实例87　拉格朗日插值

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
 
double LAG(int,double *,double *,double);
 
void main()
{
    int n;
    double *x,*y,t,lag;
    t = 0.15;
    n = 6;
    x = (double*)calloc(n,sizeof(double));
    if(x == NULL)
    {
        printf("内存分配失败\n");
        exit(1);
    }
    y = (double*)calloc(n,sizeof(double));
    if(y == NULL)
    {
        printf("内存分配失败\n");
        exit(1);
    }
    x[0] = 0;
    x[1] = 0.1;
    x[2] = 0.195;
    x[3] = 0.3;
    x[4] = 0.401;
    x[5] = 0.5;
 
    y[0] = 0.39894;
    y[1] = 0.39695;
    y[2] = 0.39142;
    y[3] = 0.38138;
    y[4] = 0.36812;
    y[5] = 0.35206;
 
    lag = LAG(n,x,y,t);
    printf("拉各朗日插值后得到的结果是：\n");
    printf("f(%.2f)=%e\n",t,lag);
    free(x);
    free(y);
}
 
double LAG(n,x,y,t)
int n;
double *x;
double *y;
double t;
{
    int i,j;
    double p,s;
    s = 0;
    for(i=0;i<n-1;i++)
    {
        p = 1;
        for(j=0;j<n-1;j++)
            if(i!=j)
                p*=(t-x[j])/(x[i]-x[j]);
            s+=p*y[i];
    }
    return (s);
}

trust100@ubuntu:~/test/clanguage$ ./a.out 
拉各朗日插值后得到的结果是：
f(0.15)=3.944728e-01