problem 33

/*
The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, 30/50 = 3/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
*/
#include <iostream>

using namespace std;

void same_digit(int numerator, int denominator){
  // how much real  division is  equal then  slice  numbers
  double divisor = static_cast<double>(numerator) / static_cast<double>(denominator);
  int first_numerator = numerator % 10;
  int second_numerator = (numerator % 100 - first_numerator) / 10;
  int first_denominator = denominator % 10;
  int second_denominator = (denominator % 100 - first_denominator) / 10;

  // first case
  if (first_numerator == second_denominator){
    double x = denominator % 10;
    double solution =  static_cast<double>((numerator / 10)) / x;
    if (solution == divisor &&  solution < 1){
      cout <<numerator << " : " << denominator << " : " <<  solution <<endl;
    }
  }


  //second case
  if(second_numerator == first_denominator){
    double y = numerator % 10;
    double solution = y / static_cast<double>((denominator / 10));
    if (solution == divisor && solution < 1){
       cout <<numerator << " : " << denominator << " : " <<  solution <<endl;
      }
  }
}

int main() {
  for(int i = 10; i < 100; i++){
    for(int j = i + 1; j < 100; j++){
     if(i % 10 == 0 || j % 10 == 0){
      }
      else{
        same_digit(i, j);
      }
    }
  }
  return 0;
}