- num = 905;
- def is_Palindrome(num):
- """
- 判断一个数字是不是回文数,这里有些取巧了
- :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算法计算指定数字的回文数
- :param num:
- :return:
- """
- count = 0;
- while True:
- if True == is_Palindrome(num):
- output = "这是一个回文数:%d"%num + "\\r\\n总共次数为%d"%count;
- print(output);
- break;
- else:
- num = add(num);
- count += 1;
- def add(num):
- """
- num 与自己倒序的数字相加
- :param num:
- :return:
- """
- temp = "%d"%num;
- str = temp[::-1];
- return int(temp) + int(str);
- print(create_Palindrome(num));
- #该片段来自于http://www.codesnippet.cn/detail/1203201511884.html
来源: http://www.codesnippet.cn/detail/1203201511884.html