#Find out if a given year is a "leap" year.

#In the Gregorian calendar, three criteria must be taken into account to identify leap years:
#The year must be evenly divisible by 4;
#If the year can also be evenly divided by 100, it is not a leap year; unless...
#The year is also evenly divisible by 400. Then it is a leap year.
#According to these rules, the years 2000 and 2400 are leap years, while 1800, 1900, 2100, 2200, 2300, and 2500 are not the leap years.
#Write a Python program that prints such as "2020 is a leap year" if the given year by the user is a leap year, prints such as "2019 is not a leap year" otherwise.



leap_year = int(input("Enter a year:"))

if leap_year % 4 == 0 :
    if (leap_year % 100 == 0) and (leap_year % 400 == 0):
        print(f"{leap_year} is a leap year.")
    elif (leap_year % 4 == 0) and (leap_year % 100 != 0):
        print(f"{leap_year} is a leap year.")
    else :
        print(f"{leap_year} is not a leap year.")
else :
    print(f"{leap_year} is not a leap year.")


