105.多项式乘法

  

   

    

     

    
    

     

      #include <stdio.h>
     
    
   

    

     

    
    

     

      #include <math.h>
     
    
   

    

     

    
    

     

      #define MAX 50
     
    
   

    

     

    
    

     

      /* 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,j,m,n;
     
    
   

    

     

    
    

     
	double  px[MAX],qx[MAX],rx[MAX];
     
    
   

    

     

    
    

     
	float x;
     
    
   

    

     

    
    

     
	clrscr();
     
    
   

    

     

    
    

     
	for(i=0;i<MAX;i++)
     
    
   

    

     

    
    

     

      		rx[i]=0;
     
    
   

    

     

    
    

     
	puts(" This is a polynomial multiplication program.");
     
    
   

    

     

    
    

     
	puts("It calculate the product of two polynomials: P(x) and Q(x)");
     
    
   

    

     

    
    

     
	puts(" P(x)=Pm-1*x^(m-1)+Pm-2*x^(m-2)+...+P1*x+P0");
     
    
   

    

     

    
    

     
	puts(" Q(x)=Qn-1*x^(n-1)+Qn-2*x^(n-2)+...+Q1*x+Q0");
     
    
   

    

     

    
    

     
	printf("\n >> Please input m (>=1): ");
     
    
   

    

     

    
    

     
	scanf("%d",&m);
     
    
   

    

     

    
    

     
	printf(" >> Please input P%d, P%d, ... P1, P0 one by one:\n",m-1,m-2);
     
    
   

    

     

    
    

     
	for(i=0;i<m;i++)
     
    
   

    

     

    
    

     
		scanf("%f",&px[i]);
     
    
   

    

     

    
    

     
	printf("\n >> Please input n (>=1): ");
     
    
   

    

     

    
    

     
	scanf("%d",&n);
     
    
   

    

     

    
    

     
	printf(" >> Please input Q%d, Q%d, ... Q1, Q0 one by one:\n",n-1,n-2);
     
    
   

    

     

    
    

     
	for(i=0;i<n;i++)
     
    
   

    

     

    
    

     
		scanf("%f",&qx[i]);
     
    
   

    

     

    
    

     
	npmul(p,m,q,n,rx);
     
    
   

    

     

    
    

     
	printf("\nThe product of two polynomials R(x) is :\n");
     
    
   

    

     

    
    

     
	for (i=m+n-1,j=0;i>=1;i--)					/*逐行逐项打印出结果多项式*/
     
    
   

    

     

    
    

     

      	{
     
    
   

    

     

    
    

     
		printf(" (%f*x^%d) + ",rx[m+n-1-i],i-1);
     
    
   

    

     

    
    

     
		if(j==2)
     
    
   

    

     

    
    

     

      		{
     
    
   

    

     

    
    

     
			printf("\n");
     
    
   

    

     

    
    

     

      			j=0;
     
    
   

    

     

    
    

     

      		}
     
    
   

    

     

    
    

     
		else
     
    
   

    

     

    
    

     

      			j++;
     
    
   

    

     

    
    

     

      	}
     
    
   

    

     

    
    

     
	printf("\n");
     
    
   

    

     

    
    

     
	printf("Input the value of x: ");
     
    
   

    

     

    
    

     
	scanf("%f",&x);
     
    
   

    

     

    
    

     
	printf("\nThe value of the R(%f) is: %13.7f",x,compute(rx,m+n-1,x));
     
    
   

    

     

    
    

     
	puts("\n Press any key to quit...");
     
    
   

    

     

    
    

     
	getch();
     
    
   

    

     

    
    

     

      }
     
    
  
 

文章来源: blog.csdn.net，作者：程序员编程指南，版权归原作者所有，如需转载，请联系作者。

原文链接：blog.csdn.net/weixin_41055260/article/details/124558650