Python program to check leap year by using the calendar module # importing the moduleimportcalendar# input the yearyear=int(input('Enter the value of year: '))leap_year=calendar.isleap(year)# checking leap yearifleap_year:# to check conditionprint('The given year is a leap year.')else:...
calendar.isleap(year) returns a boolean if the year is a leap year. One thing you need to know is that the function depends on the calendar module, So you must import the calendar module at the start of your Python script or program Also Read: Basic Python Programs 3) Divisibility Meth...
6.程序实现的功能:输人某年某月某日,判断这一天是这一年的第几天,Python代码如下: d e f leap(year):$$ l e a p = 0 $$$ i f ( y e a r \% 4 0 0 = = 0 ) o r ( ( y e a r \% 4 = = 0 ) a n d ( y e a r \% 1 0 0 ! = 0 ) ) : $$$ l e a ...
Enter a year: 40004000 qualifies as a leap year Determine Leap Year Through theCalendarModule in Python Python’scalendarmodule is one of the reliable tools to perform computations involving dates. It follows the European convention (i.e., displays Monday as the first day of the week) and ser...
#In this leap year python program, the user to asked enter a year. The program checks whether the entered year is a leap year or not. 1 2 3 4 5 6 7 8 9 10 11 year = int(input("Enter a year: ")) if (year % 4) == 0: ...
if is_leap(year): print("您输入的年份是闰年。") else: print("您输入的年份不是闰年。") 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 2、定义一个count_str_char(my_str)函数,该函数返回参数字符串中包含多少个数字,多少个字母,多少个空白字符,多少...
Python程序LeapYear源码 import math def valid(month,day,year):if day>31:return False elif month==2:if leap(year):if day <=29:return True else:return False else:if day<=28:return True else:return False elif month==4 or 6 or 9 or 11:if day>31:return False else:return True else:r...
#The test case is passing 4/6. I don't know what is wrong with my code, please help year = int(input()) if year % 4 == 0: if year % 100 == 0: if year
1 TypeError: is_leap_year() takes 0 positional arguments but 1 was given 问题分析: 报错的意思是:is_leap_year()这个函数不需要参数,但是函数却被传递了一个参数。 掉头检查代码,会发现: 1 def is_leap_year(): 这里出了问题,我们对其进行修改: 1 def is_leap_year(year): 再次运行代码,成功...
Given a year, determine whether it is a leap year. If it is a leap year, return the Boolean True, otherwise, returnFalse. Note that the code stub provided reads from STDIN and passes arguments to theis_leapfunction. It is only necessary to complete theis_leapfunction. ...