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# Fizz Buzz

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)

2 Antworten

+ 2

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):

+ 1

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.