title: Recursively Break Number 3 Parts Get Maximum Sum
signature: function recursively_break_number_3_parts_get_maximum_sum(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(39, 41);
      @printf("TEST-0...%s\n",result);
      
      result = driver(79, 87);
      @printf("TEST-1...%s\n",result);
      
      result = driver(7, 7);
      @printf("TEST-2...%s\n",result);
      
      result = driver(76, 87);
      @printf("TEST-3...%s\n",result);
      
      result = driver(48, 57);
      @printf("TEST-4...%s\n",result);
      
      result = driver(18, 19);
      @printf("TEST-5...%s\n",result);
      
      result = driver(58, 63);
      @printf("TEST-6...%s\n",result);
      
      result = driver(17, 17);
      @printf("TEST-7...%s\n",result);
      
      result = driver(36, 41);
      @printf("TEST-8...%s\n",result);
      
      result = driver(5, 5);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def recursively_break_number_3_parts_get_maximum_sum(n):
      if ((n == 0) or (n == 1)):
          return n
      return max(((recursively_break_number_3_parts_get_maximum_sum((n // 2)) + recursively_break_number_3_parts_get_maximum_sum((n // 3))) + recursively_break_number_3_parts_get_maximum_sum((n // 4))), n)
source_cpp: int recursively_break_number_3_parts_get_maximum_sum(int n) {
    if ( n == 0 || n == 1 ) return n;
    return max ( ( recursively_break_number_3_parts_get_maximum_sum ( n / 2 ) + recursively_break_number_3_parts_get_maximum_sum ( n / 3 ) + recursively_break_number_3_parts_get_maximum_sum ( n / 4 ) ), n );
  }