蓝桥杯练习题 —— 闰年判断（python）

代码一

# encoding = utf-8
# 开发者：xxx
# 开发时间： 20:55 
# "Stay hungry，stay foolish."

y = int(input())
is_leap_year = (y % 4 == 0 and y % 100 != 0) or (y % 400 == 0)

if is_leap_year:
    print("yes")
else:
    print("no")

简化代码

n = int(input())
is_leap_year = (n % 4 == 0 and n % 100 != 0) or (n % 400 == 0)
print('yes' if is_leap_year else 'no')