In a calendar year, it is exactly 365.25 days. But, eventually, this will lead to confusion because humans normally count by exact divisibility of 1 and not with decimal points. So, to avoid the latter, it was decided to add up all 0.25 days every four-year cycle and give that year 366 days (including February 29 as an intercalary day) and call it a leap year. The other three years in the four-year cycle would only contain 365 days and wouldn't be leap years.
In this challenge, though quite repetitive, we'll take it to a new level, where you are to determine if it's a leap year or not without the use of the datetime class, if blocks, if-elif blocks, or conditionals (a if b else c). You also may not use the logical operators AND (and) or OR (or), with the exemption of the NOT (not) operator.
Return True if it's a leap year, False otherwise.
leap_year(1979) ➞ False
leap_year(2000) ➞ True
leap_year(2016) ➞ True
leap_year(1521) ➞ False
leap_year(1996) ➞ True
leap_year(1800) ➞ False
You can't use the datetime class, if statements in general, the conditional, or the logical operators (and, or).