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 ...
#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
Python- Leap Year Hi all, This does not bode too well but I have become stuck on the Leap YearPythonpractice test. Can you please let me know why the below code is not working? This one has got the better of me. year = int(input()) if year % 4 == 0: if year % 100 == ...
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...
Logic Of Code Explained: So basically, this code uses the“isleap” function provided by the calendar module in Python. 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...
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...
= 0 and year % 4 == 0: return True if year % 100 == 0 and year % 400 ==0: return True return False def which_day(date): time_str = date.split("-") leap_year = is_a_leap_year(int(time_str[0])) if leap_year: day_list = [0,31,29,31,30,31,30,31,31,30,31,30...
经典题目:输入年份判断闰年 year = int(input('请输入年份')) leap_year = year % 4 == 0 && year % 100 != 0 || year % 400 == 0 print(leap_year) 输入:2000 结果:True 三、分支结构 在python中分支结构使用if-elif-else 代码语言:javascript 代码运行次数:0 运行 AI代码解释 题目:百分制...
is_leap(2017) Output: bash True False As excepted we get the desired result. Solution-4: Using the Python calendar module Python is known for its various built-in modules. Instead of writing the code for leap year manually using the conditions, we can simply use the Python module to check...
if year%4==0 and year%100!=0: return False 1. 2. 3. 4. 5. 这个代码也是在之后的题目中判断闰年的重要方法。 二、题目解析 1.一周中的第几天(1185) class Solution: def dayOfTheWeek(self,day:int,month:int,year:int)->str: res=["Friday","Saturday","Sunday","Monday","Tuesday","We...