HDU 2031 进制转换

这个题是数据结构教材上的一个经典例题，有两种方法，即数组或链表，这里我采用链表做的

#include <stdio.h>
#include <stdlib.h>
typedef char Elemtype;
typedef struct Lnode
{
        Elemtype e;
        struct Lnode *next;
}Ln,*Link;
void change ( Link L,int n,int r )
{
     Link s;
     L -> next = NULL;//必须的，否则上一次的数据下一次还保留着 
     if ( n < 0 )
       {
             L -> e = 0;
             n = -n;
             }
     else 
        L -> e = 1;
     while ( n )
     {
           s = ( Link ) malloc ( sizeof ( Ln ) );
           s -> next = L -> next;
           L -> next = s;
           s -> e = n % r ;
           n = n / r;
           switch ( s -> e )
           {
                  case 10:
                          s -> e = 'A';
                          break;
                  case 11:
                          s -> e = 'B';
                          break;
                  case 12:
                          s -> e = 'C';
                          break;
                  case 13:
                          s -> e = 'D';
                          break;
                  case 14:
                          s -> e = 'E';
                          break;
                  case 15:
                          s -> e = 'F';
                          break;
                  default:
                          s -> e += '0';
                  }
           }
 }
void print ( Link L )
{
     Link p = L -> next;
     if ( !L -> e )
        printf ( "-" );
     while ( p )
     {
           printf ( "%c",p -> e );
           p = p -> next;
           }
     puts ( "" );
 }
int main ( )
{
    int N,R;
    while ( scanf ( "%d%d",&N,&R ) == 2 )
    {
          Link L;
          L = ( Link ) malloc ( sizeof ( Ln ) );
          change ( L,N,R );
          print ( L );
          free ( L );
          }
    return 0;
}

下面是递归实现的算法

#include<stdio.h>
#include<string.h>
int n,num[15],r,count;
void print( int x )
{
     if( x <= 9 )
         printf( "%d",x );
     else
         printf( "%c",x - 10 + 'A' );
 }
void change( int n,int r )
{
     if( n )
     {
         change( n / r, r );
         print( n % r );
     }
 }
int main( )
{
    while( scanf( "%d%d",&n,&r ) != EOF )
    {
           if( n < 0 )
               printf( "-" ),n = -n;
           change( n,r ),puts( "" );
           }
    return 0;
}