C++ exercise: Recursion example -- sum up numbers

Q: Given a Sum and a string of numbers, add minimum operator "+" to make the sum of numbers equal to Sum. e.g. Sum=9, numbers: 101124, we get 9=1+01+1+1+2+4; Sum=88, numbers: 2112331453, we get 88=21+12+33+14+5+3.

#include <string>
#include <stack>
#include <iostream>

using namespace std;
 
// Returns: 0 -- Got a solution.
//         -1 -- No solution.
int Calculate(int sum, string numbers,
            int& best_op_count, string& best_solution)
{
   //cout << " Enter: " << sum << " " << numbers <<  endl;;
 
   bool gotone = false;
   int pow10 = 1;
   stack<char> tail_nums;
 
   // Pop maximum digits from tail of "numbers".
   // All the digits form a "num".
   int last_num;
   int num = 0;
   while (num < sum && numbers.size() > 0)
   {
      char last_ch = *(numbers.rbegin());
      last_num = last_ch - '0';
      if (num + last_num * pow10 > sum)
         break;
 
      tail_nums.push(last_ch);
      numbers.erase(numbers.size() - 1);
      num += last_num * pow10;
      pow10 *= 10;
   }
 
   if (numbers.size() == 0 && num == sum)
   {
      // Got one solution done.
      best_op_count = 0;
      best_solution = "";
      while (tail_nums.size() > 0)
      {
         best_solution += tail_nums.top();
         tail_nums.pop();
      }
      return 0;
   }
 
   // loop_1:
   int op_count;
   int temp_best_op_count = -1;
   string solution;
   string temp_best_solution = "";
   while ( pow10 > 0 && num > 0 && tail_nums.size() > 0)
   {
      //cout << " ... " << sum << "(=)" << numbers << "(+)" << num << endl;
 
      // Calculate result from the rest heading numbers as "subres".
      int subres = Calculate((sum - num), numbers, op_count, solution);
 
      // If "subres" == 0, we have a solution.
      if ( ! subres )
      {
         gotone = true;
 
         op_count++;  // one more op for "num" + Calculate(subsum);

         //cout << "Got one solution: " << sum << "=" << solution << "+" << num 
         //     << " <=============================" << op_count << "(+)s" << endl;
 
         // If it is a better solution, save it.
         if (op_count < temp_best_op_count || temp_best_op_count < 0)
         {
            temp_best_op_count = op_count;
            temp_best_solution = solution + "+";
            stack<char> temp_tail_nums = tail_nums;
            while (temp_tail_nums.size() > 0)
            {
               temp_best_solution += temp_tail_nums.top();
               temp_tail_nums.pop();
            }
         }
      }
                                                                                                                              
      // Remove the most significant digit from "num" back to "numbers".
      if (tail_nums.size() > 0)
      {
         char most_sig_char = tail_nums.top();
         tail_nums.pop();
         numbers += most_sig_char;
 
         pow10 /= 10;
         num -= (most_sig_char - '0') * pow10;
      }
 
      // Go to loop_1.
   }
 
   best_solution = temp_best_solution;
   best_op_count = temp_best_op_count;
 
   //cout << " Exit: " << sum << " " << numbers << " gotone?" << gotone 
   //     << "(" << best_solution << ")" << " count(" << best_op_count << ")" << endl;;
 
   if (gotone)
      return 0;
   else
      return -1;
}
 
 
int main()
{
   int mysum;
   string mystr;
   int op_cnt;
   string solution;
 
   while (cin)
   {
      cout << "Input sum: ";
      cin >> mysum;
      cout << endl;
 
      cout << "Input string: ";
      cin >> mystr;
      cout << endl;
 
      int res = Calculate(mysum, mystr, op_cnt, solution);
      if (res)
         cout << "No solution." << endl;
      else
         cout << "Solution: " << solution << endl;
 
      cout << endl;
   }

   return 0;
}