title: Maximum Number Of Squares That Can Be Fit In A Right Angle Isosceles Triangle
signature: function maximum_number_of_squares_that_can_be_fit_in_a_right_angle_isosceles_triangle(b, m)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return abs(actual - expected) < 1e-09
  end
  
  function driver(b, m, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(b, m), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(40, 74, -0.1241782322863404);
      @printf("TEST-0...%s\n",result);
      
      result = driver(38, 35, 0.04653061224489791);
      @printf("TEST-1...%s\n",result);
      
      result = driver(47, 71, -0.11188256298353501);
      @printf("TEST-2...%s\n",result);
      
      result = driver(52, 29, 0.7110582639714625);
      @printf("TEST-3...%s\n",result);
      
      result = driver(21, 9, 1.5555555555555558);
      @printf("TEST-4...%s\n",result);
      
      result = driver(50, 33, 0.3902662993572084);
      @printf("TEST-5...%s\n",result);
      
      result = driver(8, 82, -0.04402141582391434);
      @printf("TEST-6...%s\n",result);
      
      result = driver(56, 80, -0.10500000000000001);
      @printf("TEST-7...%s\n",result);
      
      result = driver(93, 5, 163.68000000000004);
      @printf("TEST-8...%s\n",result);
      
      result = driver(21, 90, -0.08944444444444444);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def maximum_number_of_squares_that_can_be_fit_in_a_right_angle_isosceles_triangle(b, m):
      return ((((b / m) - 1) * (b / m)) / 2)
source_cpp: int maximum_number_of_squares_that_can_be_fit_in_a_right_angle_isosceles_triangle(int b, int m) {
    return ( b / m - 1 ) * ( b / m ) / 2;
  }