C语言实现多项式相加算法

C语言实现多项式相加
#include
#include
typedef struct node
{
float xi;
int n;
struct node *next;
}term;
void sort(term *head)
{
term *p,*q,*s;
p=(term *)malloc(sizeof(term));
q=(term *)malloc(sizeof(term));
s=(term *)malloc(sizeof(term));
p=head;

while(p->next!=NULL)
{
s=p->next;
while(s!=NULL)
{
if(p->n>s->n)
{
q->n=p->n;
p->n=s->n;
s->n=q->n;
q->xi=p->xi;
p->xi=s->xi;
s->xi=q->xi;
}
s=s->next;
}
p=p->next;
}
}
term *polynadd(term *head1,term *head2)
{
term *p,*q,*s,*r,*head3;
float x;
p=head1->next;
q=head2->next;
head3=(term *)malloc(sizeof(term));
r=head3;
while(p!=NULL&&q!=NULL)
{
if(p->n==q->n)
{
x=p->xi+q->xi;
if(x!=0)
{
s=(term *)malloc(sizeof(term));
s->xi=x;
s->n=p->n;
r->next=s;
r=s;
}
p=p->next;
q=q->next;
}
else if(p->n>q->n)
{
s=(term *)malloc(sizeof(term));
s->n=q->n;
s->xi=q->xi;
r->next=s;
r=s;
q=q->next;
}
else
{
s=(term *)malloc(sizeof(term));
s->n=p->n;
s->xi=p->xi;
r->next=s;
r=s;
p=p->next;
}
}
while(p!=NULL)
{
s=(term *)malloc(sizeof(term));
s->n=p->n;
s->xi=p->xi;
r->next=s;
r=s;
p=p->next;
}
while(q!=NULL)
{
s=(term *)malloc(sizeof(term));
s->n=q->n;
s->xi=q->xi;
r->next=s;
r=s;
q=q->next;
}
r->next=NULL;
return head3;

}
term *createpolyn(int m)
{

term *p,*head,*q;
int i;
head=(term *)malloc(sizeof(term));
q=head;
for(i=0;i{
p=(term *)malloc(sizeof(term));
printf("请输入第%d项数的系数和指数\n",i+1);
scanf("%f%d",&p->xi,&p->n);
q->next=p;
q=p;
}
p->next=NULL;
return head;

}
dayin(term *head)
{
term *p,*q;
q=head->next;
if(q->xi==0)
printf(" ");
if(q->n==0&&q->xi>0&&q->xi!=1)
printf("%0.2f",q->xi);
if(q->n==0&&q->xi==1)
printf("1");
if(q->n==0&&q->xi<0)
printf("%0.2f",q->xi);
if(q->n==1&&q->xi==1)
printf("x");
if(q->n==1&&q->xi!=1&&q->xi>0)
printf("%0.2fx",q->xi);
if(q->n==1&&q->xi!=1&&q->xi<0)
printf("%0.2fx",q->xi);
if(q->n!=1&&q->n!=0&&q->xi>0&&q->xi!=1)
printf("%0.2fx^%d",q->xi,q->n);
if(q->n!=1&&q->n!=0&&q->xi<0)
printf("%0.2fx^%d",q->xi,q->n);
if(q->n!=0&&q->n!=1&&q->xi==1)
printf("x^%d",q->n);
p=q->next;
while(p!=NULL)
{
if(p->xi==0)
printf(" ");
if(p->n==0&&p->xi>0&&p->xi!=1)
printf("+%0.2f",p->xi);
if(p->n==0&&p->xi==1)
printf("+1");
if(p->n==0&&p->xi<0)
printf("%0.2f",p->xi);
if(p->n==1&&p->xi==1)
printf("+x");
if(p->n==1&&p->xi!=1&&p->xi>0)
printf("+%0.2fx",p->xi);
if(p->n==1&&p->xi!=1&&p->xi<0)
printf("%0.2fx",p->xi);
if(p->n!=1&&p->n!=0&&p->xi>0&&p->xi!=1)

printf("+%0.2fx^%d",p->xi,p->n);
if(p->n!=1&&p->n!=0&&p->xi<0)
printf("%0.2fx^%d",p->xi,p->n);
if(p->n!=0&&p->n!=1&&p->xi==1)
printf("+x^%d",p->n);
p=p->next;
}
printf("\n");
}
main()
{
int i,j;
term *head1,*head2,*head3;
printf("请输入第一个多项式的项数：\n");
scanf("%d",&i);
head1=createpolyn(i);
sort(head1);
printf("\n");
dayin(head1);
printf("\n");
printf("请输入第二个多项式的项数：\n");
scanf("%d",&j);
head2=createpolyn(j);
sort(head2);
printf("\n");
dayin(head2);
printf("\n\n");
head3=polynadd(head1,head2);
sort(head3);
printf("\n");
dayin(head3);
printf("\n");
}