Three ways to judge whether it is a leap year in Python

‘’’
An integer that meets the following two conditions can be called a leap year:
(1) Ordinary leap year: it can be divided by 4 but cannot be divided by 100 (such as 2004 is an ordinary leap year);
(2) Century leap year: can be divided by 400 (such as 2000 is a leap year in the century, and 1900 is not a leap year in the century);

Make judgment based on if condition:
year%4==0 and year%100！=0 or year %400=0
‘’’

#Method 1
#single branch if
try:
year=int(input("Please enter a year:"))
if (year%4==0) and (year%100 !=0) or (year%400)==0:
print("{} year is a leap year".format(year))
else:
print("{} year is not a leap year".format(year))
except:
print("You entered incorrectly!")

#Method 2
#Nested if statement to judge
try:
year=int(input('Please enter a year:'))
if (year%4)==0:
if(year%100)!=0:
if (year%400)==0:
print(’{} year is a leap year’.format(year))
else:
print(’{} year is not a leap year’.format(year))
else:
print(’{} year is a leap year’.format(year))
else:
print(’{} year is not a leap year’.format(year))
except:
print("You entered incorrectly!")

Method 3

The isleap() method encapsulated in the calendar library determines whether it is a leap year

try:
year = int(input('Please enter a year:'))
import calendar
year = int(input("Please enter year:"))
check_year = (year)
if check_year == True:
print(’{} year is a leap year’.format(year))
else:
print(’{} year is not a leap year’.format(year))
except:
print("You entered incorrectly!")