Write a function (Python 3)
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# Write a function (Python 3)

• 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.
• 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

• 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
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 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)) ```
```1 2 3 4 1994 False ```
```1 ```