HEU 1031 Basic Remains

/**//**************************************
Problem: HEU 1031 Basic Remains
Time: 0.0070 s
Memory: 536 k 
Accepted Time: 2009-05-05 21:44:38
Tips: 转化成十进制，求余，在转化为X进制 
**************************************/
#include <stdio.h>
#include <string.h>
int main()
{
    int n;
    while(scanf("%d",&n),n!=0)
    {
        char a[1009],b[15];
        scanf("%s%s",a,b);
        int lena=strlen(a),lenb=strlen(b),i;
        long long at=0,bt=0;
        for(i=0;i<lenb;i++)bt=bt*n+b[i]-'0';
        for(i=0;i<lena;i++)
        {
            at=at*n+a[i]-'0';
            if(at>=bt)at%=bt;
        }
        if(at==0)
        {
            printf("0\n");
            continue;
        }
        int c[12];
        for(i=11;at!=0;i--)
        {
            c[i]=at%n;
            at=at/n;
        }
        for(i++;i<12;i++)printf("%d",c[i]);
        printf("\n");
    }
    return 0;
}