两个大数相加的问题

[lcmidnight830118]

有两个非常大的数字M和N，存储形式为string，将其相加后得到的数字也存为string类型并返回
这个程序怎么写

[delectation]

数字是什么类型的，要考虑溢出。

[CGabriel]

一年级学的竖式加法没有忘记吧。 M 和 N 都从最后一位（个位）向前遍历， 取每个字符相加，中间准备个进位就是。

无论多大的数都可以加起来

[yangls06]

按照加法的法则来就行了

[yuanleijiwei]

string  m;
  string n;
  string sum =Convert .ToString( Convert.ToInt64(m)+Convert.ToInt64(n));

[CSharpEx]

用大数值型class，网络上可以下载到的

[showjim]

[Quote=引用 2 楼 cgabriel 的回复:]
一年级学的竖式加法没有忘记吧。 M 和 N 都从最后一位（个位）向前遍历， 取每个字符相加，中间准备个进位就是。

无论多大的数都可以加起来
[/Quote]
简单的正解

[kjiwu]

自己写的，试了试可以。

        public class MyFucntions
        {
            private static Boolean TwoBigNumbersAdd(string num1, string num2, out string result)
            {
                string bigone = "";
                string tmpStr = "";
                string smallone = "";
                int len = 0;
                if (num1.Length < num2.Length)
                {
                    len = num1.Length;                    
                    bigone = num2;
                    smallone = num1;
                }
                else
                {
                    len = num2.Length;
                    bigone = num1;
                    smallone = num2;
                }

                tmpStr = bigone.Substring(bigone.Length - len);//取出len这么长的低位部分
                bigone = bigone.Substring(0, bigone.Length - len); //取出大数剩余的高位部分               

                int tmp = 0; //进位
                result = "";
                
                //低位部分的计算
                for (int i = len-1; i >= 0; i--)
                {
                    try
                    {
                        int a = Convert.ToInt32(Convert.ToString( smallone[i]));
                        int b = Convert.ToInt32(Convert.ToString(tmpStr[i]));
                        int sum = a + b + tmp; 

                        result = Convert.ToString( sum % 10 ) + result;
                        tmp = sum / 10;
                    }
                    catch (Exception e)
                    {
                        result = "error";
                        return false;
                    }
                }

                //剩余高位部分的计算
                if (tmp != 0)
                {
                    for (int i = bigone.Length - 1; i >= 0; i--)
                    {
                        try
                        {
                            int a = Convert.ToInt32(Convert.ToString(bigone[i]));
                            int sum = a + tmp;

                            result = Convert.ToString(sum % 10) + result;
                            tmp = sum / 10;
                        }
                        catch (Exception e)
                        {
                            result = "error";
                            return false;
                        }
                    }
                }
                else
                {
                    result = bigone + result;
                }

                return true;
            }
            
            public static Boolean BigNumbersAdd(string[] nums, out string Result)
            {
                string firstnum = nums[0];
                try
                {
                    for (int i = 1; i < nums.Length; i++)
                    {
                        if (!TwoBigNumbersAdd(firstnum, nums[i], out firstnum))
                        {
                            Result = "Error";
                            return false;
                        }
                    }
                } 
                catch (Exception e)
                {
                    Result = "Error";
                    return false;
                }

                Result = firstnum;
                return true;
            }
        }       

[angelfeng9]

楼上好强

[啵比]

[Quote=引用 7 楼 kjiwu 的回复:]
自己写的，试了试可以。
C# codepublicclass MyFucntions
{privatestatic Boolean TwoBigNumbersAdd(string num1,string num2,outstring result)
{string bigone="";string tmpStr="";string smallone="";int len=0;if (num1.Length< num2.Length)
{
len= num1.Length;
bigone= num2;
smallone= num1;
}else
{
len= num2.Length;
bigone= num1;
smallone= num2;
}

tmpStr= bigone.Substring(bigone.Length- len);//取出len这么长的低位部分 bigone= bigone.Substring(0, bigone.Length- len);//取出大数剩余的高位部分int tmp=0;//进位 result="";//低位部分的计算for (int i= len-1; i>=0; i--)
{try
{int a= Convert.ToInt32(Convert.ToString( smallone[i]));int b= Convert.ToInt32(Convert.ToString(tmpStr[i]));int sum= a+ b+ tmp;

result= Convert.ToString( sum%10 )+ result;
tmp= sum/10;
}catch (Exception e)
{
result="error";returnfalse;
}
}//剩余高位部分的计算if (tmp!=0)
{for (int i= bigone.Length-1; i>=0; i--)
{try
{int a= Convert.ToInt32(Convert.ToString(bigone[i]));int sum= a+ tmp;

result= Convert.ToString(sum%10)+ result;
tmp= sum/10;
}catch (Exception e)
{
result="error";returnfalse;
}
}
}else
{
result= bigone+ result;
}returntrue;
}publicstatic Boolean BigNumbersAdd(string[] nums,outstring Result)
{string firstnum= nums[0];try
{for (int i=1; i< nums.Length; i++)
{if (!TwoBigNumbersAdd(firstnum, nums[i],out firstnum))
{
Result="Error";returnfalse;
}
}
}catch (Exception e)
{
Result="Error";returnfalse;
}

Result= firstnum;returntrue;
}
}
[/Quote]
正解!

[zgke]

自己算....


private void button2_Click(object sender, EventArgs e)
{
string _Value1 = "1231231231231231298379842391214331221361524398124391824311322131234918249";
string _Value2 = "9999999999999999999999999999999991232345234523452304952345230234523452349";


string _Text = AddNumbText(_Value1, _Value2);

}

public string AddNumbText(string p_Value1, string p_Value2)
{
if (p_Value2.Length < p_Value1.Length)
{
p_Value2 = p_Value2.PadLeft(p_Value1.Length, '0');

}
else
{
p_Value1 = p_Value1.PadLeft(p_Value2.Length, '0');
}

char[] _Test = new char[p_Value1.Length];
bool _Add = false;
int _Temp = 0;

for (int i = p_Value1.Length - 1; i != -1; i--)
{
byte _Int1 = Convert.ToByte(p_Value1[i].ToString(), 10);
byte _Int2 = Convert.ToByte(p_Value2[i].ToString(), 10);
_Temp = _Int1 + _Int2;
if (_Add) _Temp++;

if (_Temp > 10)
{
_Add = true;
_Test[i] = _Temp.ToString()[1];
}
else
{
_Add = false;
_Test[i] = _Temp.ToString()[0];
}
}

if (_Add)
{
return "1" + new string(_Test);
}
else
{
return new string(_Test);
}
}

[Deathsign]

答案出来了吧

高精度计算

[chen_dian_dian]

class LargeNum
{
string numValue;
public int Length
{
get { return numValue.Length; }
}
/// <summary>
/// 获取和设置当前大数的制定位的数字值
/// </summary>
/// <param name="index">指定位的索引</param>
/// <returns>指定位的数值</returns>
public int this[int index]
{
get { return Convert.ToInt32(numValue[index].ToString()); }
}
/// <summary>
/// 构造函数
/// </summary>
/// <param name="numValue">大数的值</param>
public LargeNum(string numValue)
{
this.numValue = numValue;
}
/// <summary>
/// 计算两个指定大数的和
/// </summary>
/// <param name="num1">一个数</param>
/// <param name="num2">另一个数</param>
/// <returns></returns>
public static LargeNum operator +(LargeNum num1, LargeNum num2)
{
LargeNum numLong;
LargeNum numShort;
//根据两个数字的长度，分别设置为长的数和短的数
if (num1.Length < num2.Length)
{
numShort = num1;
numLong=num2;
}
else
{
numShort = num2;
numLong =num1 ;
}
int[] result = new int[numLong .Length+ 1];

//计算结果的后几位，从最后一位摆竖式开始算，计算到短的数那么多位
for (int i =numLong.Length;i> numLong.Length - numShort.Length; i--)
{
result[i] = numLong[i-1] + numShort[i - numLong.Length + numShort.Length-1]+result[i];
if (result[i] >= 10)
{
result[i-1] = 1;
result[i] = result[i] - 10;
}
}
//计算结果的前几位，把长的数剩下的几位加上
for (int i = numLong.Length - numShort.Length ; i >0; i--)
{
result[i] = numLong[i-1]+result[i];
if (result[i] >= 10)
{
result[i - 1] = 1;
result[i] = result[i] - 10;
}
}
return new LargeNum( ConvertIntArrayToString(result)/*将计算结果转换成字符串*/);//将计算结果转换成大数，返回
}

public override string ToString()
{
return numValue;
}
/// <summary>
/// 将指定的整型数组转换成字符串
/// </summary>
/// <param name="intArray">整型数组</param>
/// <returns>对应字符串，每个整数对应数组的一位</returns>
private static string ConvertIntArrayToString(int[] intArray)
{
string result="";
for (int i = 1; i < intArray.Length; i++)
{
result += intArray[i].ToString();
}
if (intArray[0] >0)
{
result =intArray[0]+ result;
}
return result;
}
}

[大Y]

学习了