用链表实现两大数相乘

#include<iostream>
#include<string>
#include<list>
using namespace std;

//大整数相加 
list<char> long_add(list<char> num1,list<char>num2);
//大整数相减 
list<char> long_sub(list<char>num1,list<char>num2);
//大整数相乘
list<char> long_mutliply(list<char>num1,list<char>num2);
//输出结果
void print(list<char> rs);

int main(void)
{
string str1,str2;
cout<<"Please input the value of a&& b"<<endl;
cout<<" a>     ";
cin>>str1;
cout<<" b>     ";
cin>>str2;
cout<<"input the operation"<<endl;

cout<<"operation: >";
char ch;
cin>>ch;

//存放操作数和结果
list<char>num1,num2;
list<char> rs;
for(int i=0;i<str1.size();i++)
{
     num1.push_back(str1[i]);
}
for( i=0;i<str2.size();i++)
{
     num2.push_back(str2[i]);
}
switch(ch){
//test the add function
case '+':
     rs=long_add(num1,num2);
     break;
case '-':
     rs=long_sub(num1,num2);
     break;
case'*':
     rs=long_mutliply(num1,num2);
     break;
}
print(rs);
getchar();
getchar();
return 0;
}

//the definition of     long_add
list<char> long_add(list<char> num1,list<char>num2)
{
list<char> rs;

//两个数符号不同
if((num1.front()=='-')&&(num2.front()!='-'))
{
     num1.erase(num1.begin());
     rs=long_sub(num1,num2);
     return rs;
}
if((num1.front())!='-'&&(num2.front()=='-'))
{
     num2.erase(num2.begin());
     rs=long_sub(num2,num1);
     return rs;
}

//两个数均为负数
if((num1.front()=='-')&&(num2.front()=='-'))
{
     num1.erase(num1.begin());
     num2.erase(num2.begin());
     rs=long_add(num1,num2);
     rs.push_front('-');
     return rs;
}

//两个数均为正数 
if((num1.front())!='-'&&(num2.front())!='-')
{
     int size1=num1.size();
     int size2=num2.size();
     //补全位数并确定size值
     if(size1<size2)
     {
       for(int i=0;i<(size2-size1);i++)
       num1.push_front('0');
     }
     if(size2<size1)
     {
         for(int i=0;i<(size1-size2);i++)
       num2.push_front('0');
     }
     int c_high=0;//表示进位
     list<char>::iterator iter1=num1.end();
     iter1--;
     list<char>::iterator iter2=num2.end();
     iter2--;
     while(iter1!=num1.begin()&&iter2!=num2.begin())
     {
      int m=(*iter1)-'0';
      int n=(*iter2)-'0';
      int temp=m+n+c_high;
      rs.push_front(temp%10+'0');
      c_high=temp/10;
      iter1--;
      iter2--;
     }
     //最高位的计算
        int m=(*iter1)-'0';
      int n=(*iter2)-'0';
      int temp=m+n+c_high;
      rs.push_front(temp%10+'0');
      c_high=temp/10;
   
      //判断最高位是否有进位
      if(c_high!=0)
      {
       rs.push_front(c_high+'0');
      }
      return rs;
}
}

//the definition of long_sub
list<char> long_sub(list<char> num1,list<char> num2)
{
list<char> rs;//存储结果

//两个数符号相异
if((num1.front())=='-'&&(num2.front())!='-')
{
     num1.erase(num1.begin());
     rs=long_add(num1,num2);
     return rs;
}
if((num1.front())!='-'&&(num2.front())=='-')
{
     num2.erase(num2.begin());
     rs=long_add(num1,num2);
     rs.push_front('-');
     return rs;
}

//两个数均为负数
if((num1.front())=='-'&&(num2.front())=='-')
{
     num1.erase(num1.begin());
     num2.erase(num2.begin());
     rs=long_sub(num2,num1);
     return rs;
}

//两个数均为正数
if((num1.front())!='-'&&(num2.front())!='-')
{
     int size1=num1.size();
     int size2=num2.size();
     int flag=0;//表示结果的符号 flag=0     为正数，flag=1 为负数
     if(size1<size2)
     {
      flag=0;
      for(int i=0;i<(size2-size1);i++)
       num1.push_front('0');
     }
     if(size2<size1)
     {
      flag=1;
      for(int i=0;i<(size1-size2);i++)
       num2.push_front('0');
     }
     if(size1==size2)
     {
      list<char>::iterator iter1=num1.begin();
      list<char>::iterator iter2=num2.begin();
      while((*iter1)==(*iter2)&&iter1!=num1.end())
      {
       iter1++;
       iter2++;
      }
      if(iter1==num1.end())
      {
       rs.push_front('0');
       return rs;
      }
      else
       if((*iter1)>(*iter2))
        flag=1;
       else
        flag=0;
     }
  
     //结果为负
     if(flag==1)
     {
      rs=long_sub(num2,num1);
      rs.push_front('-');
      return rs;
     }
     if(flag==0)
     {
   
      list<char>::iterator iter1=num1.end();
      iter1--;
      list<char>::iterator iter2=num2.end();
      iter2--;
      int c_high=0;//表示借位 c_high=0不借位 c_high=1 借位
      while(iter1!=num1.begin()&&iter2!=num2.begin())
      {
       int m=(*iter1)-'0';
       int n=(*iter2)-'0';
       int temp=n-(m+c_high);
       if(temp<0)
       {
        c_high=1;
        temp=temp+10;
        rs.push_front(temp+'0');
       }
       else
       {
        rs.push_front(temp+'0');
        c_high=0;
       }
       iter1--;
       iter2--;
      }
         //处理最高位的减法，同时根据判断，借位为0
       int m=(*iter1)-'0';
       int n=(*iter2)-'0';
       int temp=n-(m+c_high);
       rs.push_front(temp+'0');  
       return rs;
     }
}
}

list<char> long_mutliply(list<char> num1,list<char>num2)
{
list<char> rs;//存储结果

//如果两数符号相异，结果为负
if(num1.front()=='-'&&num2.front()!='-')
{
     num1.erase(num1.begin());
     rs=long_mutliply(num1,num2);
     rs.push_front('-');
     return rs;
}
if(num2.front()=='-'&&num1.front()!='-')
{
     num2.erase(num2.begin());
     rs=long_mutliply(num1,num2);
     rs.push_front('-');
     return rs;
}

//两数均为负数
if(num1.front()=='-'&&num2.front()=='-')
{
     num1.erase(num1.begin());
     num2.erase(num2.begin());
     rs=long_mutliply(num1,num2);
     return rs;
}

//两数均为正数
if(num1.front()!='-'&&num2.front()!='-')
{
  
     int size1=num1.size();
     int size2=num2.size();
     //只剩下一位
     if(size1==1&&size2==1)
     {
      int m=(num1.front())-'0';
      int n=(num2.front())-'0';
      int temp=m*n;
      int high=temp/10;
      int low=temp%10;
      if(high!=0)
      rs.push_back(high+'0');
      rs.push_back(low+'0');
      return rs;
     }

     if(size1==1&&size2!=1)
     {
      list<char> num2_high;//num2的高位部分
      list<char> num2_low;//num2的低位部分
      list<char>::iterator iter2;
      int i;
      for(i=0,iter2=num2.begin();i<size2/2;i++,iter2++)
      {
       num2_high.push_back(*iter2);
      }
      for(i=size2/2;i<size2;i++,iter2++)
      {
       num2_low.push_back(*iter2);
      }
      list<char> rs_high;//高位的积
      list<char> rs_low;//低位的积
      rs_high=long_mutliply(num1,num2_high);
      rs_low=long_mutliply(num1,num2_low);
      //高位移位
      for( i=0;i<num2_low.size();i++)
       rs_high.push_back('0');
      rs=long_add(rs_high,rs_low);
      return rs;
     }

     if(size1!=1&&size2==1)
      return long_mutliply(num2,num1);//根据乘法交换律 
//********************************************************************************************
//********************************************************************************************
//********************************************************************************************
     if(size1!=1&&size2!=1)
     {
      list<char> num1_high;
      list<char> num1_low;
      list<char> num2_high;
      list<char> num2_low;
      list<char>::iterator iter1;
      list<char>::iterator iter2;
      int i;
      //分别找出num1和num2的低位和高位
      for(i=0,iter1=num1.begin();i<size1/2;i++,iter1++)
      {
       num1_high.push_back(*iter1);
      }
      for(;iter1!=num1.end();iter1++)
      {
       num1_low.push_back(*iter1);
      }
      for(i=0,iter2=num2.begin();i<size2/2;i++,iter2++)
      {
       num2_high.push_back(*iter2);
      }
   
      for(;iter2!=num2.end();iter2++)
      {
       num2_low.push_back(*iter2);
      }
      list<char> rs_high;//存储高位乘积
      list<char> rs_low;//存储低位乘积
      list<char> rs_high_low;
      list<char> rs_low_high;
      rs_high=long_mutliply(num1_high,num2_high);
      rs_low=long_mutliply(num1_low,num2_low);
   
      //num1和num2的高位进行移位
      for( i=0;i<num1_low.size();i++)
      {
       num1_high.push_back('0');
      }
      for( i=0;i<num2_low.size();i++)
      {
       num2_high.push_back('0');
      }
      rs_high_low=long_mutliply(num1_high,num2_low);
      rs_low_high=long_mutliply(num1_low,num2_high);
      //高位乘积移位
      for( i=0;i<num1_low.size()+num2_low.size();i++)
       rs_high.push_back('0');
   
      //把四项结果进行相加
              rs=long_add(rs_high,rs_low);
        rs=long_add(rs,rs_high_low);
        rs=long_add(rs,rs_low_high);
      return rs;
     }
        return rs;
}
return rs;
}


void print(list<char> rs)
{
list<char>::iterator iter=rs.begin();
cout<<(*iter);
iter++;

//如果结果是0 ，则输出0值
while((*iter)=='0'&&iter!=rs.end())
{
     iter++;
}
if(iter==rs.end())
     cout<<0;
else
{
     while(iter!=rs.end())
     {
      cout<<(*iter);
         iter++;
     }
}
}
 

/***********描述:实现两个大整数的加、减、乘的运算，其中乘法运算
*********************利用分治的方法解决：假设X*Y可以分解为
********************* (X1*(10^n/2)+X2)*(Y1*(10^n/2)+Y2)=******************
********************X1*Y1*10^n+((X1-X2)(Y2-Y1)+X1*Y1+X2*Y2)*10^(n/2)+X2*Y2
********************此算法降低了乘法的次数***********************/