高精度减法

#include<iostream>
using namespace std;
string a,b,str;
int main()
{
    cin>>a;
    cin>>b;
    int sign = 0;
    if(a==b)
    {
        cout<<"0"<<endl;
        return 0;
    }
    if((a < b && a.size()==b.size()) || (a.size()<b.size()))
    {
        sign = 1;
        swap(a,b);
    }
    int len1 = a.length();
    int len2 = b.length();
    int tep = len1 - len2;
    int cf = 0;
    for(int i=len2-1;i>=0;i--)
    {
        if(a[i+tep]<b[i]+cf)
        {
            str = char(a[i+tep]-b[i]-cf+'0'+10) + str;
            cf = 1;
        }else{
            str = char(a[i+tep]-b[i]-cf+'0')+str;
            cf = 0;
        }
    }
    for(int i =tep-1;i>=0;i--)
    {
        if(a[i]-cf>='0')
        {
            str = char(a[i]-cf)+str;
            cf = 0;
        }else{
            str = char(a[i]-cf+10)+str;
            cf = 1;
        }
    }
    str.erase(0,str.find_first_not_of('0'));
    if(sign == 1) cout<<"-";
    cout<<str<<endl; 
} 

此高精度只能计算输入的两个整数都为正数。