python小demo之判断闰年
#_*_ coding=utf-8__author__ = 'kysida'
#定义函数
def leap_year(year):
if year % 4 == 0:
if (year / 100) % 4 == 0:
return "the year is a leap year"
return "the year is not a leap year"
#主程序
if __name__ == '__main__':
year = int(raw_input("Please input the year:"))
print leap_year(year)
页:
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