유한소수 판별하기 - Python

def solution(a, b):
    divisor = 2
    while divisor <= min(a, b):
        if a % divisor == 0 and b % divisor == 0:
            a //= divisor
            b //= divisor
            divisor = 2
        else:
            divisor += 1
    while b != 1:
        if b % 2 == 0:
            b //= 2
        elif b % 5 == 0:
            b //= 5
        else:
            return 2
    return 1


# Test Cases
print(solution(7, 20))
print("="*50)
print(solution(11, 22))
print("="*50)
print(solution(12, 21))
print("="*50)

  정직하게 풀었습니다.

 

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from math import gcd
def solution(a, b):
    b //= gcd(a,b)
    while b%2==0:
        b//=2
    while b%5==0:
        b//=5
    return 1 if b==1 else 2

  math 라이브러리를 쓰면 이렇게 되네요.

 

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• 프로그래머스: https://school.programmers.co.kr/learn/courses/30/lessons/120878

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