1079. 延迟的回文数 (20)——Python

Easy,判断回文时,将列表翻转,翻转后与原先的相等则说明是回文,麻烦点就在于类型的转换。

def judge_palindrome(my_number):
	#判断是否为回文
	list_number = list(my_number)
	list_number.reverse()
	if list_number == list(my_number):
		#print("OK")
		return True
	else:
		return False
	
if __name__ == "__main__":
	my_number = input()
	flag = judge_palindrome(my_number)
	for i in range(10):
		if flag:
			print(my_number + " is a palindromic number.")
			break
		else:
			list_number = list(my_number)
			list_number.reverse()
			sum_number = int(my_number) + int(''.join(list_number))
			sum_number = str(sum_number)
			print(my_number + " + " + ''.join(list_number) + " = " + sum_number)
			my_number = sum_number
			flag = judge_palindrome(my_number)
	if not flag:
		print("Not found in 10 iterations.")

猜你喜欢

转载自blog.csdn.net/baidu_38271024/article/details/79533278