The year is called a leap year in python if this condition is met. If the year is not divisible by 400, it means it’s not a leap year. If the year is not divisible by 100 (as per the second condition), it directly returns True, indicating a leap year. If the year is not div...
For Leap Year : Condition: Year % 100 != 0 1st Jun 2021, 6:38 PM Tejas Raut - 1 year = int(input()) if(year%4 == 0): if(year%100 == 0): if(year%400 == 0): print("Leap year") else: print("Not a leap year") else: print("Leap year") else: print ("Not a le...
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: 20242024 is a leap year Output Evaluating the 3rd Condition: 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...
Another way is to use the nested if statements where each if statement contains one condition. Let us use the nested if statements to create a function that checks whether the given year is a leap year or not. python # creating the functiondefis_leap(year):# variable to check the leap ...
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:...
LI HANthat's because it's notpythonbut C 1st Jun 2021, 6:54 PM ChaoticDawg + 1 TejasIf it passed, then it will be purely circumstantial. Because it should be; if(year % 400 == 0 || (year % 4 == 0 && year % 100 != 0) {} There is no condition in which; year % 400 ...
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...
#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: ...
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...