求多个数的最小公倍数，个数最小公倍数,import math＂

求多个数的最小公倍数，个数最小公倍数,import math"

import math"""    获得多个数的最小公倍数    isPrime(n)    getMutiPrime(n)    getLeastCommonMutible(numList)"""def isPrime(n):    """    判断一个数是否是质数    """    if n==2:        return True    if n==1 or n%2==0:        return False    p = int(math.sqrt(n))+1    i = 3    while i<p:        if n%i==0:            return False        i+=2    return Truedef getMutiPrime(n):    """get muti num of n    n=num1*num2...*numx for num1...numx are all prime numbers    将数n进行质因数分解 prime factorization    """    if n==2:        resultlist = [2]    else:        list1 = [i for i in range(2,int(math.sqrt(n))+1) if isPrime(i)]        #print(list1)        resultlist=[]        tmp = n        while not isPrime(tmp) and tmp!=1:            for i in list1:                if tmp%i==0:                    resultlist.append(i)                    tmp=tmp//i                    #print(tmp)                    #print(resultlist)        if tmp!=1:            resultlist.append(tmp)    return resultlistdef isMutillistFill(mutilLsit):    """        判断一个二维列表中的每一个列表是否都为空        mutilLsit=[[],[],[]] 返回False        mutilLsit=[[1],[],[]] 返回True"""    for row in mutilLsit:        if len(row)>0:            return True    return Falsedef getLeastCommonMutible(numList):    """    numList 待求数的列表，    返回它们的最小公倍数的质因数列表    """    primelists=[]    set1=set({})    mutilResult = []    for i in numList:        primelists.append(getMutiPrime(i))        set1.update(set(i1 for i1 in getMutiPrime(i)))    #print(primelists)    flag = True    flag1 = False    while flag:        for i in set1:            for row in primelists:                if i in row:                    row.remove(i)                    flag1 = True            if flag1:                mutilResult.append(i)                flag1 = False        flag = isMutillistFill(primelists)    #print(mutilResult)            return mutilResultdef getMutil(numList):    result = 1    for i in numList:        result*=i    return resultif __name__=='__main__':    print(getMutil(getLeastCommonMutible([8,12,20])))#获取8,12,20的最小公倍数    print(getMutil(getLeastCommonMutible([i for i in range(2,21)])))#获取1到20的所有数的最小公倍数#该片段来自于http://byrx.net