求1,20的最小公倍数,十分简单,
def gcd(x, y): if x < y: x, y = y, x while y: x, y = y, x % y return x def lcm(x, y): return x * y / gcd(x, y) ans = 1 for x in range(2,21): ans = lcm(ans, x) print(ans)