多项式相加（C语言）

/*多项式相加*/
#include  <stdio.h> 
#include  <malloc.h> 
#define NULL 0 
struct poly                                       //设置结构体 
{ 
      int xi; 
      int zhi; 
      struct poly *next; 
}; 

struct poly *jianli(void)                         //建立链表 
{ 
      struct poly *p1, *head1, *p2; 
      p1 = ( struct poly* )malloc( sizeof(struct poly ) );
      printf( "请输入两个整数:/n" );
      scanf( "%d %d",&p1->xi,&p1->zhi ); 
      head1 = p1; 
      while( p1->xi != 0)  
      {   
          p2 = ( struct poly* )malloc( sizeof(struct poly) );
          printf( "input two integers:/n" ); 
          scanf("%d %d",&p2->xi,&p2->zhi); 
          p1->next=p2; 
          p1=p2; 
      } 
      p1->next = NULL;
      return head1; 
}

int list(struct poly *head)                             //求链表长度
{
     struct poly *p;
     int L=0;
     p=head;
     while(p->next!=NULL)
     {
          p=p->next;
          L++;
      }
      return(L);
}

struct poly* paixu(struct poly *head)                   //对链表进行排序冒泡法
{
     struct poly *p,*r,*q;
     int i,j,L;
     L=list(head);
     r=p=q=head;
     for(i=0;i<L;i++)
     {
          r=p=q=head;
          for(j=0;j<(L-i);j++)
          {
              p=r->next;
              if(p->zhi>r->zhi)
              {
                  if(r==head)
                 {
                      head=p;
                      r->next=p->next;
                      head->next=r;
                      p=r;
                      r=head;
                  }
                  else
                  {
                      q->next=p;
                       r->next=p->next;
                       p->next=r;
                       p=r;
                       r=q->next;
                  }
              }
             q=r;
              r=p;
          }
     }
  return(head);
}

struct poly* jisuan(struct poly *head1,struct poly *head2 )                  //多项式的相加
{
     struct poly *p1,*p2,*r1,*r2;
     p1=r1=head1;
     p2=r2=head2;
     while((p1->next!=0)||(p2->next!=0))                         
     {    
          if(p1->zhi==p2->zhi)                                     
          {
               p1->xi=(p1->xi)+(p2->xi); 
               r1=p1; 
               p2=p2->next; 
               free(r2); 
               r2=p2; 
               p1=p1->next;  
           }
           else if(p1->zhi<p2->zhi)                                
           {
               if(head1==p1)
               {
                   r2=r2->next;
                   p2->next=p1;
                   head1=p2;
                   r1=p2;
                   p2=r2;
               }
               else
               {                
                    r2=r2->next;
                    r1->next=p2;
                    p2->next=p1;
                    r1=p2;
                    p2=r2;
                } 
            }
            else if(p1->zhi>p2->zhi)
            {   
                if(p1->next!=NULL)
                {
                    r1=p1; 
                    p1=p1->next;
                 }
                 else
                 if(p1->next==NULL)
                 {
                      r1=p1;
                      p1->next=p2;
                      r1->next=p2;
                  }
              } 
        }
      return head1;
}

void print(struct poly* head)                             //输出函数
{

    struct poly* p;
    p=head;
    while(p->next!=NULL)
    {
        printf("%d %d   ",p->xi,p->zhi);
        p=p->next;
        free(head);
        head=p;
     }
     free(head);
}

int main() 
{ 
     struct poly *head1,*head_one,*head2,*head_two,*p;
     printf("请输入多项式的每一项的系数与指数并以0 0为结束标志/n"); 
     head1=jianli();
     head_one=paixu(head1);

     printf("/n请输入另一个多项式的每一项的系数与指数并以0 0为结束标志/n"); 
     head2=jianli();
     head_two=paixu(head2);

     printf("相加后的多项式是:/n");
     p=jisuan(head_one,head_two); 
     print(p);
     return 0;
}