@[华为算法题](正整数A和正整数B 的最小公倍数是指 能被A和B整除的最小的正整数值，设计一个算法，求输入A和B的最小公倍数。)

python实现

这是按照我个人思路实现的。

实现一：

import datetime


def get_com_val():
    a = int(input("请输入a："))
    b = int(input("请输入b："))
    begin_time = datetime.datetime.now()
    if a>b:
        n_min = b
        n_max = a
    else:
        n_min = a
        n_max = b
    if n_max%n_min == 0:
        return n_max
    dict = {}
    for i in range(2, n_max):
        if i > n_max:
            break
        if n_max % i == 0:
            cout = 0
            while n_max % i == 0:
                n_max = int(n_max/i)
                cout += 1
            if i in dict.keys():
                if dict[i] < cout:
                    dict[i] = cout
            else:
                dict[i] = cout
        if n_min % i == 0:
            cout = 0
            while n_min % i == 0:
                n_min = int(n_min/i)
                cout += 1
            if i in dict.keys():
                if dict[i] < cout:
                    dict[i] = cout
            else:
                dict[i] = cout
    val = n_min * n_max
    for key, k_val in dict.items():
        val *= key**k_val
    end_time = datetime.datetime.now()
    d_time = end_time - begin_time
    print("运行时间:%f" % d_time.total_seconds())
    return val


if __name__ == '__main__':
    print("当前最小公倍数:{}".format(get_com_val()))

实现二：

def m_split(num):
    dict = {}
    i = 1
    while num > i:
        i += 1
        if num % i == 0:
            count = 0
            while num % i == 0:
                num = int(num / i)
                count += 1
            dict[i] = count
    else:
        dict[num] = 1
    return dict


def get_com_val():
    n = input().split()
    a = int(n[0])
    b = int(n[1])
    mul = a * b
    a_dict = m_split(a)
    b_dict = m_split(b)
    c_dict = {}
    for a_k, a_v in a_dict.items():
        if a_k in b_dict.keys():
            if a_v < b_dict[a_k]:
                c_dict[a_k] = a_dict[a_k]
            else:
                c_dict[a_k] = b_dict[a_k]
    m_v = 1
    for k, k_v in c_dict.items():
        m_v *= k ** k_v
    return int(mul / m_v)

print(get_com_val())

实现三：

import datetime


def get_com_val():
    n = input().split()
    a = int(n[0])
    b = int(n[1])
    dict = {}
    for i in range(2, a):
        if a % i == 0:
            cout = 0
            while a % i == 0:
                a = int(a/i)
                cout += 1
            if i in dict.keys():
                if dict[i] < cout:
                    dict[i] = cout
            else:
                dict[i] = cout
        if b % i == 0:
            cout = 0
            while b % i == 0:
                b = int(b/i)
                cout += 1
            if i in dict.keys():
                if dict[i] < cout:
                    dict[i] = cout
            else:
                dict[i] = cout
    val = a * b
    for key, k_val in dict.items():
        val *= key**k_val
    return val


print(get_com_val())

本文地址：https://blog.csdn.net/weixin_41964616/article/details/107125343