题解 | #高精度整数加法#

a[i]+b[j]+up-'0'；up=0;

a[i]+b[j]+up-'0'-10;up=1;

/*HJ57高精度整数加法*/
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    string a;
    string b;
    string c="";
    int up=0;
    cin>>a>>b;
    int i=a.length()-1;
    int j=b.length()-1;   
    while(i+1&&j+1)
    {
        if(a[i]+b[j]+up-'0'-'0'<=9)
        {
            c+=a[i]+b[j]+up-'0';
            up=0;
        }
        else if(a[i]+b[j]+up-'0'-'0'>=10)
        {
            c+=a[i]+b[j]+up-'0'-10;
            up=1;
            if(i==0&&j==0) c+='1';            
        }
        i--;
        j--;
    }
    while(i>=0)//处理长数据
    {
        if(a[i]+up-'0'<=9)
        {
            c+=a[i]+up;
            up=0;
        }
        else if(a[i]+up-'0'>=10)
        {
            c+=a[i]+up-10;
            up=1;
        }
        
        i--;
    }
    while(j>=0)
    {
        if(b[j]+up-'0'<=9)
        {
            c+=b[j]+up;
            up=0;
        }
        else if(b[j]+up-'0'>=10)
        {
            c+=b[j]+up-10;
            up=1;
        }

        j--;
        
    }
    reverse(c.begin(),c.end());
    cout<<c<<endl;
}


///*HJ57高精度整数加法*/
//#include<iostream>
//using namespace std;
//int main()
//{
//    string a;
//    string b;
//    long na=0,nb=0;
//    cin>>a>>b;
//    for(int i=0;i<a.length();i++)
//    {
//        na=na*10+a[i]-'0';
//    }
//    for(int i=0;i<b.length();i++)
//    {
//        nb=nb*10+b[i]-'0';
//    }
//    cout<<na<<endl;
//}//超大数据会GG