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  1. Prepare
  2. Python
  3. Introduction
  4. Python If-Else
  5. Discussions

Python If-Else

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  • Diapolo10
     + 133 comments

    Fairly straightforward even without using if-else statements:

    n = int(input().strip())
check = {True: "Not Weird", False: "Weird"}

print(check[
        n%2==0 and (
            n in range(2,6) or 
            n > 20)
    ])

    I defined a simple dictionary. If the value it gets is true, the program prints "Not Weird", and if it's false "Weird" is printed. Then I simply created a bunch of statements that evaluate to either True or False.

    Sure, one could argue that this is against the rules because the task was all about if-else statements, but I wanted to see if an alternative solution was better.

    EDIT: Replaced the list comprehension with a simple boolean expression

  • TheCodeHere
     + 10 comments

    I am pretty proud of my code, i think this is the best option:

    n = int(raw_input())

if n%2==0 and (n in range(2,6) or n>20 ):
    print "Not",

print "Weird"

  • Jitender98
     + 22 comments

    Question specifically asks to use "If-Else",So I did.

    n = int(input())
if n % 2 == 0:
    if n in range(2,6):
        print("Not Weird")

    elif n in range(6,21):
        print("Weird")

    elif n > 20:
        print("Not Weird")
else:
    print("Weird")

  • lucasisaacbr
     + 5 comments
    if __name__ == '__main__':
    n = int(raw_input())
    if n % 2 != 0 or n > 5 and n < 21 :
        print("Weird")
    else: 
        print("Not Weird")

  • zereiji
     + 11 comments

    My solution :),

    if __name__ == '__main__':
  n = int(input())
  print(("Not " if n%2==0 and (n<=4 or n>20) else "") + "Weird")
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