Python Program to Check Leap Year

easyif / elsePython

Leap year looks simple but has a trap — century years. That is exactly why examiners like it.

The question

Read a year and print Leap year or Not a leap year. A year is a leap year if it is divisible by 4 but not by 100, or if it is divisible by 400.

program.py

y = int(input())
if (y % 4 == 0 and y % 100 != 0) or y % 400 == 0:
    print("Leap year")
else:
    print("Not a leap year")

Leap year

1A year divisible by 4 is usually a leap year…
2…except century years (divisible by 100), which are not…
3…unless they are also divisible by 400. So 2000 is a leap year but 1900 is not.
4All three rules fit in one condition: (y % 4 == 0 and y % 100 != 0) or y % 400 == 0.

Checking only y % 4 == 0 — it wrongly calls 1900 and 2100 leap years.
Missing brackets around the and part, which makes the condition hard to read and easy to break.

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