以下是引用轻风无言在2013-1-6 01:31:51的发言：

#include
#include
 
typedef struct student
 {
     int num;
     char name[20];
     struct student *next;
 }Code;
 
Code *Creat(int length);
Code *Delete(Code *ahead,Code *bhead);
void Print(Code *head);

void main()
 {
     
     Code *ahead,*bhead;
     int a_length,b_length;
     printf("input length of list_a:\n");
     scanf("%d",&a_length);
     ahead=Creat(a_length);
     Print(ahead);
     printf("input length of list_b:\n");
     scanf("%d",&b_length);
     bhead=Creat(b_length);
     Print(bhead);
     ahead=Delete(ahead,bhead);
     Print(ahead);
 }
 
//建立链表函数
 Code *Creat(int length)
 {
     Code *p1,*p2,*head;
     int i = 1;
     printf("Now,input the students' information:\n");
      head = ( Code * )malloc( sizeof( Code ) );
        head->next = NULL;   
      p1 = p2 = head;
      while( i <= length ){
            p1=(Code *)malloc(sizeof(Code));
            scanf("%d %s",&p1->num,p1->name);
            p2->next = p1;
            p2 = p1;
            i++;
      }
      p1->next = NULL;
     return head;
 }
 
//输出链表函数
 void Print(Code *head)
 {
     Code *p;
     p=head->next;
     if( p !=NULL)
         do
         {
             printf("number:%d  name:%s\n",p->num,p->name);
             p=p->next;
         }while(p!=NULL);
 }
 
//删除相同学号函数
 Code *Delete(Code *ahead,Code *bhead)
 {
     Code *pa1,*pa2,*pb1;
     pb1=bhead->next;
 
    while(pb1!=NULL)             //让b表中的每个结点和a表比较
     {
         pa1=ahead->next;
            pa2 = ahead;
        
        while((pa1!=NULL)&&(pb1->num !=pa1->num))
         {
         
         
             pa1=pa1->next;
                 pa2 = pa2->next;
         
         }
         printf("运行\n");
            
        
        if(  pa1 != NULL)         //两链表中学号相同
         {   
              pa2->next = pa1->next;              
              free( pa1 );
              pa1 = pa1->next;
            
            
          }
            pb1 = pb1->next;
 }
     return ahead;
 }
//我刚刚在你的程序上改了下，用的带头结点的链表，运行了下可以，不知道结果符不符合要求，你试试
              pa2->next = pa1->next;              
              free( pa1 );
              pa1 = pa1->next;