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 <iostream>
       
#include <set>
       
 
       
// generate sum of all divisor's of x (where x > 1)

       
unsigned int getSum(unsigned int x)
       
{
       
  // find all factors:
       
  // look only for the "smaller" divisors <= sqrt(x)
       
  // and they have a "bigger" brother x / divisor
       
 
       
  // 1 is always a divisor, but not the number x itself
       
  unsigned int divisorSum = 1;
       
  // therefore start at 2
       
  for (unsigned int divisor = 2; divisor * divisor <= x; divisor++)
       
    if (x % divisor == 0)
       
    {
       
      divisorSum += divisor;
       
 
       
      // add the "bigger brother"
       
      auto otherDivisor = x / divisor;
       
      // except square numbers
       
      if (otherDivisor != divisor)
       
        divisorSum += otherDivisor;
       
    }
       
 
       
  return divisorSum;
       
}
       
 
       
int main()
       
{
       
  // contain all numbers which are part of an amicable pair
       
  std::set<unsigned int> amicables;
       
 
       
  // precomputation step:
       
  // find all amicable numbers <= 100000
       
  const unsigned int MaxAmicable = 100000;
       
  for (unsigned int i = 2; i <= MaxAmicable; i++)
       
  {
       
    auto sibling = getSum(i);
       
 
       
    // found a pair ?
       
    if (i == getSum(sibling) && i != sibling)
       
    {
       
      amicables.insert(i);
       
      amicables.insert(sibling);
       
    }
       
  }
       
 
       
  // and now start processing input
       
  unsigned int tests;
       
  std::cin >> tests;
       
  while (tests--)
       
  {
       
    unsigned int x;
       
    std::cin >> x;
       
 
       
    // just look up all suitables numbers
       
    unsigned int sum = 0;
       
    for (auto i : amicables)
       
    {
       
      // discard those that are too big
       
      if (i > x)
       
        break;
       
      // note: an set::set is sorted ascendingly by default
       
 
       
      // yes, accept that amicable number
       
      sum += i;
       
    }
       
 
       
    std::cout << sum << std::endl;
       
  }
       
  return 0;
       
}

This solution contains 11 empty lines, 17 comments and 2 preprocessor commands.