回数的判断

#! /usr/bin/env python
count = 0
for i in range(1000,3000):
    set = str (i)
    if (set[::-1]) == (set[0::]):
        count += 1
        print(i,"count=",count)
#第二种方法
# def fn(n):
#     if str(n)[::-1] == str(n):
#         return True
#     else:
#         return False
#
# output = filter(fn,range(1000,3000))
# print("所有的回数：",list(output))