任意进制的任意位数字转换

直接从m进制转换成n进制，以防用例有大数所以用字符串处理。编写字符串除法和字符串求余实现。。。

    #include <iostream>
    #include <cstdio>
    #include <stack>
    #include <string>
    #include <cstring>
    #include <vector>
    #include <algorithm>
    using namespace std;
int getvalue(char c){
    if(c>='0' && c<='9'){
        return c-'0';
    }
    else{
        return c-'A'+10;
    }
}

char getcharacter(int n){
if(n<10)
    return '0'+n;
   else
    return 'A'+n-10;
}

char qiuyu(string s,int n,int m){//在字符串除法的基础上很容易得到求余函数
    int yushu=0;
    int current;
    for(int i=0;i<s.size();i++){
        current=yushu*m+getvalue(s[i]);
        s[i]=getcharacter(current/n);
        yushu=current%n;
    }
   if(yushu<10)
    return '0'+yushu;
   else
    return 'a'+yushu-10;
}



string divide(string s,int n,int m){//字符串除法
    int yushu=0;
    int current;
    for(int i=0;i<s.size();i++){
        current=yushu*m+getvalue(s[i]);
        s[i]=getcharacter(current/n);
        yushu=current%n;
    }
    while(s[0]=='0' && s.size()>1)
        s=s.substr(1,s.size()-1);
        return s;
}

 int main(){
   int m,n;
   string s;
   vector<char> nnum;
   while(cin>>m>>n){
    cin>>s;
    while(s!="0"){
        nnum.push_back(qiuyu(s,n,m));
        s=divide(s,n,m);
    }
    for(int i=nnum.size()-1;i>=0;i--)
        cout<<nnum[i];
        cout<<endl;
        nnum.clear();
   }
    return 0;
}