알고리즘
1. 최대공약수와 최소공배수
풀이:
def solution(n, m):
for i in range(min(n, m), 0, -1):
if(n % i == 0) and (m % i == 0):
a = i
break
for j in range(max(n, m), (n*m)+1):
if j % n == 0 and j % m == 0:
b = j
break
return [a, b]