数据结构问题
编写将两个多项式相加的函数.不要毁坏输入数据,用一个链表实现.
参考答案:#include<stdio.h>
#include<malloc.h>
main()
{
int i=0;
struct xianshu{
int x;
int p;
struct xianshu *next;
}xianshu;
struct xianshu *p1,*p2,*p3,*head1,*head2,*head3;
head1=p1=(struct xianshu *)malloc(sizeof(struct xianshu));
printf("Input x:" );
scanf("%d",&p1->x);
printf("Input p:");
scanf("%d",&p1->p);
while(p1->x!=0){
p2=p1;
p1=(struct xianshu *)malloc(sizeof(struct xianshu));
p2->next=p1;
printf("Input x:" );
scanf("%d",&p1->x);
printf("Input p:");
scanf("%d",&p1->p);
}
p2->next=NULL;
printf("\n");
head2=p1=(struct xianshu *)malloc(sizeof(struct xianshu));
printf("Input x:" );
scanf("%d",&p1->x);
printf("Input p:");
scanf("%d",&p1->p);
while(p1->x!=0){
p2=p1;
p1=(struct xianshu *)malloc(sizeof(struct xianshu));
p2->next=p1;
printf("Input x:" );
scanf("%d",&p1->x);
printf("Input p:");
scanf("%d",&p1->p);
}
p2->next=NULL;
printf("\n");
head3=p3=(struct xianshu *)malloc(sizeof(struct xianshu));
while((head1->x!=0)||((head2->x!=0))){
p2=(struct xianshu *)malloc(sizeof(struct xianshu));
if(head1->x==head2->x){
p3->x=head1->x;
p3->p=(head1->p)+(head2->p);
p1=head1->next;
head1=p1;
p1=head2->next;
head2=p1;
printf("yi%d\n",p3->x);
p3->next=p2;
p3=p2;
}
else if(head1->x==0){
p3->x=head2->x;
p3->p=head2->p;
p1=head2->next;
head2=p1; printf("san%d\n",p3->x);
p3->next=p2;
p3=p2;
}
else if(head2->x==0){
p3->x=head1->x;
p3->p=head1->p;
p1=head1->next;
head1=p1; printf("er%d\n",p3->x);
p3->next=p2;
p3=p2;
}
else if(head1->x<head2->x){
p3->x=head1->x;
p3->p=head1->p;
p1=head1->next;
head1=p1; printf("er%d\n",p3->x);
p3->next=p2;
p3=p2;
}
else if(head1->x>head2->x){
p3->x=head2->x;
p3->p=head2->p;
p1=head2->next;
head2=p1; printf("san%d\n",p3->x);
p3->next=p2;
p3=p2;
}
}
p3->next=NULL;
while(head3->next!=NULL){
printf("x:");
printf("%d",head3->x);
printf(" ");
printf("p%d:",i++);
printf("%d:",head3->p);
printf("\n");
p1=head3;
head3=p1->next;
}
getch();
}
x就是x的指数而p，q就是项数