A29695.高精度B进制加法

#include<iostream>
#include<cstring>
using namespace std;
string jzadd(string A,string B,int jz){
	string C="";
	int a[200],b[200],sum[201],c[201],lena=A.length(),lenb=B.length(),i=0,max=(lena>lenb?lena:lenb),x=0,digs=max;
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	memset(sum,0,sizeof(c));
	for(;i<max;i++){
		if(i<lena){ 
            if(A[lena-i-1]>='0'&&A[lena-i-1]<='9'){
            	a[i]=A[lena-i-1]-'0';
            }else{
                a[i]=10+A[lena-i-1]-'A';
            }
        }
		if(i<lenb){ 
            if(B[lenb-i-1]>='0'&&B[lenb-i-1]<='9'){
            	b[i]=B[lenb-i-1]-'0';
            }else{
                b[i]=10+B[lenb-i-1]-'A';
            }
        }
		sum[i]=a[i]+b[i]+x;
		x=sum[i]/jz;
		sum[i]%=jz;
	}
	digs+=x;
	if(x){
		sum[i]=a[i]+b[i]+x;
		x=sum[i]/jz;
		sum[i]%=jz;
	}
	for(int i=0;i<digs;i++){
		c[i]=sum[digs-i-1];
        if(c[i]>=0&&c[i]<=9){
            C+=c[i]+'0';
		}else{
            C+=c[i]-10+'A';
        }
	}
	return C;
}
int main(){
	int jz;
    string A,B;
    cin>>jz>>A>>B;
    cout<<jzadd(A,B,jz);
	return 0;
}