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【数据结构】 实现两个多项式的相加、减。要求:输出两个多项式原式,以及计算结果。

答案:1  悬赏:0  手机版
解决时间 2021-03-10 10:06
  • 提问者网友:锁深秋
  • 2021-03-10 01:16
用C语言……
最佳答案
  • 五星知识达人网友:刀戟声无边
  • 2021-03-10 01:49
#include "stdio.h"


typedef struct LNode{
double coef;
int exp;
struct LNode *next;
}LNode;


LNode *Init()
{
LNode *head = (LNode*)malloc(sizeof(LNode));
head->next = NULL;
return head;
}


void AddNode(LNode *head, double coef, int exp)
{
LNode *pre = head->next;
LNode *temp;
temp = (LNode*)malloc(sizeof(LNode));
temp->coef = coef;
temp->exp = exp;
temp->next = NULL;

if(pre == NULL)
{
head->next = temp;
return;
}

for(; pre->next!=NULL; pre=pre->next);

pre->next = temp;
}


void List(LNode *head)
{
LNode *curr;
printf("All nodes : ");
for(curr=head->next; curr!=NULL; curr=curr->next)
{
printf("(%lf, %d)\t", curr->coef, curr->exp);
}
printf("\n");
}


int Size(LNode *head)
{
int len = 0;
LNode *curr;
for(curr=head->next; curr!=NULL; curr=curr->next,len++);

return len;
}

LNode *Add(LNode *headA, LNode *headB)
{
LNode *currA, *currB, *headC;
double sum;
currA = headA->next;
currB = headB->next;
headC = Init();
while(currA!=NULL && currB!=NULL)
{
if(currA->exp > currB->exp)
{
AddNode(headC, currA->coef, currA->exp);
currA = currA->next;
}
else if(currA->exp < currB->exp)
{
AddNode(headC, currB->coef, currB->exp);
currB = currB->next;
}
else
{
sum = currA->coef + currB->coef;
if(sum != 0)
{
AddNode(headC, sum, currA->exp);
}
currA = currA->next;
currB = currB->next;
}
}
while(currA != NULL)
{
AddNode(headC, currA->coef, currA->exp);
currA = currA->next;
}
while(currB != NULL)
{
AddNode(headC, currB->coef, currB->exp);
currB = currB->next;
}

return headC;
}

void main()
{
LNode *headA, *headB, *headC;
headA = Init();
headB = Init();
AddNode(headA, 1.0, 5);
AddNode(headA, -1.0, 3);
AddNode(headA, 1, 0);
AddNode(headB, 0.5, 5);
AddNode(headB, 1.0, 4);
AddNode(headB, 1.0, 3);

List(headA);
List(headB);

headC = Add(headA, headB);

List(headC);

}
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