C語言 習題1-1-a 請用Lanrange內差法寫出下列諸點的方程式

C語言 習題1-1-a 請用Lanrange內差法寫出下列諸點的方程式

x       f(x)
============
0.0   0.00
1.0  -3.00
2.0   0.00
34.0  15.00
============

並計算 P[0.5 , 1.5 .2.5  ,3.5] 之值並與 f(x)=x^3 -4 x 的誤差


程式
/* ex1-4.c: Lagrange Interpolation Algorithm
 * Read in data file of ex1-4.dat which has n point values
 * and the value of interpolating point xa. Based on Lagrange
 * Interpolation algorithm to compute p(xa) and output its value.
 * (x[i],f[i]):given points and n+1 are number of points
 * Ln,k(x)=l=summation of (x-x[i])/(x[k]-x[i]).
 * p(x)=ff=L(x)*f(x[k])
 */
#include <stdio.h>
//#include <conio.h>
//#include <math.h>

int main()
{
   double x[30],f[30],l,ff,xa[30],xb;
   int i,k,n ,n1 ,j ;
 
   scanf("n=%d n1=%d",&n,&n1);
   getch();
 
   for(k=0;k<=n;k++)
   {
      scanf("%lf %lf",&x[k],&f[k]);
      getch();
   }
 
   for(k=0;k<=n1;k++)
   {
      scanf("%lf",&xa[k]);
      getch();
      //printf("The value of p(%.4lf)\n",xa[k]);
   }
 
   for (j=0;j<=n1;j++)
   {
        xb=xa[j];
        //printf("The value of p(%.4lf)\n",xb);
        ff=0.0;
        for(k=0;k<=n;k++)
        {
            l=1.0;
            for(i=0;i<=n;i++)
            {
           if(i !=k)
           {
              l=l*(xb-x[i])/(x[k]-x[i]);
               getch();
           }
            }
        ff=ff+l*f[k];
        }
        printf("The value of p(%.4lf)=%.4lf\n",xb,ff);
        printf("The value of f(%.4lf)=%.4lf\n",xb,(xb*xb*xb-4*xb));
        printf("The value of | p(%.4lf)-f(%.4lf)|=%.6lf\n",xb,xb, ((xb*xb*xb-4*xb)-ff) );
   }
   return 0;
}

輸入資料
n=3 n1=3
0.0  0.0
1.0  -3.0
2.0  0.0
3.0  15.0
0.5
1.5
2.5
3.5

輸出畫面
The value of p(0.5000)=-1.8750
The value of f(0.5000)=-1.8750
The value of | p(0.5000)-f(0.5000)|=0.000000
The value of p(1.5000)=-2.6250
The value of f(1.5000)=-2.6250
The value of | p(1.5000)-f(1.5000)|=0.000000
The value of p(2.5000)=5.6250
The value of f(2.5000)=5.6250
The value of | p(2.5000)-f(2.5000)|=0.000000
The value of p(3.5000)=28.8750
The value of f(3.5000)=28.8750
The value of | p(3.5000)-f(3.5000)|=0.000000