输入两个长整型数字利用字符串方式相加？

Question

RT，假定数字位很长 整形位不够用 利用字符串相加方式 怎么用c#实现：
方法写到类里面；

Answer 1

用decimal进行处理，
decimal d1 =0;
decimal  d2 =5;
decimal d3;
d3 =d1+d2;

Answer 2

这个什么意思？？decimal类型够长？

Answer 3

http://www.google.com.hk/search?hl=zh-CN&newwindow=1&safe=strict&biw=1260&bih=837&q=%E9%AB%98%E7%B2%BE%E5%BA%A6%E5%A4%A7%E6%95%B0%E8%BF%90%E7%AE%97&aq=f&aqi=&aql=&oq=

Answer 4

参考
http://topic.csdn.net/u/20090701/09/918423b3-fc8b-4446-b8c2-67a9488bc377.html

Answer 5

long a="121212121212121";
long b="212121212121212";
String c=a.toString()+b.toString();

Answer 6

///我蛋疼写一个，没考虑有负数的情况
            string a0 = "14536544";
            string a2 = "647811351545481";            string a1=a0.PadLeft(a2.Length, '0');
            List<int> result = new List<int>();            int head = 0;//进位
            for (int i = a2.Length - 1; i >= -1; i--)
            {
                if (i == -1)
                {
                    if(head!=0)result.Add(head);
                    break;
                }
                int ha = int.Parse(a1[i].ToString());
                int xi = int.Parse(a2[i].ToString());
                if (ha + xi + head < 10)
                {
                    result.Add(ha + xi + head);
                    head = 0;
                }
                else
                {
                    result.Add(ha + xi + head - 10);
                    head = 1;
                }
            }
            result.Reverse();
           string strResult= result.Aggregate<int, string>("",(agg,next)=>agg+next);
           Console.WriteLine(strResult);

Answer 7

长整型用double，长度比decimal大得多...整型不需要小数精度...再长.NET 4.0有专门的长整型System.Numerics.BigInteger结构，理论上支持无限大...

Answer 8

以前发的贴，之后就忘了再看了，谢谢大家的回复啊；
现在我把之后自己写出来的代码贴出来 如果有需要的童鞋可以看看；
包括字符相加减（小数和负数均可），乘法没弄出来 有待大虾指教！using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;namespace jiafa
{
    class Program
    {
        static void Main(string[] args)
        {
            Shuju sj = new Shuju(); //定义一个Shuju类对象sj；
            Yuansuan js = new Yuansuan();    //定义一个Yuansuan类对象js；
            Console.WriteLine("求两个长精度型的字符运算结果:");
            do
            {
                Console.WriteLine("请输入运算符:加法（“+”）减法（“-”）乘法（“*”）");
                char x=Convert.ToChar(Console.ReadLine());
                Console.WriteLine("输入第一个数据：");
                sj.S1 = Console.ReadLine();
                Console.WriteLine("输入第二个数据：");
                sj.S2 = Console.ReadLine();
                switch (x)
                {
                    case '+':
                        Console.Write("{0}+{1}=", sj.S1, sj.S2);
                        sj.Change();
                        Console.WriteLine(js.Jsresult(sj.S1, sj.S2, '+'));
                        break;
                    case '-':
                        Console.Write("{0}-{1}=", sj.S1, sj.S2);
                        sj.Change();
                        Console.WriteLine(js.Jsresult(sj.S1, sj.S2, '-'));
                        break;
                    case '*':
                        Console.WriteLine("由于技术原因乘法运算尚待完成！");
                        break;
                    default:
                        Console.WriteLine("符号输入不正确！");
                        break;
                }
                Console.WriteLine("继续（y）；结束（n）");
                string b = Console.ReadLine();
                if (b != "y") { break; }
            }while(true);
            Console.WriteLine("请按任意键结束！");
            Console.Read();
        }
    }
    class Shuju
    {
        private string _s1;     
        private string _s2;
        public string S1
        {
            get { return _s1; }
            set { _s1 = value; }
        }
        public string S2
        {
            get { return _s2; }
            set { _s2 = value; }
        }
        public void Change() //调整两个字符串的位数
        {
            int m = Math.Abs(S1.Length - S2.Length);
            int d1 = S1.IndexOf('.');
            int d2 = S2.IndexOf('.');
            #region 两个都是整数
            if (d1 == -1 && d2 == -1)
            {
                if (S1.Length > S2.Length)
                {
                    S2=ChPoint(S2,S1.Length-S2.Length);
                }
                else if (S1.Length < S2.Length)
                {
                    S1 = ChPoint(S1, S2.Length - S1.Length);
                }            }
            #endregion
            #region 两个都是小数
            else if (d1 != -1 & d2 != -1)
            {
                string la = S1.Substring(0, d1);
                string ra = S1.Substring(d1 + 1);
                string lb = S2.Substring(0, d2);
                string rb = S2.Substring(d2 + 1);
                if (la.Length > lb.Length)
                {
                    lb = ChPoint(lb,la.Length - lb.Length);
                }
                else if(la.Length<lb.Length)
                {
                    la = ChPoint(la,lb.Length - la.Length);
                }                if(ra.Length>rb.Length)
                {
                    rb=ChPoint1(rb,ra.Length-rb.Length);
                }
                else if(ra.Length<rb.Length)
                {
                    ra=ChPoint1(ra,rb.Length-ra.Length);
                }                S1=la+"."+ra;
                S2=lb+"."+rb;
            }
            #endregion
            #region 一个是小数
            else if (d1 != -1 & d2 == -1)
            {
                string la = S1.Substring(0, d1);
                string ra = S1.Substring(d1 + 1);
                string lb = S2;
                string rb = "";
                if (la.Length > lb.Length)
                {
                    lb = ChPoint(lb, la.Length - lb.Length);
                }
                else if (la.Length < lb.Length)
                {
                    la = ChPoint(la, lb.Length - la.Length);
                }
                rb = ChPoint1(rb, ra.Length - rb.Length);                S1 = la + "." + ra;
                S2 = lb + "." + rb;
            }
            else if (d1 == -1 & d2 != -1)
            {
                
                string la = S1;
                string ra = "";
                string lb = S2.Substring(0,d2);
                string rb = S2.Substring(d2+1);
                if (la.Length > lb.Length)
                {
                    lb = ChPoint(lb, la.Length - lb.Length);
                }
                else if (la.Length < lb.Length)
                {
                    la = ChPoint(la, lb.Length - la.Length);
                }
                ra = ChPoint1(ra, rb.Length - ra.Length);                S1 = la + "." + ra;
                S2 = lb + "." + rb;
                
            }
            #endregion
        }
        private string ChPoint(string a,int b)
        {  
            for (int i = 0; i < b; i++)
            {
                a = a.Insert(0, "0");
            }
            return a;
        }  //小数点前面凑0
        private string ChPoint1(string a,int b)
        {
            for (int i = 0; i < b; i++)
            {
                a = a.Insert(a.Length,"0");
            }
            return a;
        } //小数点后面凑0
    }    public class Yuansuan
    {
        public string Jsresult(string SS1, string SS2, char y) 
        {
            //int m= S1.Length > S2.Length ? S1.Length : S2.Length;
            char[] ch1 = SS1.ToCharArray();
            char[] ch2 = SS2.ToCharArray();
            string result = "";
            int flag=0;            #region 加法运算
            if (y == '+')
            {
                for (int i = ch1.Length - 1; i >= 0; i--)
                {
                    if (ch1[i].ToString() != ".")
                    {
                        int a = Convert.ToInt32(ch1[i].ToString());
                        int b = Convert.ToInt32(ch2[i].ToString());
                        int s = a + b + flag;
                        if (s >=10)
                        {
                            s = s -10;
                            flag = 1;
                        }
                        else { flag = 0; }
                        result = s + result;
                        //int j=(s>=10?(s/10):0);
                    }
                    else
                    {
                        result = '.' + result;
                    }
                }
                if (flag == 1)
                {
                    result = 1 + result;
                }
                while (result[result.Length - 1] == '0' || result[result.Length - 1] == '.')
                {
                    if (result[result.Length - 1] != '.')
                    {
                       result=result.Remove(result.Length-1,1);
                    }
                    else if (result[result.Length - 1] == '.')
                    {
                        result = result.Remove(result.Length-1, 1);
                        break;
                    }
                }
            }
            #endregion
            #region 减法运算
            else if (y == '-')
            {
                int j=0,s=0;
                do
                {
                    if (ch1[j] != '.')
                    {
                        s = Convert.ToInt32(ch1[j].ToString()) - Convert.ToInt32(ch2[j].ToString());
                    }
                        j++;
                } while (s == 0 && j < ch1.Length);
                if (s >= 0)
                {
                    result = Plus(ch1, ch2);
                }
                else if (s < 0)
                {
                    result = "-" + Plus(ch2, ch1);
                }
            }
            #endregion
            #region 乘法运算
            #endregion
            return result;
        }//方法；计算两个凑0后的字符串运算结果
        private  string Plus(char[] x,char[] y) 
        {
            string re = "";
            int flag = 0;
            for (int i = x.Length - 1; i >= 0; i--)
            {
                if (x[i].ToString() != ".")
                {
                    int a = Convert.ToInt32(x[i].ToString());
                    int b = Convert.ToInt32(y[i].ToString());
                    int s = a - b + flag;
                    if (s < 0)
                    {
                        s = s + 10;
                        flag = -1;
                    }
                    else { flag = 0; }
                    re = s + re;
                    //int j=(s>=10?(s/10):0);
                }
                else
                {
                    re = '.' + re;
                }
            }
            while (re[0] == '0'&&re[1]!='.') 
            {
                re = re.Remove(0, 1);
            }
            while (re[re.Length - 1] == '0' || re[re.Length - 1] == '.')
            {
                if (re[re.Length - 1] != '.')
                {
                    re = re.Remove(re.Length - 1, 1);
                }
                else if (re[re.Length - 1] == '.')
                {
                    re = re.Remove(re.Length - 1, 1);
                    break;
                }
            }
            return re;
        }  // 两个字符数组相减方法
    }
}

Answer 9

你这是啥回答啊？告诉人家 用GOOGLE搜 高精度大数运算！
鄙视你，还微软MVP呢

Answer 10

 TO caozhy：你看人家jiang_zm2009多实在，一上来就很生猛，代码都贴上了

Answer 11

你还是考虑下.net4.0里现成的方法吧

Answer 12

神马数据那么长。long 整形都不够吗？