自己编的程序，链接两个链表。但只输出第一个链表。请你们个看看解决方案

自己编的程序，链接两个链表。但只输出第一个链表。请你们个看看
此程序分3个函数，一个是申请地址creat，一个是链接连个链表join，一个是输出output.输入输出结果正如下。只输出第一个链表，不知道哪里错了。我想很可能是join()的错。

#include <stdio.h>
#include <malloc.h>
#define   LEN   sizeof(struct   student)


struct   student
{
int   num;
int   score;
struct   student   *next;
};
int   n=0;


struct   student   *creat(void)
{
struct   student   *head,*p1,*p2;
p1=p2=(struct   student   *)malloc(LEN);
printf( "input   num,score\n ");
scanf( "%d,%d ",&p1-> num,&p1-> score);
head=NULL;
while(p1-> num!=0)
{n=n+1;
if(n==1)
    head=p1;
else
    p2-> next=p1;
p2=p1;
p1=(struct   student   *)malloc(LEN);
printf( "input   num,score\n ");
scanf( "%d,%d ",&p1-> num,&p1-> score);
}
p2-> next=NULL;
return(head);
}


struct   student   *join(struct   student   *p1,struct   student   *p2)/*把p2连接到p1的尾部*/
{
struct   student   *s1;
for(s1=p1;s1-> next!=NULL;s1=s1-> next)
;
s1-> next=p2;
return(p1);
}


void   output(struct   student   *p)
{
struct   student   *cur;
for(cur=p;cur!=NULL;cur=cur-> next)
printf( "No.%d   score   %d\n ",cur-> num,cur-> score);
}

int   main()
{
struct   student   *lista,*listb,*list;
printf( "the   one   list\n ");
lista=creat();
printf( "the   other   list\n ");
listb=creat();
list=join(lista,listb);
printf( "\nthe   join   list   is:\n ");
output(list);
}



输入：
the   one   list
input   num,score
1,97
input   num,score
2,45
input   num,score
3,89
input   num,score
0,0
the   other   list
input   num,score
5,69
input   num,score
6,93
input   num,score
0,0

输出：
the   join   list   is:
No.1   score   97
No.2   score   45
No.3   score   89
Press   any   key   to   continue



[解决办法]
#include <stdio.h>
#include <malloc.h>
#define LEN sizeof(struct student)


struct student
{
int num;
int score;
struct student *next;
};


struct student *creat(void)
{
struct student *head,*p1,*p2;
p1=p2=(struct student *)malloc(LEN);
printf( "input num,score\n ");
scanf( "%d,%d ",&p1-> num,&p1-> score);
head=NULL;
int n=0;
while(p1-> num!=0)
{
n=n+1;
if(n==1)
head=p1;
else
p2-> next=p1;

p2=p1;
p1=(struct student *)malloc(LEN);
printf( "input num,score\n ");
scanf( "%d,%d ",&p1-> num,&p1-> score);
}
p2-> next=NULL;
return(head);
}


struct student *join(struct student *p1,struct student *p2)/*把p2连接到p1的尾部*/
{
struct student *s1;
for(s1=p1;s1-> next!=NULL;s1=s1-> next)
;
s1-> next=p2;
return(p1);

}


void output(struct student *p)
{
struct student *cur;
for(cur=p;cur!=NULL;cur=cur-> next)
printf( "No.%d score %d\n ",cur-> num,cur-> score);
}

int main()
{
struct student *lista,*listb,*listx;
printf( "the one list\n ");
lista = creat();
printf( "the other list\n ");
listb = creat();
listx = join(lista,listb);
printf( "\nthe join list is:\n ");
output(listx);
}


[解决办法]
#include <stdio.h>
#include <malloc.h>

struct student
{
int num;
int score;
struct student *next;
};

#define LEN sizeof(struct student)

struct student *creat(void)
{
struct student *head,*p1,*p2;
char end = 'e ';
int n=0; //注意啊
p1=p2=(struct student *)malloc(LEN);
printf( "input num,score\n ");
head=NULL;
while(scanf( "%d,%d ",&p1-> num,&p1-> score))//建议这样写
{
n=n+1;
if(n==1)
head=p1;
else
p2-> next=p1;
p2=p1;
p1=(struct student *)malloc(LEN);
printf( "input num,score\n ");
}
p2-> next=NULL;
scanf( "%c ",&end);
return(head);
}


struct student *join(struct student *p1,struct student *p2)/*°&Ntilde;p2&Aacute;&not;&frac12;&Oacute;&micro;&frac12;p1&micro;&Auml;&Icirc;&sup2;&sup2;&iquest;*/
{
struct student *s1;
for(s1=p1;s1-> next!=NULL;s1=s1-> next)
;
s1-> next=p2;
return(p1);
}


void output(struct student *p)
{
struct student *cur;
for(cur=p;cur!=NULL;cur=cur-> next)
printf( "No.%d score %d\n ",cur-> num,cur-> score);
}

void main()
{
struct student *lista,*listb,*list;
lista = listb = list = NULL;
printf( "the one list\n ");
lista=creat();
printf( "the other list\n ");
listb=creat();
list=join(lista,listb);
printf( "\nthe join list is:\n ");
output(list);
}

[解决办法]
cur=p应写到第一个for循环内部