Question

3. Определить, является ли год, который ввел пользователем, високосным или обычным. (4)​

Answer 1

Решение:

Python

year = int(input())

if year == 365:

   print('Обычный.')

else:

   print('Високосный.')

Answer 2

Determining if a Year is Leap or Common

To determine if a year is leap or common, we need to consider the rules for leap years. In the Gregorian calendar, which is the most widely used calendar system today, the following rules apply:

1. A year is a leap year if it is divisible by 4, but not divisible by 100, unless it is also divisible by 400.

Based on these rules, we can write a simple algorithm to determine if a year is leap or common.

Here's an example in Python:

def is_leap_year(year):
    if year % 4 == 0:
        if year % 100 == 0:
            if year % 400 == 0:
                return True
            else:
                return False
        else:
            return True
    else:
        return False

You can use this function by passing the year as an argument, and it will return `True` if the year is a leap year, and `False` if it is a common year.

For example:

print(is_leap_year(2020))  # Output: True
print(is_leap_year(2021))  # Output: False
print(is_leap_year(2000))  # Output: True
print(is_leap_year(1900))  # Output: False

Please note that this algorithm is based on the Gregorian calendar rules. Different calendar systems may have different rules for determining leap years.

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