4.3.2 연습 문제: 내일의 날짜 구하기(1)

4.3.2 연습 문제: 내일의 날짜 구하기(1) - 왕초보를 위한 Python: 쉽게 풀어 쓴 기초 문법과 실습

4.3.2 연습 문제: 내일의 날짜 구하기(1)

 

# 내일의 날짜 구하기
def dayTomorrow(y, m, d):
  
  day = f"{int(d):02d}"
  month = f"{int(m):02d}"
  today = [month, day, y]
  print('/'.join(today))

  day = d
  month = m
  year = y
  if (month == "2" and int(d) == 28):
    month = "03"
    day = "01"
  elif (d == "31" and (month == "1" or month == "3" or month == "5" or month == "7" or month == "8" or month == "10")):
    month = str(int(m) + 1)
    day = "01"
  elif (d == "30" and (month == "4" or month == "6" or month == "9" or month == "11")):
    month = str(int(m) + 1)
    day = "01"
  elif (month == "12" and int(d) == 31):
    month = "01"
    year = str(int(y) + 1)
    day = "01"
  else:
    day = f"{int(d)+1:02d}"
    month = f"{int(m):02d}"
    
  tomorrow = [month, day, year]
  print('/'.join(tomorrow))


year, month, day = input().split()
dayTomorrow(year, month, day)

 

 

# 리펙토링한 내일 날짜 구하기
from datetime import datetime, timedelta

def dayTomorrow(y, m, d):
    today = datetime(int(y), int(m), int(d))
    tomorrow = today + timedelta(days=1)

    print(today.strftime("%m/%d/%Y"))
    print(tomorrow.strftime("%m/%d/%Y"))

year, month, day = input().split()
dayTomorrow(year, month, day)