URL : https://www.hackerrank.com/challenges/write-a-function/problem

An extra day is added to the calendar almost every four years as February 29, and the day is called a leap day. It corrects the calendar for the fact that our planet takes approximately 365.25 days to orbit the sun. A leap year contains a leap day.

- In the Gregorian calendar, three conditions are used to identify leap years:
- The year can be evenly divided by 4, is a leap year, unless:
- The year can be evenly divided by 100, it is NOT a leap year, unless:
- The year is also evenly divisible by 400. Then it is a leap year.

- The year can be evenly divided by 100, it is NOT a leap year, unless:

- The year can be evenly divided by 4, is a leap year, unless:
This means that in the Gregorian calendar, the years 2000 and 2400 are leap years, while 1800, 1900, 2100, 2200, 2300 and 2500 are NOT leap years. Source

- Task
- Given a year, determine whether it is a leap year. If it is a leap year, return the Boolean True, otherwise return False.

Note that the code stub provided reads from STDIN and passes arguments to the is_leap function. It is only necessary to complete the is_leap function.

- Input Format
- Read year, the year to test.

- Constraints
- 1900 <= year <= 10^5

- Output Format
- The function must return a Boolean value (True/False). Output is handled by the provided code stub.

#### 문제풀이

- 윤년인지 확인하는 함수 작성
- 윤년은 4년마다 한번씩 오지만, 100년단위로 끊어지면 (ex 1900)은 윤년이 아님.
- 하지만 100년 단위 중에서 400년 마다 1번씩 윤년임
- 아래의 조건들을 순차적으로 하나씩 필터링 하게 하는 함수를 만듬
- year를 400으로 나누어 0이 되면 윤년
- year를 100으로 나누어 0이 되면 윤년이 아님
- year를 4로 나누어 0이되면 윤년
- 위의 조건이 모두 맞지않으면 윤년이 아님

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def is_leap(year):
leap = False
# Write your logic here
if year % 400 == 0:
leap = True
elif year % 100 == 0:
leap = False
elif year % 4 == 0:
leap = True
return leap
year = int(input())
print(is_leap(year))

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1994
False

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