简单多项式求值c语言编程(简单多项式求值c语言编程实例)
C语言中一维多项式求值
计算多项式 p(x)=a(n-1)x(n-1)+a(n-2)x(n-2)+.....a1x+a0;
在指定点x处的函数值。
算法：
首先将多项式表述成如下嵌套的方式：
p(x)=(...((a(n-1)+a(n-2))x+a(n-3))x+....a1)x+a0;
然后依次从里向外算（因为x是已知的么），得到递推公式：
U(n-1)=a(n-1)
U(k)=U(k+1)x+a(k); K=n-2,n-3......1,0;
那当算到k=0时，得到的U(0)就是要求的值。
下面是用C语言实现的：
double plyv( double a[],double x,int n) //a[]是多项式的系数,n是数组长度。
{
double u;//一直存放递归结果；
Int i;
for(i=n-2;i=0;i--)
{
u=u*x+a[i];
}
return u;
}
#include
int main()
{
double a[3]={2,3,4};//根据多项式的形式定义数组长度以及个数，如果有的x项没有，则视系数为0；
double s;
double x;
s=plyv(a,x,3);//此为最后结果；
printf("%f",s);
return 0;
}
此题的解题重点在于：找到求解的递归关系，然后依据递归关系求解。
C语言 简单多项式的求值 题目是：对用户输入的任一整数，输出以下多项式 y=2x的平方+x
#include?stdio.h
int?main(){
int?x?=?0,?y?=?0;
scanf("%d",?x);
y?=?2?*?x?*?x?+?x?+?8;
printf("%d\n",?y);
return?0;
}
执行结果:
多项式计算 C语言编程
这个其实很简单，需要3个数组（暂时考虑int数组），长度都是10，分别保存多项式1、2和计算结果。初始化为全0。输入就按照你的假设吧。输入后三个数组分别为：
多项式1：[7, 0, -5, 2, 0, 0, 0, 0, 0, 0]（x的0次幂系数是7，x的1次幂系数是2，以此类推，下同）
多项式2：[-8, 1, 3, 0, 0, 0, 0, 0, 0, 0]
计算结果：[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]（还没算呢，当然都是0）
加法减法很好算，不赘述。乘法怎么算呢，你按照真实的数学计算步骤推一遍就知道了，你会把3x2、x、-8分别乘以2x3-5x2+7，最后把结果加起来。转换到程序中，就是把若干个数组加起来：
[-56, 0, 40, -16, 0, 0, 0, 0, 0, 0]
[0, 7, 0, -5, 2, 0, 0, 0, 0, 0]
[0, 0, 21, 0, -15, 6, 0, 0, 0, 0]
加起来就可以了。
至于提高水平，这个题目出得不好，因为多项式相除结果不唯一。比如说2x2 + 1除以x2 + 1，你可以说2x2 + 1 = 2(x2 + 1) - 1，也可以说2x2 + 1 = 1(x2 + 1) + x2。这样的题目数学上就意义不大，用程序去实现也达不到锻炼水平的作用。也许我理解有误？
C语言程序题：编写程序实现多项式计算
#include?stdio.h
#include?stdlib.h
#include?math.h
#define?EPS?1E-6
typedef?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(1)?{
printf("系数?幂次(0?0结束)?:?");
scanf("%lf%d",coe,pwr);
if(fabs(coe)?=?EPS??!pwr)?break;
p-next?=?(pItem)malloc(sizeof(item));
p-next-coefficient?=?coe;
p-next-power?=?pwr;
p?=?p-next;
}
p-next?=?NULL;
return?head;
}
void?Sort(POLYNOMIAL?head)?{?//?按幂次降排序
pItem?pt,q,p?=?head;
while(p-next)?{
q?=?p-next;
while(q-next)?{
if(p-next-power??q-next-power)?{
pt?=?p-next;
p-next?=?q-next;
q-next?=?p-next-next;
p-next-next?=?pt;
}
else?q?=?q-next;
}
p?=?p-next;
}
}
void?Show(POLYNOMIAL?head)?{?//?显示多项式
POLYNOMIAL?p?=?head-next;
int?flag?=?1;
if(p?==?NULL)?return;
while(p)?{
if(flag)?{
if(fabs(p-coefficient)?=?EPS)?{
if(p-power?==?0)?printf("%.2lf?",p-coefficient);
else?if(p-power?==?1)?{
if(p-coefficient?==?1.0)?printf("x?");
else?if(p-coefficient?==?-1.0)?printf("-x?");
else?printf("%.2lfx?",p-coefficient);
}
else?if(p-coefficient?==?1.0)?printf("x^%d?",p-power);
else?if(p-coefficient?==?-1.0)?printf("-x^%d?",p-power);
else?printf("%.2lfx^%d?",p-coefficient,p-power);
flag?=?0;
}
}
else?if(p-coefficient??0.0??fabs(p-coefficient)?=?EPS)?{
if(p-power?==?0)?printf("+?%.2lf?",p-coefficient);
else?if(p-power?==?1)?{
if(p-coefficient?==?1.0)?printf("+?x?");
else?printf("+?%.2lfx?",p-coefficient);
}
else?if(p-coefficient?==?1.0)?printf("+?x^%d?",p-power);
else?printf("+?%.2lfx^%d?",p-coefficient,p-power);
}
else?if(p-coefficient??0.0??fabs(p-coefficient)?=?EPS)?{
if(p-power?==?0)?printf("-?%.2lf?",-p-coefficient);
else?if(p-power?==?1)?{
if(p-coefficient?==?-1.0)?printf("-?x?");
else?printf("-?%.2lfx?",-p-coefficient);
}
else?if(p-coefficient?==?-1.0)?printf("-?x^%d?",p-power);
else?printf("-?%.2lfx^%d?",-p-coefficient,p-power);
}
p?=?p-next;
}
printf("\n");
}
double?Power(double?x,int?n)?{
double?value?=?1.0;
int?i;
for(i?=?0;?i??n;?++i)?value?*=?x;
return?value;
}
double?Value(POLYNOMIAL?head,double?x)?{?//?多项式求值
POLYNOMIAL?p;
double?value?=?0.0;
for(p?=?head-next;?p;?p?=?p-next)
value?+=?p-coefficient?*?Power(x,p-power);
return?value;
}
POLYNOMIAL?Copy(POLYNOMIAL?A)?{
POLYNOMIAL?head,t,p;
head?=?t?=?(pItem)malloc(sizeof(item));
for(p?=?A-next;?p;?p?=?p-next)?{
t-next?=?(pItem)malloc(sizeof(item));
t-next-coefficient?=?p-coefficient;
t-next-power?=?p-power;
t?=?t-next;
}
t-next?=?NULL;
return?head;
}
POLYNOMIAL?Additive(POLYNOMIAL?A,?POLYNOMIAL?B)?{?//?多项式相加
POLYNOMIAL?head,p,q,t;
head?=?Copy(A);
for(p?=?B;?p-next;?p?=?p-next)?{
q?=?head;
while(q-next)?{
if(p-next-power?==?q-next-power)?{
q-next-coefficient?+=?p-next-coefficient;
if(fabs(q-next-coefficient)?=?EPS)?{
t?=?q-next;
q-next?=?t-next;
free(t);
}
break;
}
q?=?q-next;
}
if(q-next?==?NULL)?{
q-next?=?(pItem)malloc(sizeof(item));
q-next-coefficient?=?p-next-coefficient;
q-next-power?=?p-next-power;
q-next-next?=?NULL;
}
}
Sort(head);
return?head;
}
POLYNOMIAL?Subtract(POLYNOMIAL?A,?POLYNOMIAL?B)?{?//?多项式相减
POLYNOMIAL?head,p,q,t;
head?=?Copy(A);
for(p?=?B;?p-next;?p?=?p-next)?{
q?=?head;
while(q-next)?{
if(p-next-power?==?q-next-power)?{
q-next-coefficient?-=?p-next-coefficient;
if(fabs(q-next-coefficient)?=?EPS)?{
t?=?q-next;
q-next?=?t-next;
free(t);
}
break;
}
q?=?q-next;
}
if(q-next?==?NULL)?{
q-next?=?(pItem)malloc(sizeof(item));
q-next-coefficient?=?-p-next-coefficient;
q-next-power?=?p-next-power;
q-next-next?=?NULL;
}
}
Sort(head);
return?head;
}
POLYNOMIAL?Multiplication(POLYNOMIAL?A,?POLYNOMIAL?B)?{?//?多项式相乘
POLYNOMIAL?head,t,p,q;
head?=?t?=?(pItem)malloc(sizeof(item));
for(p?=?A-next;?p;?p?=?p-next)?{?//?完成相乘过程
for(q?=?B-next;?q;?q?=?q-next)?{
t-next?=?(pItem)malloc(sizeof(item));
t-next-coefficient?=?p-coefficient?*?q-coefficient;
t-next-power?=?p-power?+?q-power;
t?=?t-next;
}
}
t-next?=?NULL;
Sort(head);?//?排序
p?=?head;
while(p-next)?{?//?合并同类项
q?=?p-next;
while(q-next)?{
if(p-next-power?==?q-next-power)?{
p-next-coefficient?+=?q-next-coefficient;
t?=?q-next;
q-next?=?t-next;
free(t);
}
else?q?=?q-next;
}
p?=?p-next;
}
return?head;
}
void?FreeMemory(POLYNOMIAL?head)?{
POLYNOMIAL?q,p?=?head;
while(p)?{
q?=?p;
p?=?q-next;
free(q);
}
}
int?main()?{
printf("创建多项式A:\n");
POLYNOMIAL?A?=?Create();
Sort(A);
printf("A(x)?=?");Show(A);
printf("创建多项式B:\n");
POLYNOMIAL?B?=?Create();
Sort(B);
printf("B(x)?=?");Show(B);
POLYNOMIAL?C?=?Additive(A,B);
printf("C(x)?=?");Show(C);
POLYNOMIAL?D?=?Subtract(A,B);
printf("D(x)?=?");Show(D);
POLYNOMIAL?E?=?Multiplication(A,B);
printf("E(x)?=?");Show(E);
printf("A(%.2lf)?=?%.4lf\n",2.0,Value(A,2.0));
printf("B(%.2lf)?=?%.4lf\n",2.0,Value(B,2.0));
printf("C(%.2lf)?=?%.4lf\n",2.0,Value(C,2.0));
printf("D(%.2lf)?=?%.4lf\n",2.0,Value(D,2.0));
printf("E(%.2lf)?=?%.4lf\n",2.0,Value(E,2.0));
FreeMemory(A);
FreeMemory(B);
FreeMemory(C);
FreeMemory(D);
FreeMemory(E);
return?0;
}