SPOJ #429 Simple Numbers Conversion

This is simply a human work simulation - exactly reproducing how you do it by hand. Nothing special. You'd better put each step on a paper to make everything clear. Reference: http://blog.csdn.net/rappy/article/details/1737671

My code passed all 6 test cases. Yay..

// 429

#include <iostream>
#include <cstring>
#include <vector>
#include <cmath>
using namespace std;

char GetDigitChar(int val)
{
    if(val >= 0 && val <= 9)        return val + '0';
    else if(val >= 10 && val < 36)    return val - 10 + 'A';    
    return '';
}

int GetDigitVal(char c)
{
    c = toupper(c);
    if( c >= '0' && c <= '9')        return c - '0';
    else if(c >= 'A' && c <= 'Z')    return c - 'A' + 10;
    return 0;
}

void num_conv(string &str, int r, int s)
{
    int strLen = str.length();
    
    //    Convert string to val first
    vector<int> origVal;
    origVal.reserve(strLen);
    for(int i = 0; i < strLen; i ++)
    {
        origVal.push_back(GetDigitVal(str[i]));
    }
    
    //    Go
    vector<char> ret;    
    
    int currStartInx = 0;    
    bool bAllDone = false;
    while(currStartInx < strLen && !bAllDone)
    {
//        cout << "Start from " << currStartInx << endl;
        for(int i = currStartInx; i < strLen;i++)    
        {            
//            cout << "	 Curr Digit: " << origVal[i] << endl;
            int quo =  origVal[i] / s;
            int rem =  origVal[i] % s;            
//            cout << "	 Quo: " << quo << " Rem: " << rem << endl;
            
            origVal[i] = quo;            
            //    The digit to record
            if(i == strLen - 1)
            {
                ret.push_back(GetDigitChar(rem));
//                cout << "!" << GetDigitChar(rem) << endl;
                bAllDone = (currStartInx == (strLen - 1) && quo == 0 )? true: false;
                break;
            }
            
            //    Add remainer to next digit
            if(rem > 0)
            {    
//                cout << "	Adding rem "    << r * rem << " = ";
                origVal[i+1] += r * rem;            
//                cout << origVal[i+1] << endl;                
            }            
            
            if(i == currStartInx)
            {
                currStartInx += quo == 0? 1 : 0;
            }
        }// for                    
    }// while
    
    //    Output
    for(int i = ret.size() - 1; i >=0; i --)            
    {
        cout << ret[i];
    }
    cout << endl;
}

int main() 
{
    int runcnt = 0;
    cin >> runcnt;
    while(runcnt--)
    {
        string strnum;
        int r, s;
        cin >> strnum >> r >> s;
        num_conv(strnum, r, s);
    }
    return 0;
}