关于润年问题及其时间问题

#_author:"yongjia"
#Time : 2019/4/8 16:52

import calendar
input_y=int(input("年份："))
input_m=int(input("月份："))
def year_month(input_y,input_m):
    sign=calendar.isleap(input_y)
    if sign==True:
        date=[31,29,31,30,31,30,31,30,31,30,31,30]
        year="润年,"
    else:
        date=[31,28,31,30,31,30,31,30,31,30,31,30]
        year="不是润年,"
    print(input_y,year,input_m,"月份天数=",date[input_m-1])
year_month(input_y,input_m)

输入时间则输出下一秒的时间

h=int(input("hours："))
m=int(input("min:"))
s=int(input("seconds"))
def time_h_m_s(h,m,s):
    s=s+1
    if s==60:
       m=m+1
       s=0
       if m==60:
           h=h+1
           m=0
    if h>24 or s>60 or m>60:
        print("时间范围输入有误")
    else:
        print("下一秒时间为",h,"时:",m,"分:",s,"秒")
time_h_m_s(h,m,s)