基础算法-最小公倍数最大公约数

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闲的蛋疼,刷下初级题....基础很重要 

   //1、分解素因数法:把每个数分别分解素因数,再把各数中的全部公有素因数提取出来连乘,
        //所得的积就是这几个数的最大公约数;先把这几个数的质因数写出来,最小公倍数等于它们所有的质因数的乘积;
        static void GreatestCommonDivisorAndLeastCommonMultiple(int a,int b)
        {
            List<int> zhishuA = new List<int>();
            List<int> zhishuB = new List<int>();
            for (int i = 2; i <=a; i++)
            {
                if (a%i==0)
                {
                    a /= i;
                    zhishuA.Add(i);
                    i--;//防止相同 
                }
            }
            for (int i = 2; i <=b; i++)
            {
                if (b%i==0)
                {
                    a /= i;
                    zhishuB.Add(i);
                    i--;
                }
            }
            int GCD = 1;
            for (int i = 0; i <zhishuA.Count; i++)
            {
                for (int j = 0; j < zhishuB.Count; j++)
                {
                    if (zhishuA[i]==zhishuB[j])
                    {
                        GCD *= zhishuA[i];
                    }           
                }
            }          
        }

最大公约数 递归

 /// <summary>
        /// 辗转相除法
        /// </summary>
        /// <param name="a"></param>
        /// <param name="b"></param>
        /// <returns></returns>
        static float GCD(int a, int b)
        {
            int c = a % b;
            if (c==0)
            {
                return b;
            }
            else
            {
                a = b;
                b = c;
                return GCD(a, b);
            }

        }

 最小公倍数

   public static void LCM(int a, int b)
        {
            int temp = Math.Max(a, b);
            b = Math.Min(a, b);
            a = temp; //a 放最大 b最小
            int product = a * b;

            while (b!=0) //让b 始终是 最小
            {
                a = a > b ? a : b;
                int m = a % b;
                a = b;
                b = m;
            }
            Console.Write(product / a);
        }

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转载自blog.csdn.net/qq2512667/article/details/82320467