weixin_39533896的博客如何用Python编写一个判断回文数的程序123456789101112131415def ishuiweinum(num): if not isinstance(num,int): return False if num0: numlist.append(num%10) num/=10 reverselist=numlist[:] reverselist....
原来是赵先生!的博客用Python计算1-100000以内的回文数 vim test.py #!/usr/bin/env python2.7 for i in range(1,10001): if len(str(i)) > 2: a=len(str(i)) if str(i)[0:a] == str(i)[::-1]: print i python test.py