Fizz Buzz

Question

in the test for module 3 in Python Core, it was proposed to create code that would not display even numbers. I solved this problem by simply inserting the line if not x%2==0:. But in the description it was written that you can use "continue".
What prompted me to think that I cheated somewhere. Now it became interesting, what other solutions exist?

n = int(input())

for x in range(1, n):
if not x%2==0:
if x % 3 == 0 and x % 5 == 0:
print("SoloLearn")
elif x % 3 == 0:
print("Solo")
elif x % 5 == 0:
print("Learn")
else:
print(x)

Answer 1

The rewrite with continue can remove nesting from your if condition.
That way the code will skip to the next value of the loop.

for x in range(1, n):
if x%2==0:
continue
if x % 3 == 0 and x % 5 == 0:
print("SoloLearn")
elif x % 3 == 0:
print("Solo")
elif x % 5 == 0:
print("Learn")
else:
print(x)

Another even simpler way would be to use a range with step. So it would only process every second number (leave out the evens)

for x in range(1, n, 2):

Answer 2

Павел Сидельников the lowest bit of an integer can indicate whether the number is even or odd. If the bit is 0 then it is even. If it is 1 then it is odd. The Bitwise And (&) operator can mask out the other bits and leave only that bit for testing:

if x & 1:
#process the odd values

This is a much faster method of testing than modulo (%).

Still, faster yet is no testing at all by using Tibor Santa's tip of using a step of 2 in the range.