       
 
       
    // subdivide 64 bit integer into 10 "digit counters", each 6 bits wide
       
    // => each digit may occur up to 2^6=64 times, more than enough ...
       
    const auto BitsPerDigit = 6;
       
    result += 1ULL << (BitsPerDigit * digit);
       
  }
       
  return result;
       
}
       
 
       
int main()
       
{
       
  unsigned int maxCube = 10000;
       
  unsigned int numPermutations = 5;
       
  std::cin >> maxCube >> numPermutations;
       
 
       
  // [fingerprint] => [list of numbers, where number^3 produced that fingerprint]
       
  std::map<unsigned long long, std::vector<unsigned int>> matches;
       
  for (unsigned int i = 1; i < maxCube; i++)
       
  {
       
    // find fingerprint
       
    auto cube = (unsigned long long)i * i * i;
       
    // add current number to the fingerprint's list
       
    matches[fingerprint(cube)].push_back(i);
       
  }
       
 
       
  // extract all smallest cube, std::set is sorting them
       
  std::set<unsigned long long> smallest;
       
  for (auto m : matches)
       
    // right number of permutations ?
       
    if (m.second.size() == numPermutations)
       
      smallest.insert(m.second.front());
       
 
       
  // print in ascending order
       
  for (auto s : smallest)
       
    std::cout << (s*s*s) << std::endl;
       
 
       
  return 0;
       
}

This solution contains 7 empty lines, 13 comments and 4 preprocessor commands.