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# Python program to check if year is a leap year or not year = 2000 # To get year (integer input) from the user # year = int(input("Enter a year: ")) # divided by 100 means century year (ending with 00) # century year divided by 400 is leap year if (year % 400 == 0) and (year % 100 == 0): print("{0} is a leap year".format(year)) # not divided ...
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8 de feb. de 2024 · In this example, the lambda function `is_leap_year_lambda` checks if a given year is a leap year using a concise expression with the modulo operator. In the provided example, the year 2023 is checked, and the result (True or False) is printed indicating whether it is a leap year or not.
24 de jul. de 2012 · yy = 2100 lst = [ (yy % 4 == 0) , (yy % 400 == 0) , (yy % 100 == 0) ] if lst.count(True) in [0,2]: print('Not Leap Year') else: print('Leap Year') Output : Not Leap Year
#Centenary year is a leap year divided by 400 if (year % 400 == 0) and (year % 100 == 0): print("{0} is a leap year".format(year)) #If it is not divisible by 100 then it is not a centenary year #A year divisible by 4 is a leap year elif (year % 4 ==0) and (year % 100 != 0): print("{0} is a leap year.".format(year)) #If the Year is not divisible ...
11 de sept. de 2023 · The Quick Answer. To check if a year is a leap year in Python, you can use this function: def is_leap(year): return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0) Here are some example calls to the function: print(is_leap(2000)) # --> True. print(is_leap(1900)) # --> False. print(is_leap(2016)) # --> True.
31 de ene. de 2024 · In Python, the calendar module provides functions to determine if a year is a leap year and to count leap years in a specified period. calendar — General calendar-related functions — Python 3.12.1 documentation. Contents. The algorithm for leap years. Determine if a year is a leap year: calendar.isleap()
13 de feb. de 2024 · Step 1: Import the calendar module. Step 2: Use the isleap (year) function to check if the year is a leap year. Example: import calendar. year = int(input("Enter a year: ")) if calendar.isleap(year): print(f"{year} is a leap year.") else: print(f"{year} is not a leap year.") Notes: This method is very concise and efficient.
