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标题:多项式加法:
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我的手机哦
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已结贴  问题点数:5 回复次数:3 
多项式加法:
   
#include"stdio.h"
#include"stdlib.h"
struct node
{
  int cof,exp;
  struct node *next;
};
struct node *create()
{
  struct node *head,*p,*q;
  int i,j;
  head=(struct node *)malloc(sizeof(struct node));
  q=head;
  printf("input i,j:");
  scanf("%d%d",&i,&j);
  while(i!=-999&&j!=-999)
  {
    p=(struct node *)malloc(sizeof(struct node));
    p->cof=i;
    p->exp=j;
    q->next=p;
    q=p;
    scanf("%d%d",&i,&j);
  }
  q->next=NULL;
  return(head);
}
void outline(struct node *head)
{
   struct node *p;
   p=head->next;
   while(p!=NULL)
   {
     printf("%5d%5d",p->cof,p->exp);
     p=p->next;
 
}
void lineadd(struct node *head1,struct node *head2)
{
  struct node *p,*q,*r;
  int x;
  int  cof,exp;
  p=head1->next;
  q=head2->next;
  r=head1;
  while(p!=NULL&&q!=NULL)
  {
    if(p->exp==q->exp)
    {
     x=p->cof+q->cof;
     if(x!=0)
     {
       p->cof=x;
       r->next=p;
       r=p;
     }
       p=p->next;
       q=q->next;

    }
    else if(p->exp>q->exp)
     {
       r->next=p;
       r=p;
       p=p->next;
     }
     else
     {
      r->next=q;
      r=q;
      q=q->next;
     }
    }
    if(p==NULL)
      r->next=q;
     if(q==NULL)
      r->next=p;
}

main()
{
 struct node *head1,*head2;
 head1=(struct node *)malloc(sizeof(struct node));
 head2=(struct node *)malloc(sizeof(struct node));
 head1=create();
 head2=create();
lineadd(head1,head2);
 outline(head1);
}
高手指点我的程序有没有可能可以改的更好一点。
搜索更多相关主题的帖子: include outline return create 多项式 
2011-10-20 23:22
pauljames
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得分:5 
《数据结构基础》里面有多项式算法,你可以参考一下

经常不在线不能及时回复短消息,如有c/单片机/运动控制/数据采集等方面的项目难题可加qq1921826084。
2011-10-21 07:08
柏霖
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得分:0 
回复 2楼 pauljames
我看了。但是里面没有代码
2016-06-28 19:12
柏霖
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#include<stdio.h>
#include<stdlib.h>
typedef  struct NODE
{
    int a;
    int n;
    struct NODE *next;
}Node;
Node *create();
void output(Node*);
Node* Add(Node*, Node*);
void Paixu(Node *);
int main()
{
    Node *p, *q,*h;
    p = create();
    q = create();
    Paixu(q); Paixu(p);
    printf("第一个多项式为:\n");
    output(p); printf("\n");
    printf("第二个多项式为:\n");
    output(q); printf("\n");
    h=Add(p, q);
    printf("二个多项式之和为:\n");
    output(h);printf("\n");
    return 0;
}
Node *create()
{
    int  n,i;
    Node *p, *h = NULL, *q = NULL;
    printf("请输入多项式项数:");
    scanf("%d", &n);
    printf("请分别输入系数与次数:\n");
    for ( i = 0; i < n; i++)
    {
        p = (Node*)malloc(sizeof(Node));
        scanf("%d%d", &p->a, &p->n);
        if (h == NULL) h = p;
        else q->next = p;
        q = p;
    }
    q->next = NULL;
    return h;
}
void output(Node *h)
{
    if (h != NULL)
    {
        if (h->next != NULL) printf("%dx^%d + ", h->a, h->n);
        else printf("%dx^%d", h->a, h->n);
        output(h->next);

    }
}
Node* Add(Node *a, Node*b)
{
    Node *p, *q, *pre, *t,*h1,*h2;
    int s;
    h1 = (Node*)malloc(sizeof(Node)); h1->next = a;
    h2 = (Node*)malloc(sizeof(Node)); h2->next = b;
    p = a; q = b;
    pre = h1;
    while (p != NULL&&q != NULL)
    {
        if (p->n < q->n)
        {
            pre->next = p;
            pre = pre->next;
            p = p->next;
        }
        else if (p->n == q->n)
        {
            s = p->a+q->a;
            if (s != 0)
            {
                p->a = s;
                pre->next = p;
                pre = pre->next;
                p = p->next;
                t = q;
                q = q->next;
                free(t);
            }
            else
            {
                t = p->next;
                free(p);
                p = t;
                t = q->next;
                free(q);
                q = t;
            }
        }
        else
        {
            pre->next = q;
            pre = pre->next;
            q = q->next;
        }
    }
    if (p != NULL)
        pre->next = p;
    else
        pre->next = q;
    return h1->next;
}
void Paixu(Node *h)
{
    int p, t,m;
    Node *q, *r;
    do
    {
        p = h->n; q = h; r = h;
        do
        {

            if (r->n < p) { p = r->n; q = r; }
            r = r->next;
        } while (r != NULL);
        if (r != h) { t = h->n; m = h->a; h->n = q->n; h->a = q->a; q->n = t; q->a = m; }
        h = h->next;
    } while (h != NULL);
}
2016-06-28 19:48
快速回复:多项式加法:
数据加载中...
 
   



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