num = 905;def is_Palindrome(num):  """  判斷一個數(shù)字是不是回文數(shù)，這里有些取巧了  :param num:  :return:  """  """  :param num:  :return:  """  temp = "%d"%num;  str = temp[::-1];  if temp == str:    return True;  else:    return False;def create_Palindrome(num):  """  用196算法計算指定數(shù)字的回文數(shù)  :param num:  :return:  """  count = 0;  while True:    if True == is_Palindrome(num):      output = "這是一個回文數(shù):%d"%num + "/r/n總共次數(shù)為%d"%count;      print(output);      break;    else:      num = add(num);      count += 1;def add(num):  """  num 與自己倒序的數(shù)字相加  :param num:  :return:  """  temp = "%d"%num;  str = temp[::-1];  return int(temp) + int(str);print(create_Palindrome(num));