npmul.c

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#include <stdio.h>
#include <math.h>
/* 下面的两个数组可以根据具体要求解的多项式来决定其值*/
static double p[6]={4,-6,3,1,-1,5};	/*表示多项式4x^5 - 6x^4 + 3x^3 + x^2 - x + 5 */
static double q[4]={3,2,-5,1};		/*表示多项式3x^3 + 2x^2 - 5x + 1 */
static double result[9]={0,0,0,0,0,0,0,0,0};		/*存放乘积多项式*/

void npmul(p,m,q,n,s)
int m,n;
double p[],q[],s[];
{
	int i,j;
	for (i=0; i<=m-1; i++)
	for (j=0; j<=n-1; j++)
		s[i+j]=s[i+j]+p[i]*q[j];		/*迭带计算各项系数*/
	return;
}

double compute(s,k,x)					/*计算所给多项式的值*/
double s[];
int k;
float x;
{
	int i;
	float multip = 1;
	double sum = 0;
	for (i=0;i<k;i++)
		multip = multip * x;			/*先求出x的最高次项的值*/
	for (i=k-1;i>=0;i--)
	{
		sum = sum + s[i] * multip;		/*依次从高到低求出相对应次项的值*/
		if (x<>0)
		multip = multip / x;
	}
	return sum;
}

void main()
{
	int i;		
	float x;
	npmul(p,6,q,4,result);
	printf("Result is :\n");
	for (i=9;i>=1;i--)					/*逐行逐项打印出结果多项式*/
		printf(" (%f*x^%d) + \n",result[9-i],i-1);
	printf("\n");
	printf("Input the value of x:");
	scanf("%f",&x);
	printf("\nThe value of the x-multinomial is: %13.7f",compute(result,9,x));
	getch();
}