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์œค๋…„ ๊ณ„์‚ฐํ•˜๊ธฐ

by rindev 2020. 12. 5.

 

www.hackerrank.com/challenges/write-a-function/problem

 

Write a function | HackerRank

Write a function to check if the given year is leap or not

www.hackerrank.com

 

ํ•ด์ปค๋žญํฌ ๊ธฐ์›ƒ๊ฑฐ๋ฆฌ๋‹ค๊ฐ€...

๋ฌธ์ œ๊ฐ€ ์˜์–ด์—ฌ์„œ ์ฝ๋Š”๋ฐ ๋” ์˜ค๋ž˜๊ฑธ๋ฆฌ๋Š”๋“ฏ๐Ÿ˜ญ ์•„์ง๊นŒ์ง„ print ์ˆ˜์ค€์ด๋ผ ๋ฌธ์ œ๊ฐ€ ์–ด๋ ค์šด๊ฑด ์•„๋‹Œ๋ฐ....

 

์•”ํŠผ ๋ฌธ์ œ์—์„œ๋Š”......

 

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.

...๋ผ๊ณ  ํ•œ๋‹ค!๐Ÿ™„  

 

 

๐Ÿ’์œค๋…„์ด ๋˜๋ ค๋ฉด.. 

  • 4๋กœ ๋‚˜๋ˆ„์–ด๋–จ์–ด์ง€๋ฉด ๊ธฐ๋ณธ ์œค๋…„
  • but 4๋กœ ๋‚˜๋ˆ„์–ด๋–จ์–ด์ง€๋Š”๋ฐ, ๊ทผ๋ฐ 100์œผ๋กœ๋„ ๋‚˜๋ˆ ๋–จ์–ด์ง€๋ฉด ํ‰๋…„
  • but 4๋กœ ๋‚˜๋ˆ ๋–จ์–ด์ง€๊ณ  + 100์œผ๋กœ ๋‚˜๋ˆ ๋–จ์–ด์ง€๊ณ  + ๊ฒŒ๋‹ค๊ฐ€! 400์œผ๋กœ๋„ ๋‚˜๋ˆ ๋–จ์–ด์ง€๋ฉด ์œค๋…„

์ •๋ฆฌํ•˜์ž๋ฉด

3๋ฒˆ์€ 400์œผ๋กœ ๋‚˜๋ˆ ๋–จ์–ด์ง€๋ฉด ์œค๋…„์ด๋ผ๋Š” ๋œป์ด๊ณ 

1์ด๋ž‘ 2๋ฅผ ๋ณด๋ฉด (4๋กœ๋Š” ๋–จ์–ด์ง + 100์œผ๋กœ๋Š” ์•ˆ๋–จ์–ด์ง) -> ์œค๋…„ ์ด๋ผ๋Š” ๋œป์ด๋‹ค

 

 

๐Ÿƒ‍โ™€๏ธ ์ž‘์„ฑํ•œ ์ฝ”๋“œ

๋”๋ณด๊ธฐ

 

 

def is_leap(year):
    leap = False
    
    if year%400==0 or (year%4==0 and year%100!=0):
        leap = True 
    
    return leap

year = int(input())
print(is_leap(year))

๋Œ“๊ธ€