As we know the condition to check the given year is a leap year or not. So, here we will implement the condition and try to write the Python program. Python program to check leap year by checking the condition # input the yeary=int(input('Enter the value of year: '))# To check f...
• The century year is a leap year only if it is perfectly divisible by 400. For example, 2000 is a leap year. Python Program to Check Leap Year #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....
If a “ValueError exception” is found, it means February 29 does not exist in the given year, indicating it’s not a leap year. The function then returns False. Conclusion: I hope you get a clear understanding of the Leap Year Program in Python and get the answer on Is 2024 a Leap...
Python Leap Year #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 % 400 == 0: print("Leap year") else: print("Not a leap year") else: print("Not a leap 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...
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 ...
In the above program, we have imported thedatetime moduleand taken the date in the form ofd,m,ywheredmeans day,mmeans month andymeans year. Since some date input provided by the user may be valid or not that's why initially in the try block we arechecking the date validationand if it...
Thecalendar.isLeap()method takes an integer year value and returnstrueorfalseif the argument is a leap year or not. Therefore, we can apply decision statements to display our personalized messages. Let’s test this code by running the program for different years: ...
year”,反之输出“xxxisnotleapyear”,并⼀直循环,改正其中的三处错误,使之正常运⾏,输出如下: theyear:2018 isnotleapyear theyear:2020 isleapyear theyear: 中位于每个//ERROR***found***下的语句⾏有错误,请加以更正,使它能得出正确的结果。 :只能修改每个//ERROR***found***下的那⼀...
if int(year)% 100 != 0 and int(year) % 4 == 0: return True else: return False data = input("请输入年份:") if is_leap(data): print("是闰年") else: print("不是闰年") 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.