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:输入某年某月某日,判断这一天是这一年的第几天,Python代码如下def leap(year):leap=0if(year%400==0)or((year%4==0)and(year%100!=0)leap=1①y=int(input("请输入年份:"))m=int(input("请输人月份:"))d=int(input("请输入日:"))months=(0,31,59,90,120,151,181,212,...
1.1 定义一个is_leap(year)函数,该函数可判断year是否为闰年。若是闰年,则返回True;否则返回False。 def is_leap(year): 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...
答案:defis_leap_year(year):if(year%4==0andyear%100!=0)or(year%400==0):returnTrueelse:returnFalsedefmain():try:year=int(input("Enterayear:"))ifis_leap_year(year):print(f"{year}isaleapyear.")else:print(f"{year}isnotaleapyear.")exceptValueError:print("Invalidinput.Pleaseentera...
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Entertheyear:2018 2018isnotleapyear Entertheyear:2020 2020isleapyear Entertheyear: 程序中位于每个//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。 注意:只能修改每个//ERROR***found***下的那⼀⾏,不要改动程序中的其他内容。 参考答案: 【第⼀错误】 没有添加应有的空格...
Tocheck the given date is valid or not, we will use thedatetime modulein the program by using the import function and also we will use the try-except statement. For example date 31-02-2020 is invalid because we know that February month has only 28 days in an ordinary year and a leap...
I do not consider myself as a programmer. I create these little programs as experiments to play with Python, or to solve problems for myself. I would gladly accept pointers from others to improve, simplify, or make the code more efficient. If you would like to make any comments then ple...
fromdatetimeimportdatetimeexec_date='2024-02-29'current_date=datetime.strptime(exec_date,"%Y-%m-%d")end_date=current_date.replace(year=current_date.year-2,month=current_date.month,day=current_date.day) We are trying to get the date of two years from now in Python datetime. ...
4 or -1 if that char is not part of string.''' 5 if char not in string: 6 return -1 7 else: 8 leng_str = len(string) 9 leng_chr = len(char) 10 for i in range(leng_str - leng_chr + 1): 11 if string[i : i + leng_chr] == char: ...