• C#版 大数计算器加法


    ///
     

       /// 基类,抽象类,定义常用方法
        ///
        abstract class BigCalculate
        {
            public abstract string Oper(string num1, string num2);

            //反转数字顺序
            protected string InvertOrder(string str)
            {
                char[] chTemp = new char[str.Length];

                for (int i = 0; i < str.Length; i++)
                {
                    chTemp[i] = str[str.Length - i-1];
                }

                return new string(chTemp.ToArray());
            }

            //交换数字
            protected void ChangeNum(ref string num1,ref string num2)
            {
                string strTemp = num1;
                num1 = num2;
                num2 = strTemp;
            }
            //获得较大数
            protected string Max(string num1, string num2)
            {
                if (num1.Length > num2.Length)
                {
                    return num1;
                }
                else if (num1.Length==num2.Length)
                {
                    for (int i = num1.Length-1; i >=0 ; i--)
                    {
                        if (num1[i].CompareTo(num2[i])<0)
                        {
                            return num2;
                        }
                    }
                    return num1;
                }
                return num2;
            }
        }
    --------------------------------------------------------------------------
        class BigAdd:BigCalculate
        {
            public override string Oper(string num1, string num2)
            {
                //获得较大数的位数
                int len = num1.Length > num2.Length ? num1.Length : num2.Length;
                //将较小数差位补0
                string tempNum1 = new string('0', len - num1.Length) + num1;
                string tempNum2 = new string('0', len - num2.Length) + num2;

                int flag = 0;//进位符
                List liCh = new List();//存放运算后数字
                for (int i = len - 1; i >= 0; i--)
                {
                    liCh.Add((char)(tempNum1[i] + tempNum2[i] + flag - '0'));
                    flag = (liCh[liCh.Count - 1] - '0') / 10;
                    liCh[liCh.Count - 1] = (char)(((liCh[liCh.Count - 1] - '0') % 10) + '0');
                }

                //若运算完成后进位符不为0,及最高位进位,则再加一位
                if (flag > 0)
                {
                    liCh.Add((char)(flag + '0'));
                }

                return InvertOrder(new string(liCh.ToArray()));
            }
        }
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  • 原文地址:https://www.cnblogs.com/qiuyeyaozhuai/p/2719081.html
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