HDU 1019 Least Common Multiple Python实现

a=int(input())
group=[]
def gcd(m,n):#最大公因数
   if not n:#n是否为0
        return m
   else:
        return gcd(n, m%n)
def LCM(m, n):#最小公倍数
   return int(m*n/gcd(m, n))
for i in range(0,a):
    k=input().split(" ")
    for t in range(0,len(k)):
        k[t]=int(k[t])
    for s in range(1,len(k)):
        k[s]=LCM(k[s],k[s-1])
    group.append(k[len(k)-1])
for m in range(0,len(group)):
    print (group[m])