微乳笔试题-Large Integers,请赐教

[kernelkoder]

Large Integers
A 4-byte unsigned integer can only represent values between 0 and 4294967295.
1. Define a class UInteger encapsulating arbitrary non-negative integers, e.g.
123456789123456789123456789123456789.
2. Define and implement a constructor that takes a string containing the hexadecimal representation of an
integer:
UInteger(string numberInHex);
You may assume that the string contains only hex digits and you do not have to check the string’s
validity.
3. Define and implement all relational operators: ==, !=, <, <=, >, >=.
4. Define and implement the addition (+) operator for UIntegers.

[欢乐的小猪]

 class UInteger
    {
        private   string numberInHex;
       

       public  UInteger(string numberInHex)
        {
            this.numberInHex = numberInHex;
         
        }

       public static bool operator ==(UInteger a, UInteger b)
       {
          
           if (System.Object.ReferenceEquals(a, b))
           {
               return true;
           }
        
           if (((object)a == null) || ((object)b == null))
           {
               return false;
           }

           return a.numberInHex.ToLower() == b.numberInHex.ToLower();
       }

       public static bool operator  != (UInteger a, UInteger b)
       {

           return !(a==b);
       }

       public static bool operator >(UInteger a, UInteger b)
       {
           if(a.numberInHex.Count()>b.numberInHex.Count())
               return true;
           if(a.numberInHex.Count()<b.numberInHex.Count())
               return false;
           for(int i=0;i<a.numberInHex.Count();i++)
               if (a.numberInHex.ToLower().ElementAt(i) > b.numberInHex.ToLower().ElementAt(i))
                   return true;
         
           return false;
       }

       public static bool operator <=(UInteger a, UInteger b)
       {
           return !(a > b);
       }

       public static bool operator <(UInteger a, UInteger b)
       {
           if (a.numberInHex.Count() > b.numberInHex.Count())
               return false;
           if (a.numberInHex.Count() < b.numberInHex.Count())
               return true;
           for (int i = 0; i < a.numberInHex.Count(); i++)
               if (a.numberInHex.ToLower().ElementAt(i) < b.numberInHex.ToLower().ElementAt(i))
                   return true;

           return false;
       }

       public static bool operator >=(UInteger a, UInteger b)
       {
           return !(a < b);
       }


       public static UInteger operator +(UInteger a, UInteger b)
       {
           UInteger u = new UInteger(a.numberInHex + b.numberInHex);
          
           return u;
       }
    }

[小笨和漂向北方]

水猪，我书读读少，你能告诉我，operator + override 里两个string相加，是如何实现大数加法的吗？string default operator +不是做concatenation吗？

[欢乐的小猪]

数太大算不了，换个服务器呗

引用 2 楼 micropentium6 的回复:

水猪，我书读读少，你能告诉我，operator + override 里两个string相加，是如何实现大数加法的吗？string default operator +不是做concatenation吗？

[欢乐的小猪]

   public static UInteger operator +(UInteger a, UInteger b)
        {
            //进位为1
            UInt32 C = 0;
            string result = string.Empty;

            for (int i = 0; i < Math.Max(a.numberInHex.Length, b.numberInHex.Length); i++)
            {
                UInt32 n1 = a.numberInHex.Length - 1 - i < 0 ? 0 : UInt32.Parse(a.numberInHex.Substring(a.numberInHex.Length - 1 - i, 1), NumberStyles.HexNumber);
                UInt32 n2 = b.numberInHex.Length - 1 - i < 0 ? 0 : UInt32.Parse(b.numberInHex.Substring(b.numberInHex.Length - 1 - i, 1), NumberStyles.HexNumber);
                result = ((n1 + n2 + C) % 16).ToString("X") + result;
                C = (n1 + n2 + C) / 16;
            }

            UInteger u = new UInteger(C == 0 ? result : "1" + result);

            return u;
        }

引用 2 楼 micropentium6 的回复:

水猪，我书读读少，你能告诉我，operator + override 里两个string相加，是如何实现大数加法的吗？string default operator +不是做concatenation吗？

[cnmhx]

应该是将大数字转换为字符串，进行相应的运算之后再转换回来的过程。