day 3-5 python (윤년, Leap Year)

윤년인가 아닌가를 알아보자

💪This is a Difficult Challenge 💪

Instructions

Write a program that works out whether if a given year is a leap year. A normal year has 365 days, leap years have 366, with an extra day in February. The reason why we have leap years is really fascinating, this video does it more justice:

https://www.youtube.com/watch?v=xX96xng7sAE

This is how you work out whether if a particular year is a leap year.

on every year that is evenly divisible by 4 

**except** every year that is evenly divisible by 100 

**unless** the year is also evenly divisible by 400

e.g. The year 2000:

2000 ÷ 4 = 500 (Leap)

2000 ÷ 100 = 20 (Not Leap)

2000 ÷ 400 = 5 (Leap!)

So the year 2000 is a leap year. : 4의 배수이면서 100의 배수는 아니고 400의 배수이면 윤년이다.

 

But the year 2100 is not a leap year because:

2100 ÷ 4 = 525 (Leap)

2100 ÷ 100 = 21 (Not Leap)

2100 ÷ 400 = 5.25 (Not Leap)

Warning your output should match the Example Output format exactly, even the positions of the commas and full stops.

Example Input 1

2400

Example Output 1

Leap year.

Example Input 2

1989

Example Output 2

Not leap year.

e.g. When you hit run, this is what should happen:

 

나의 코드1

#윤년

# 🚨 Don't change the code below 👇
year = int(input("Which year do you want to check? "))
# 🚨 Don't change the code above 👆

#Write your code below this line 👇

if year % 4 == 0 or year % 100 !=0 and year % 400 == 0:
    print("Leap year.")
else:
    print("Not leap year")

Java 의 논리곱 논리합이 파이썬에서는 그냥 or, and 라고 해줘서 좋은거 같다.

ㅇㅋ

 

다른 방식으로 접근한다면?

 

나의 코드2

## or 또는 and 없이 코드를 짠다면?

if year % 4 == 0:
    print("Leap year.")
elif year % 100 == 0:
    print("Not leap year.")
elif year % 400 == 0:
    print("Leap year.")
else:
    print("Not leap year.")

ㅇㅋ

# 다른 풀이

if year % 4 == 0:
    if year % 100 == 0:
        if year % 400 ==0:
            print("Leap year.")
        else:
            print("Not leap year.")
    else:
        print("Leap year.")
else:
    print("Not leap year")