How to Solve the Leap Year Puzzle in Python?

Determining Leap Years: A Python Conundrum

Embarking on a mission to discern the elusiveness of leap years, you stumble upon a Python conundrum. Armed with the knowledge that leap years adhere to a specific set of divisibility rules, you meticulously craft a function that attempts to unveil this enigma.

However, upon executing your code, the resounding silence of None greets you. Undeterred, you persist in your quest, seeking enlightenment.

Leap Year Definition Revisited

As defined by the annals of time, a year qualifies as a leap year if it satisfies one of two criteria:

1. It is divisible by 4, but not by 100.
2. It is divisible by 400.

Examining the Code

Let us scrutinize the code you initially devised:

<code class="python">def leapyr(n):
    if n%4==0 and n%100!=0:
        if n%400==0:
            print(n, "is a leap year.")
    elif n%4!=0:
        print(n, "is not a leap year.")

print(leapyr(1900))</code>

Though the intention is clear, the issue arises in the handling of conditions. When a year is divisible by 4 and not by 100, it should be declared a leap year immediately. However, your code continues to check whether it is also divisible by 400, which is unnecessary.

Seeking a Simpler Solution

Luckily, the Python standard library offers a more elegant solution through the calendar module. The isleap function within this module takes a year as its argument and returns True if it is a leap year, and False otherwise.

Utilizing calendar.isleap

The following code snippet illustrates the simplicity of using calendar.isleap:

<code class="python">import calendar
print(calendar.isleap(1900))</code>

When executed, this code will correctly output the expected result:

False

With this enhanced understanding, you can now confidently navigate the intricacies of leap years in your Pythonic endeavors.

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