C++如何使用string的大数加法运算

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
#define n 10
string dezero(string a)//用来去掉正数前面的0，也就是说可以输入000001类似这样的数字
{
 long int i;
 for(i=0;i<a.length();i++)
 {
 if(a.at(i)>48) break;
 }
 if(i==a.length()) return "0";
 a.erase(0,i);
 return a;
}
string add(string a,string b)//自然数加法
{
 a=dezero(a);
 b=dezero(b);
 string c;
 string d="0";
 long int kmin,kmax,i;
 reverse(a.begin(),a.end());
 reverse(b.begin(),b.end());
 if(a.length()>b.length()) {kmin=b.length();kmax=a.length();c=a;}
 else {kmin=a.length();kmax=b.length();c=b;}
 c.insert(c.length(),d);
 for(i=0;i<kmin;i++)
 {
 if(a.at(i)>=48&&a.at(i)<=57) a.at(i)-=48;
 if(a.at(i)>=97&&a.at(i)<=122) a.at(i)-=87;
 if(b.at(i)>=48&&b.at(i)<=57) b.at(i)-=48;
 if(b.at(i)>=97&&b.at(i)<=122) b.at(i)-=87;
 c.at(i)=a.at(i)+b.at(i);
 }
 for(i=kmin;i<kmax+1;i++)
 {
 if(c.at(i)>=48&&c.at(i)<=57) c.at(i)-=48;
 if(c.at(i)>=97&&c.at(i)<=122) c.at(i)-=87;
 }
 for(i=0;i<kmax;i++)
 {
 if(c.at(i)>=n) 
 {
 c.at(i+1)+=c.at(i)/n;
 c.at(i)=c.at(i)%n;
 }
 }
 if(c.at(kmax)==0) 
 {
 c.erase(kmax,kmax+1);
 }
 for(i=0;i<c.length();i++)
 {
 if(c.at(i)>=10) c.at(i)+=87;
 if(c.at(i)<10) c.at(i)+=48;
 }
 reverse(c.begin(),c.end());
 return c;
}
int main()
{
 string a,b;
 while(cin>>a>>b)
 {
 cout<<add(a,b);
 }
 return 0;
}