一维数组进行多位数算数运算-加减乘

现实中，进行数值很大的数据（超过默认数据类型范围）运算时，可以用数组来完成。

例1：利用一维数组完成三位数加法（N位数加法类似；非最简解法，后同）

int[] a = {0,9,2,3}, b = {0,7,0,9}, c = new int[4];
            int jinwei = 0,t;
            for (int i = a.Length-1; i >=1; i--)
            {
                t = a[i] + b[i];
                c[i] = t % 10+jinwei;
                jinwei = t / 10;
            }
            c[0] = jinwei;
            for (int i = 0; i < a.Length; i++)
            {
                Console.Write(a[i]);
            }
            Console.Write("+");
            for (int i = 0; i < b.Length; i++)
            {
                Console.Write(b[i]);
            }
            Console.Write("=");
            for (int i = 0; i < c.Length; i++)
            {
                Console.Write(c[i]);
            }
            Console.ReadKey();

运行结果：

  

 三位数减法:

int[] a = { 4, 5, 6 }, b = { 1, 4, 7 }, c= new int[3];
            int jw=0,t;
            for (int i = a.Length-1; i >=0; i--)
            {
                t = a[i] - jw - b[i];
                if (t<0)
                {
                    t += 10;
                    jw = 1;
                }
                else
                {
                    jw = 0;
                }
                c[i] = t;
            }
            for (int i = 0; i < c.Length; i++)
            {
                Console.Write(c[i]);
            }
            Console.ReadKey();

结果：

 

例2：三位数乘法

int[] a = { 9, 2, 3 }, b = { 7, 0, 9 }, c = new int[4], d = { 0, 0, 0, 0, 0, 0 };
            int jinwei, t;
            for (int i = 0; i < b.Length; i++)
            {
                jinwei = 0;
                for (int j = 0; j < c.Length; j++)
                {
                    c[j] = 0;
                }
                for (int j = 0; j < a.Length; j++)
                {
                    t = a[a.Length - 1 - j] * b[b.Length - 1 - i];
                    c[c.Length - 1 - j] += t % 10;
                    jinwei = t / 10 + c[c.Length - 1 - j]/10;
                    c[c.Length - 1 - j] %= 10;
                    if (jinwei > 0)
                    {
                        c[c.Length - 1 - j-1] +=jinwei;
                    }
                }
                //测试中间累加的数字
                //for (int j = 0; j < c.Length; j++)
                //{
                //    Console.Write(c[j]);
                //}
                //Console.WriteLine();
                //测试结束
                jinwei = 0;
                for (int j = 0; j < c.Length; j++)
                {
                    t = d[d.Length - 1 - i - j] + c[c.Length - 1 - j];
                    d[d.Length - 1 - i - j] = t % 10;
                    jinwei = t / 10;
                    jinwei += d[d.Length - 1 - i - j] / 10;
                    d[d.Length - 1 - i - j] %= 10;
                    if (jinwei > 0)
                    {
                        d[d.Length - 1 - i - j - 1] += jinwei;
                    }
                }
            }
            for (int i = 0; i < a.Length; i++)
            {
                Console.Write(a[i]);
            }
            Console.Write("*");
            for (int i = 0; i < b.Length; i++)
            {
                Console.Write(b[i]);
            }
            Console.Write("=");
            for (int i = 0; i < d.Length; i++)
            {
                Console.Write(d[i]);
            }
            Console.ReadKey();

 运行结果：

 利用二位数组完成，会简单一些：

int[] a = { 9, 4, 7 }, b = { 3, 6, 9 };
            //M位数和N位数相乘，最多有M+N位数，N位中间结果
            int[,] c = new int[b.Length + 1, a.Length + b.Length];
            //初始化
            for (int i = 0; i < c.GetLength(0); i++)
            {
                for (int j = 0; j < c.GetLength(1); j++)
                {
                    c[i, j] = 0;
                }
            }
            //分位乘
            for (int i = b.Length - 1; i >= 0; i--)
            {
                for (int j = a.Length - 1; j >= 0; j--)
                {
                    c[i, c.GetLength(1) - 1 - (b.Length - 1 - i) - (a.Length - 1 - j)] = a[j] * b[i];
                }
            }
            //中间结果累加
            for (int i = 0; i <= c.GetLength(0) - 2; i++)
            {
                for (int j = c.GetLength(1) - 1; j >= 0; j--)
                {
                    c[c.GetLength(0) - 1, j] += c[i, j];
                }
            }
            //对累加结果的进位部分进行处理
            for (int i = c.GetLength(1) - 1; i >= 0; i--)
            {
                if (c[c.GetLength(0) - 1, i] >= 10)
                {
                    c[c.GetLength(0) - 1, i - 1] += c[c.GetLength(0) - 1, i] / 10;
                    c[c.GetLength(0) - 1, i] %= 10;
                }
            }
            //输出结果
            for (int i = 0; i < c.GetLength(1); i++)
            {
                Console.Write(c[c.GetLength(0) - 1, i]);
            }
            Console.ReadKey();

运行结果：