C语言求救求救啊！~~~

[CloudStrifers]

通过键盘输入一组多项式的系数和指数，用尾插法建立一元多项式的链表；
系统用COEF,指数用EXP，指针是NEXT;
方法如下，为什么运行不起来？我用F10调试时候在进入子函数就运行不起了，这怎么回事？
#include <stdio.h>
#include <malloc.h>
typedef struct seqlist
{
int coef;
int exp;
seqlist *next;
}seqlist;
int pai(seqlist *no);
int main()
{
seqlist *head;
head=(seqlist *)malloc(sizeof(seqlist));
pai(head);
return 0;
}
int pai(seqlist *no)
{
int flag=1,a,b;
seqlist *p,*q;
q=no;
while(flag)
{
scanf("%d,%d",&a,&b);
if(a!=0)
{
p->coef=a;
p->exp=b;
q->next=p;
q=p;
p=p->next;
}
else
{
flag=0;
p->next=NULL;
}
}
return 0;
}

[尘缘udbwcso]

#include <stdio.h>
#include <malloc.h>
typedef struct seqlist
{
	int coef;
	int exp;
	struct seqlist *next;
}seqlist;
/*
*功能:输入多项式的系数和指数
*返回值:多项式的首地址
*
*/
seqlist *inputMult()
{
	seqlist *head, *p1, *p2;
	head = (seqlist*)malloc(sizeof(seqlist));
	p1 = head;	
	printf("请输入多项式(当系数和指数均为0时结束):\n");	
	while(1)
	{
		printf("系数: ");
		scanf("%d", &p1->coef);
		printf("指数: ");
		scanf("%d", &p1->exp);
		if(p1->coef == 0 && p1->exp == 0)
			break;
		p2 = p1;
		p1 = (seqlist*)malloc(sizeof(seqlist));
		p2->next = p1;
	}
	p2->next = NULL;
	free(p1);
	return head;
}
/*
*功能:输出多项式
*参数:多项式的首地址
*
*/
void display(seqlist *head)
{
	while(head != NULL)
	{
		printf("%dX^%d", head->coef, head->exp);
		if(head->next != NULL)
			printf(" + ");
		head = head->next;
	}
	printf("\n");
}

int main()
{
	seqlist *head;
	head = inputMult();
	display(head);
	return 0;
}

[高性能架构探索]

head=(seqlist *)malloc(sizeof(seqlist));

head->next = NULL;



if(a!=0)
{
p->coef=a;
p->exp=b;
q->next=p;
q=p;
p=p->next;
}

p都没分配空间，你就直接操作其成员了
建议
p = (seqlist *)malloc(sizeof(seqlist));
p->next = NULL;