使用c中的链表进行多项式乘法答案

【问题标题】：Polynomial multiplication using linked list in c使用c中的链表进行多项式乘法
【发布时间】：2013-08-12 07:25:42
【问题描述】：

我正在开发一个程序，使用 c 中的链表对多项式执行加法、减法、乘法和微分运算。除了乘法一，其他操作都工作正常。这是代码

#include<stdio.h>
#include<malloc.h>
#include<conio.h>
struct link{
   int coeff;
   int pow;
   struct link *next;
   };
struct link *poly1=NULL,*poly2=NULL,*poly=NULL;
void create(struct link *node)
{
 char ch;
 do
 {
  printf("\n\nenter coeff:");
  scanf("%d",&node->coeff);
  printf("\nenter power:");
  scanf("%d",&node->pow);
  node->next=(struct link*)malloc(sizeof(struct link));
  node=node->next;
  node->next=NULL;
printf("\ncontinue(y/n):");
ch=getch();
}
while(ch=='y' || ch=='Y');
}
void show(struct link *node)
{
while(node->next!=NULL)
{

printf("%dx^%d",node->coeff,node->pow);
node=node->next;
if(node->next!=NULL)
printf(" + ");


}
}
void polyadd(struct link *poly1,struct link *poly2,struct link *poly)
{
 while(poly1->next &&  poly2->next)
 {
  if(poly1->pow>poly2->pow)
  {
   poly->pow=poly1->pow;
   poly->coeff=poly1->coeff;
   poly1=poly1->next;
   }
  else if(poly1->pow<poly2->pow)
  {
   poly->pow=poly2->pow;
   poly->coeff=poly2->coeff;
   poly2=poly2->next;
   }
  else
  {
   poly->pow=poly1->pow;
   poly->coeff=poly1->coeff+poly2->coeff;
   poly1=poly1->next;
   poly2=poly2->next;
   }
  poly->next=(struct link *)malloc(sizeof(struct link));
  poly=poly->next;
  poly->next=NULL;
 }
 while(poly1->next || poly2->next)
 {
  if(poly1->next)
  {
   poly->pow=poly1->pow;
   poly->coeff=poly1->coeff;
   poly1=poly1->next;
   }
  if(poly2->next)
  {
   poly->pow=poly2->pow;
   poly->coeff=poly2->coeff;
   poly2=poly2->next;
   }
   poly->next=(struct link *)malloc(sizeof(struct link));
   poly=poly->next;
   poly->next=NULL;
   }

}

void polysub(struct link *poly1,struct link *poly2,struct link *poly)
{
 while(poly1->next &&  poly2->next)
 {
  if(poly1->pow>poly2->pow)
  {
   poly->pow=poly1->pow;
   poly->coeff=poly1->coeff;
   poly1=poly1->next;
   }
  else if(poly1->pow<poly2->pow)
  {
   poly->pow=poly2->pow;
   poly->coeff=poly2->coeff;
   poly2=poly2->next;
   }
  else
  {
   poly->pow=poly1->pow;
   poly->coeff=poly1->coeff-poly2->coeff;
   poly1=poly1->next;
   poly2=poly2->next;
   }
  poly->next=(struct link *)malloc(sizeof(struct link));
  poly=poly->next;
  poly->next=NULL;
 }
 while(poly1->next || poly2->next)
 {
  if(poly1->next)
  {
   poly->pow=poly1->pow;
   poly->coeff=poly1->coeff;
   poly1=poly1->next;
   }
  if(poly2->next)
  {
   poly->pow=poly2->pow;
   poly->coeff=poly2->coeff;
   poly2=poly2->next;
   }
   poly->next=(struct link *)malloc(sizeof(struct link));
   poly=poly->next;
   poly->next=NULL;
   }

}

void polymul(struct link *n1, struct link *n2, struct link *n)
{
    struct link * n2beg=n2;

            while (n1) 
            {
                    struct link * temp=(struct link *)malloc(sizeof(struct link));
                    temp->next=NULL;    
                    n2=n2beg;
                    while (n2) 
                    {
                            temp->coeff = n1->coeff * n2->coeff;

                            temp->pow = n1->pow + n2->pow;

                            n2 = n2->next;
                            temp->next=(struct link *)malloc(sizeof(struct link));
                            temp=temp->next;
                            temp->next=NULL;

                    }

                    polyadd(temp,n,n);
                    n1 = n1->next;
                    free(temp);
            }

}

void diff(struct link* p1,struct link* p2)
{

while(p1->next!=NULL)
{
    p2->coeff=p1->coeff*p1->pow;
    p2->pow=p1->pow-1;
    p2->next=NULL;
    p2->next=(struct link *)malloc(sizeof(struct link));
   p2=p2->next;
   p2->next=NULL;
   p1=p1->next;
}

}
  main()
  {
  int op;
  char ch;
  do{
  poly1=(struct link *)malloc(sizeof(struct link));
  poly2=(struct link *)malloc(sizeof(struct link));
  poly=(struct link *)malloc(sizeof(struct link));
  printf("\n\nWhat do you want to do?\n1.Addition\n2.Subtraction
          \n3.Multiplication\n4.Differentiation\n0.Exit
          \nEnter your choice:");
  scanf("%d",&op);
  switch(op)
  {
        case 1:
            printf("\n\nenter 1st polynomial:");
            create(poly1);
            printf("\n\nenter 2nd polynomial:");
            create(poly2);
            printf("\n1st Polynomial:\t");
            show(poly1);
            printf("\n2nd Polynomial:\t");
            show(poly2);
            polyadd(poly1,poly2,poly);
            printf("\nAdded polynomial:\t");
            show(poly);
            break;
        case 2:
            printf("\n\nenter 1st polynomial:\t");
            create(poly1);
            printf("\n\nenter 2nd polynomial:\t");
            create(poly2);
            printf("\n\n1st Polynomial:\t");
            show(poly1);
            printf("\n\n2nd Polynomial:\t");
            show(poly2);
            polysub(poly1,poly2,poly);
            printf("\n\nSubtracted polynomial:\t");
            show(poly);
            break;  
        case 3:
            printf("\n\nenter 1st polynomial:");
            create(poly1);
            printf("\n\nenter 2nd polynomial:");
            create(poly2);
            printf("\n\n1st Polynomial:\t");
            show(poly1);
            printf("\n\n2nd Polynomial:\t");
            show(poly2);
            polymul(poly1,poly2,poly);
            printf("\n\nMultiplied polynomial:\t");
            show(poly);
            break;
        case 4:
            printf("\n\nenter polynomial:");
            create(poly1);
            printf("\n\nPolynomial:\t");
            show(poly1);
            diff(poly1,poly2);
            printf("\n\nDifferentiated Polynomial:\t");
            show(poly2);
            break;
        }

 /* printf("\n Want to continue? Y/N:");
  ch=getch();*/
  }
  while(op);

}

【问题讨论】：

（这篇文章中“社区”发生了什么变化？）除了polyadd() 之外，没有实现polysub()，而是为标量乘法执行polyscale()。（在polyadd() 中，可以将将“未配对”节点复制到结果一次 - 从所呈现的八次开始。）

标签： linked-list multiplication dev-c++ polynomial-math

【解决方案1】：

polymul() 中使用的polyadd(temp,n,n) 无法处理将n 用作源多项式和目标多项式的总和。

要么重做polyadd() 以处理指向相同多项式的参数，要么重做polymul() 中polyadd() 的调用以使用不同的多项式。建议第一个。

【讨论】：

【解决方案2】：

我没有查看您的完整代码，因为我在您的 create() 函数中发现了一个错误。请注意，您已将 poly1 作为参数传递给函数 create()。

这不正确，因为 C 遵循 按值调用。所发生的只是poly1 的值（仍然未初始化）被传递，*node 存储该值。最好将poly1 的地址作为参数传递，并使用指向指针的指针在函数中捕获该值。

【讨论】：

【解决方案3】：

#include<stdio.h>
#include <conio.h>
struct node
{
    int c,e;
    struct node *link;
}*start1=NULL,*start2=NULL,*start3=NULL,*temp1,*temp2,*temp3,*new_node;
int delete_dup(int h)
{
    struct node *cr,*prev,*run,*tmp;

    cr = start3->link;
    prev = start3;

    while(cr != NULL){

        run = start3;

        while(run != cr)
        {
             if(run->e == cr->e)
            {
                run->c+=cr->c;
                tmp = cr;
                cr = cr->link;
                prev->link = cr;
                remove(tmp);h--;
                break;
            }

            run = run->link;
        }

        if(run == cr){
            cr = cr->link;
            prev = prev->link;
        }
    }
    return h;

}

void main()
{
    int n,m,i,j,k;
    puts("Enter the number of terms in first polynomial");
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        new_node=(struct node*)malloc(sizeof(struct node));
        new_node->link=NULL;
        puts("C=");
        scanf("%d",&new_node->c);
        puts("E=");
        scanf("%d",&new_node->e);
        new_node->link=start1;
        start1=new_node;

        }
        puts("Enter the number of terms in first polynomial");
    scanf("%d",&m);
    for(i=0;i<m;i++)
    {
        new_node=(struct node*)malloc(sizeof(struct node));
        new_node->link=NULL;
        puts("C=");
        scanf("%d",&new_node->c);
        puts("E=");
        scanf("%d",&new_node->e);
        new_node->link=start2;
        start2=new_node;

        }
        temp1=start1;
        temp2=start2;
        i=0; j=0;
        while(i<m)
        {   
            j=0; temp1=start1;
            while(j<n)
            {
                new_node=(struct node*)malloc(sizeof(struct node));
                new_node->link=NULL;
                new_node->c=temp1->c*temp2->c;
                new_node->e=temp1->e+temp2->e;
                new_node->link=start3;
                start3=new_node;
                j++;
                temp1=temp1->link;
            }
            temp2=temp2->link;
            i++;
        }
        i=0;
        k=delete_dup(m*n);
        temp3=start3;
        while(i<k-1)

            {
                printf("(%dx^%d)+",temp3->c,temp3->e);
                temp3=temp3->link;
                i++;
            }
                printf("(%dx^%d)",temp3->c,temp3->e);

                    }

【讨论】：

请添加一些解释。 :)

【解决方案4】：

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
struct node{
    int coef,pow;
    struct node *next;
};

struct node* create()
{
    int c;
    struct node *new_node=NULL;
    struct node *head=NULL,*temp;



    do{
        new_node=(struct node*)malloc(sizeof(struct node));
        printf("\nEnter coef :");
        scanf("%d",&new_node->coef);
        printf("\nEnter power : ");
        scanf("%d",&new_node->pow);

        if(head==NULL)
        {
            head=new_node;
            head->next=NULL;
            temp=new_node;
        }
        else
        {
            temp->next=new_node;
            temp=temp->next;
            temp->next=NULL;
        }
        printf("\nDo you want to continue :");
        scanf("%d",&c);
    }while(c==1);
    return head;
}

void print(struct node *p)
{
    while(p!=NULL)
    {
        printf("|%d|%d|->",p->coef,p->pow);
        p=p->next;
    }
}

struct node* create_new(int p, int q)
{

    struct node *nn=(struct node*)malloc(sizeof(struct node));
    nn->coef=p;
    nn->pow=q;
    nn->next=NULL;
    return nn;
}
struct node* add(struct node *poly1,struct node *poly2)
{
    struct node *tempadd,*temp;
    struct node *temp1=poly1;
    struct node *temp2=poly2,*head=NULL;
    while(temp1!=NULL && temp2!=NULL)
    {   
        if(temp1->pow==temp2->pow)
        {
            tempadd=create_new(temp1->coef+temp2->coef,temp2->pow);
            temp1=temp1->next;
            temp2=temp2->next;
        }
        else if(temp1->pow<temp2->pow)
        {
            tempadd=create_new(temp2->coef,temp2->pow);
            temp2=temp2->next;
        }
        else
        {
            tempadd=create_new(temp1->coef,temp1->pow);
            temp1=temp1->next;
        }
        if(head==NULL)
        {
            head=tempadd;
            temp=head;
        }
        else
        {
            temp->next=tempadd;
            temp=temp->next;
        }

    }
    if(temp1!=NULL)
    {
        while(temp1!=NULL)
        {
            tempadd=create_new(temp1->coef,temp1->pow);
            temp1=temp1->next;
            temp->next=tempadd;
            temp=temp->next;
        }
    }
    else if(temp2!=NULL)
    {
        while(temp2!=NULL)
        {
            tempadd=create_new(temp2->coef,temp2->pow);
            temp2=temp2->next;
            temp->next=tempadd;
            temp=temp->next;
        }
    }

    return head;
}

void main()
{
    struct node *head1,*head2,*head3;
    head1=create();
    print(head1);
    head2=create();
    print(head1);
    printf("\n");
    print(head2);
    printf("\n");
    head3=add(head1,head2);
    printf("\n\nResult : ");
    print(head3);
    getch();
}

【讨论】：

虽然此代码 sn-p 可能会提供一些有限的即时帮助，但 proper explanation 将通过说明为什么这是解决问题的好方法并使其更加完善，从而大大提高其长期价值对有其他类似问题的未来读者有用。请edit您的回答添加一些解释，包括您所做的假设。

【解决方案5】：

#include<stdio.h>
#include<stdlib.h>
#include<malloc.h>
typedef struct node {
    struct node *previous;
    int coef;
    int pow;
    struct node *next;
}poly;
poly *locate(int,poly *);
void display(poly *);
poly *mult(poly *,poly *,poly *);
void display2(poly *);
void main()
{
char ch;
poly *s1,*s2,*s3,*head1,*head2,*head3,*s4;;
s1=(poly *)malloc(sizeof(poly));
s1->previous=NULL;
head1=s1;
s2=(poly *)malloc(sizeof(poly));
s2->previous=NULL;
head2=s2;
head3=s3;
printf("Enter first polynomial :\n");
do{                                           //Input for polynomial-1
    printf("Enter co-efficient : ");
    scanf("%d",&s1->coef);
    printf("Enter exponent : ");
    scanf("%d",&s1->pow);
    printf("Do you want to enter more terms(Y/N) : ");
    scanf(" %c",&ch);
    if(ch=='Y'){
        s1->next=(poly*)malloc(sizeof(poly));
        s1->next->previous=s1;
        s1=s1->next;
    }
    else {
        s1->next=NULL;
        break;
    }
}while(1);
printf("Enter second polynomial : \n");
do{                                     //input for polynomial-2
    printf("Enter co-efficient : ");
    scanf("%d",&s2->coef);
    printf("Enter exponent : ");
    scanf("%d",&s2->pow);
    printf("Do you want to enter more terms(Y/N) : ");
    scanf(" %c",&ch);
    if(ch=='Y'){
        s2->next=(poly*)malloc(sizeof(poly));
        s2->next->previous=s2;
        s2=s2->next;
    }
    else {
        s2->next=NULL;
        break;
    }
}while(1);
printf("Entered polynomials are : \n");
display(s1);
printf("\n");
display(s2);
s3=NULL;
s4=mult(s1,s2,s3);
printf("Resultant Polynomial after multiplication :\n");
display(s4);
}
void display(poly *a)
{
while(a!=NULL)
{
    printf("%dx^%d",a->coef,a->pow);
    if(a->previous!=NULL)
        printf(" + ");
    a=a->previous;
}
}
poly *mult(poly *s1,poly *s2,poly *s3)
{
while(s1->previous!=NULL)
    s1=s1->previous;
while(s2->previous!=NULL)
    s2=s2->previous;
while(s1)
{
    while(s2->previous!=NULL)
        s2=s2->previous;
    while(1)
    {
        if(s2->next!=NULL)
        {   
            poly *s4;
            if(s3==NULL)
            {
                s3=(poly *)malloc(sizeof(poly));
                s3->pow=s1->pow+s2->pow;
                s3->coef=s1->coef*s2->coef;
                s3->previous==NULL;
                s3->next==NULL;
            }
            else
            {
                s4=locate(s1->pow+s2->pow,s3);
                if(s4==NULL)
                {
                    s3->next=(poly *)malloc(sizeof(poly));
                    s3->next->previous=s3;
                    s3=s3->next;
                    s3->pow=s1->pow+s2->pow;
                    s3->coef=s1->coef*s2->coef;
                    s3->next==NULL;
                }
                else
                {
                    s4->coef=(s4->coef)+(s1->coef*s2->coef);    
                }
            }
            s2=s2->next;
        }
        else
        {
            poly *s4;
            if(s3==NULL)
            {
                s3=(poly *)malloc(sizeof(poly));
                s3->pow=s1->pow+s2->pow;
                s3->coef=s1->coef*s2->coef;
                s3->previous==NULL;
                s3->next==NULL;
            }
            else{
                s4=locate(s1->pow+s2->pow,s3);
                if(s4==NULL)
                {
                    s3->next=(poly *)malloc(sizeof(poly));
                    s3->next->previous=s3;
                    s3=s3->next;
                    s3->pow=s1->pow+s2->pow;
                    s3->coef=s1->coef*s2->coef;
                    s3->next==NULL;
                }
                else
                {
                    s4->coef=s4->coef+s1->coef*s2->coef;
                }
            }
            break;
        };
    }s1=s1->next;
}return s3;

}
poly *locate(int exp,poly *s3)
{
if(s3==NULL)
{
    return NULL;
}
else if(s3->pow==exp)
{
    return s3;
}
else{
    return locate(exp,s3->previous);
}
}

我考虑了第一个多项式的第一项并与第二个多项式的所有项相乘，从而创建了一个相乘多项式的主链表。作为下一段代码，我考虑了第一个多项式的下一项，然后在乘法多项式中相乘并搜索相同的索引并将结果添加到它，如果它不存在，我在乘法多项式中创建了一个新节点。快乐编码

【讨论】：