利用数组和字符型与整形转换的笨办法 
#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
int main(){
    string s;
    cin>>s;
    char t;
    int n=(s[0]-'0')*1+(s[2]-'0')*2+(s[3]-'0')*3+(s[4]-'0')*4+(s[6]-'0')*5+(s[7]-'0')*6+(s[8]-'0')*7+(s[9]-'0')*8+(s[10]-'0')*9;
    if(n%11==10)
        t='X';
     else t=n%11+'0';
    if(t==s[12]){
        cout<<"Right";
        return 0;
    }
        else for(int i=0;i<s.size()-1;i++){
            cout<<s[i];
        }
    cout<<t;
    return 0;
}