编写一段c程序,实现多项式的计算,谁能帮我呀, 谁能帮我用c语言帮我写一个一元多项式程序。

我记得数据结构书上有源代码的啊......还是帮你写了一个,整整一个小时啊,累死我了

#include
#include
#include

typedef struct _List List;
struct _List
{
int c;
int e;
List* next;
};

#define OPER_PLUS 0
#define OPER_MINUS 1

static List* list_oper(List* list1, List* list2, int oper);

/* order by e dec */
List* add_list(List* list, int c, int e);
void print_list_member(List* list);
List* list_plus(List* list1, List* list2);
List* list_minus(List* list1, List* list2);

List* list_plus(List* list1, List* list2)
{
return list_oper(list1, list2, OPER_PLUS);
}

List* list_minus(List* list1, List* list2)
{
return list_oper(list1, list2, OPER_MINUS);
}

static List* list_oper(List* list1, List* list2, int oper)
{
assert(list1 != NULL && list2 != NULL);

List* result_list = (List*) malloc(sizeof(List));
result_list->c = 0;
result_list->e = 0;

List* list1_temp = list1->next;
List* list2_temp = list2->next;

List* result_temp = result_list;
result_list->next = NULL;

while(list1_temp != NULL && list2_temp != NULL)
{
if(list1_temp->e != list2_temp->e)
{
List* add_list = (List*) malloc(sizeof(List));
add_list->next = NULL;

if(list1_temp->e > list2_temp->e)
{
add_list->e = list1_temp->e;
add_list->c = list1_temp->c;

list1_temp = list1_temp->next;
}
else
{
add_list->e = list2_temp->e;
add_list->c = list2_temp->c;

list2_temp = list2_temp->next;
}

result_temp->next = add_list;
result_temp = result_temp->next;
}
else
{
if(list1_temp->c + list2_temp->c != 0)
{
List* add_list = (List*) malloc(sizeof(List));
add_list->next = NULL;

add_list->e = list1_temp->e;
if(oper == OPER_PLUS)
{
add_list->c = list1_temp->c + list2_temp->c;
}
else
{
add_list->c = list1_temp->c - list2_temp->c;
}

result_temp->next = add_list;
result_temp = result_temp->next;
}

list1_temp = list1_temp->next;
list2_temp = list2_temp->next;
}
}/* end while */

while(list1_temp != NULL)
{
List* add_list = (List*) malloc(sizeof(List));
add_list->next = NULL;

add_list->e = list1_temp->e;
add_list->c = list1_temp->c;

result_temp->next = add_list;
result_temp = result_temp->next;
list1_temp = list1_temp->next;
}

while(list2_temp != NULL)
{
List* add_list = (List*) malloc(sizeof(List));
add_list->next = NULL;

add_list->e = list2_temp->e;
add_list->c = list2_temp->c;

result_temp->next = add_list;
result_temp = result_temp->next;
list2_temp = list2_temp->next;
}

return result_list;

}


void print_list_member(List* list)
{
assert(list != NULL);

List* print_list = list->next;
int count = 0;

while(print_list != NULL)
{
count++;
print_list = print_list->next;
}
printf("count = %d, ", count);

print_list = list->next;
count = 0;
while(print_list != NULL)
{
printf("c%d = %d, e%d = %d, ", count, print_list->c, count, print_list->e);
count++;
print_list = print_list->next;
}

printf("
");
}


List* add_list(List* list, int c, int e)
{
assert(e > 0);

List* list_temp = NULL;
List* list_add = NULL;

/* create new node */
list_add = (List*)malloc(sizeof(List));
list_add->c = c;
list_add->e = e;

/* first add */
if(list == NULL)
{
/* create head */
list = (List*)malloc(sizeof(List));
list->c = 0;
list->e = 0;

list->next = list_add;
list_add->next = NULL;
return list;
}

list_temp = list->next;
List* equel_list = list;

while(list_temp != NULL)
{
assert(list_temp->e != e);

if(list_temp->e > e)
{
equel_list = list_temp;
list_temp = list_temp->next;

if(list_temp == NULL)
{
list_add->next = NULL;
equel_list->next = list_add;
return list;
}
}
else
{
list_add->next = list_temp;
equel_list->next = list_add;
return list;
}
}

return list;
}

int main()
{
/* test */
List* list1 = NULL;
List* list2 = NULL;
list1 = add_list(list1, 3, 2);
list1 = add_list(list1, 9, 4);
list1 = add_list(list1, 1, 5);
list1 = add_list(list1, -3, 1);
list1 = add_list(list1, -1, 9);

list2 = add_list(list2, 5, 2);
list2 = add_list(list2, 2, 3);
list2 = add_list(list2, -9, 8);
list2 = add_list(list2, 1, 9);
list2 = add_list(list2, -8, 5);

// print_list_member(list1);
// print_list_member(list2);

List* list3 = list_plus(list1, list2);
List* list4 = list_minus(list1, list2);

print_list_member(list3);
print_list_member(list4);

}

#include "stdio.h"
#include "stdlib.h"
#include "string.h"
struct node{
int arr;
struct node *next;
};
typedef struct node Node;
typedef Node *Stack;
struct stacktop{
Stack top;
};
typedef struct stacktop *Top;
int jisuan(int sr1,int sr2,int sr3);
int opinion4(char *ptr,int *num);
int opinion3(char ch);
int opinion2(char ch);
int opinion(char ch);
void StackPush(int ch,Top ptop);
void StackPop(int *pi,Top ptop);
int main(void)
{

int value1;
int value2;
int value3;
char ch[100];
int R,a;
int i=0;
Top svalue;
Top sctype;
svalue=(Top)malloc(sizeof(struct stacktop));
svalue->top=NULL;
sctype=(Top)malloc(sizeof(struct stacktop));
sctype->top=NULL;
StackPush('a',sctype);
printf("请输入一个表达式：
-->");
gets(ch);
while(ch!=NULL&&ch[i]!='\0')
{
if(opinion(ch[i])==1)//如果是操作符
{
if(opinion2(ch[i])top->arr))
{
StackPop(&value1,svalue);
StackPop(&value2,svalue);
StackPop(&value3,sctype);

R=jisuan(value1,value2,value3);
StackPush(R,svalue);
}
StackPush(ch[i],sctype);
}
else if(opinion3(ch[i])==1)//不是操作符,是操作数
{
a=opinion4(ch,&i);
StackPush(a,svalue);
}
else
{
break;
}
i++;
}

while(sctype->top->arr!='a')
{
StackPop(&value1,svalue);
StackPop(&value2,svalue);
StackPop(&value3,sctype);
R=jisuan(value1,value2,value3);
StackPush(R,svalue);
}
printf("%d",R);
return 0;
}

int jisuan(int sr1,int sr2,int sr3)
{
switch((char)sr3)
{
case '+':return sr2+sr1;
case '-':return sr2-sr1;
case '*':return sr2*sr1;
case '/':return sr2/sr1;
}
}


/*将字符转换成数字*/
int opinion4(char *ptr,int *num)
{
int x;
int b=*num;
int c=0;
int d,e,f;
int brr[50];
int fang=1;
int total=0;
x=strlen(ptr);
while((opinion3(ptr[b])==1)&&b<x)
{
brr[c]=ptr[b]-48;
b++;
c++;
}

f=c-1;
if(c==1)
{
*num=b-1;
return brr[0];
}
else
{
for(d=0;d<c;d++)
{
for(e=f;e>0;e--)
fang=fang*10;
total=total+brr[d]*fang;
f--;
fang=1;
}
*num=b-1;
return total;
}
}

/*判断是否为数字*/
int opinion3(char ch)
{
if(ch>='0'&&ch<='9')
return 1;
else
return 0;
}

/*判断运算符的优先级*/
int opinion2(char ch)
{
switch(ch)
{
case '*':
case '/':return 2;
case '+':
case '-':return 1;
default:return 0;
}
}

void StackPop(int *pi,Top ptop)
{
Stack pnew;
pnew=ptop->top;
*pi=pnew->arr;
ptop->top=pnew->next;
free(pnew);
}

void StackPush(int ch,Top ptop)
{
Stack pnew;
pnew=(Stack)malloc(sizeof(Node));
if(pnew==NULL)
exit(1);
pnew->arr=ch;
if(ptop->top==NULL)
pnew->next=NULL;
else
pnew->next=ptop->top;
ptop->top=pnew;

}

/*判断是否为运算符*/
int opinion(char ch)
{
switch(ch)
{
case '+':
case '-':
case '*':
case '/': return 1;
default:
return 0;
}
}


如果有不会的加我百度HI
有什么问题可以随时问我。
yinatea@163.com

15060255288：编写一段c程序,实现多项式的计算,谁能帮我呀,
边湛答：while(c==' '||c=='\t') c=*pos++;if(isalpha(c)){ numtop++=' ';while(isalpha(c)||isdigit(c)) {*numtop++=c;c=*pos++;} if(m) return 1;m=1;continue;} else { for(i=0;opchTb1[i].code!=-1&&opchTb1[i].data!=c;i++)if(opchTb1[i].code==-1) return ...

15060255288：c语言设计 计算多项式a0+a1*x+a2*x*x+...+a9*x*x*x*x*x*x*x*x*x的...
边湛答：{ double s=0.0;double s1=1.0;double x;int i;int a[10];printf("please input x:");scanf("%f",&x);printf("please input a[1] ~a[10]:");for(i=0;i<10;i++)scanf("%d",&a[i]);for(i=0;i<10;i++){ s+=a[i]*s1;s1*=x;} printf("s=%f",s);} ...

15060255288：编写程序求多项式ax^3+bx^2+c的值
边湛答：int main() { float a, b, c, x, result;// 输入多项式系数和自变量 printf("请输入多项式系数a：");scanf("%f", &a);printf("请输入多项式系数b：");scanf("%f", &b);printf("请输入多项式系数c：");scanf("%f", &c);printf("请输入自变量x的值：");scanf("%f", &x);// ...

15060255288：多项式求和的c语言程序
边湛答：int Fluction(int);//声明实现多项式 1-1/2+1/3-1/4+1/5-1/6+...的功能函数 double sum;//定义全局变量（其实一般不推荐定义全局变量）int main(){ int m,n;//m个测试实例，求前 n项和 while(scanf("%d",&m)!=EOF){ for(int i=1;i<=m;i++)//输入 m个测试实例，所以循环...

15060255288：c语言编程求解多项式ax^3+bx^2+c的值
边湛答：以下是一个计算多项式ax^3+bx^2+c值的C语言代码：include <stdio.h> int main() { float a, b, c, x, result;printf("请输入多项式系数 a, b, c：");scanf("%f %f %f", &a, &b, &c);printf("请输入 x 的值：");scanf("%f", &x);result = a * x * x * x + b *...

15060255288：C语言程序题:编写程序实现多项式计算
边湛答：include <stdio.h>#include <stdlib.h>#include <math.h>#define EPS 1E-6typedef struct item {double coefficient;int power;struct item *next;} *POLYNOMIAL,*pItem;POLYNOMIAL Create() { // 创建多项式pItem head,p;double coe;int pwr;head = p = (pItem)malloc(sizeof(item));while...

15060255288：C语言写一个程序,用链表实现多项式相加相减
边湛答：//要求：多项式按降幂排列#include<stdafx.h>#include<iostream.h>struct node{int coef;//系数int exp;//指数node * next;};node * h1=NULL;//第一个多项式的头指针node * h2=NULL;//第二个多项式的头指针node * insert(int c,int e)//创建一个系数为c，指数为e的结点，并返回其地址{...

15060255288：C程序 输入一个实数x,计算多项式x+(x^2)/2!+(x^3)/3!+...的和,直到末...
边湛答：include<stdio.h> int main(){ double x,s=0,t;int n=2;scanf("%lf",&x);t = x;while(t>1e-5){ s+=t;t = t*x/n;n++;} printf("%.3lf\n",s);return 0;}

15060255288：C语言程序 编写一个函数,计算并输出下列多项式的值:s=1+1/1!+1/2...
边湛答：include <stdio.h> int main(void){ int n;int i, j;float sum = 1.0;float dex = 1.0;printf("输入一个数：");scanf("%d", &n);for (i = 2; i <= n; ++i){ for (j = i; j > 0; --j){ dex*=j;} sum+=1/dex;dex = 1.0;} printf("%.2f\n", sum);re...

15060255288：C语言求多项式乘法
边湛答：多项式乘法的实现,已知如下两个多项式P(x)=PX+PX+...+PX+PQ(x)=qX+qX+...+qX+q求它们的乘积多项式S(x)=sX+...+sX+s。可以定义两个一维数组p,q,按照次数从高到低存储两个多项式的系... 多项式乘法的实现,已知如下两个多项式 P(x)=P X + P X +...+ P X+P Q(x)=q X +q X +.....