Python 小目标4

Python Day 10

1. Leap Year

I have a friend Cindy, she was actually born on Feb.29th, and I remember we celebrated her birthday together 2012 in the Maths place!
Leap Year was adapted in Gregorian Calendar. There are Three Criteria:

• 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. [caption id="attachment_1848" align="alignnone" width="750"]

  image

  pixel2013 / Pixabay[/caption]

Solution:

def is_leap(year):
    if year % 4 != 0:
        print("False")
    elif year % 4 ==0 and year % 100==0 and year % 40==0:
        print("Leap Year")
    elif year % 4==0 and year % 100==0 and year % 40!=0 :
        print("False")
    else:
        print("Leap Year")

Alternative Solutions:

#Alternative Solution
def is_leap1(year):   
       return year % 4 == 0 and (year % 400 == 0 or year % 100 != 0)

Output:

print(is_leap(1900))
False

2. Second Largest Number in a List

Suppose we want to find the Second Largest Number in a List
Ex: List=[4,6,3,2,2] Output=4

Solution:

arr=[2,5,4,2]
length=len(arr)
arr.sort()
print(arr[length-2])

3. Read an integer N, try to print out 123...N

EX: N=3, Output 123

Solution:

n= int(input())
print(*range(1,n+1), sep="")

n=8
1234567

Summary:

we have been working with the Python commands Range, Length, Sort, Module (%), and If-Then(elif-then, else), dfdAs all the books suggested, the Best way to learn to use the knowledge mindfully, so I organized this note :)

Happy Practice! 🦁

Reference:
https://www.hackerrank.com/
https://www.geeksforgeeks.org/python-largest-smallest-second-largest-second-smallest-list/