Solving Problem: Leap Year

(If-elif-else Control Flow)

Problem Statement

Check if a year is a leap year.

 As per the Gregorian calendar (in 1582), the following rule is used to determine the kind of year:

• If the year number isn’t divisible by four, it’s a common year.
• Otherwise, if the year number isn’t divisible by 100, it’s a leap year.
• Otherwise, if the year number isn’t divisible by 400, it’s a common year.
• Otherwise, it’s a leap year.

The task is to determine if the given year is a leap year or not.

Output messages:

“Leap year.” if the year is a leap year.

“Common year.” if the year is common.

“Not within the Gregorian era.” If the year falls out of the Gregorian era.

Test Data:

Input: 2000

Output: Leap year.

Input: 1999

Output: Common year.

Input: 1996

Output: Leap year.

Input: 1500

Output: Not within the Gregorian era.

Solution:

year = int(input("Enter a year: "))
#
# Write your code here.
#	
if year > 1582:
    if (year%4 == 0 and year%100 != 100) or year%400 == 400:
        print("Leap year.")
    else:
        print("Common year.")
else:
    print("Not within the Gregorian era.")

Output: