拉格朗日（Lagrange）插值法（C语言）

                                                         https://blog.csdn.net/landcruiser007/article/details/79314887

题目

拉格朗日插值法流程图

(1)二次插值：

#include<stdio.h>float lagelangri(float x[],float y[],float xx,int n){    int i,j;    float *a,yy=0;    a=new float[n];    for(i=0;i<=n-1;i++)    {        a[i]=y[i];        for(j=0;j<=n-1;j++)            if(j!=i)a[i]*=(xx-x[j])/(x[i]-x[j]);        yy+=a[i];    }    delete a;    return yy;}void main(){    float x[5]={-3.0,-1.0,1.0,2.0,3.0};    float y[5]={1.0,1.5,2.0,2.0,1.0};    float xx1=-2,xx2=0,xx3=2.75,yy1,yy2,yy3;    yy1=lagelangri(x,y,xx1,3);    yy2=lagelangri(x,y,xx2,3);    yy3=lagelangri(x,y,xx3,3);    printf("x1=%-20f,y1=%f\n",xx1,yy1);    printf("x2=%-20f,y2=%f\n",xx2,yy2);    printf("x3=%-20f,y3=%f\n",xx3,yy3);}

(2)五次插值：

#include<stdio.h>float lagelangri(float x[],float y[],float xx,int n){    int i,j;    float *a,yy=0;    a=new float[n];    for(i=0;i<=n-1;i++)    {        a[i]=y[i];        for(j=0;j<=n-1;j++)            if(j!=i)a[i]*=(xx-x[j])/(x[i]-x[j]);        yy+=a[i];    }    delete a;    return yy;}void main(){    float x[6]={0.30,0.42,0.50,0.58,0.66,0.72};    float y[6]={1.04403,1.08462,1.11803,1.15603,1.19817,1.23223};    float xx1=0.46,xx2=0.55,xx3=0.60,yy1,yy2,yy3;    yy1=lagelangri(x,y,xx1,6);    yy2=lagelangri(x,y,xx2,6);    yy3=lagelangri(x,y,xx3,6);    printf("x1=%-20f,y1=%f\n",xx1,yy1);    printf("x2=%-20f,y2=%f\n",xx2,yy2);    printf("x3=%-20f,y3=%f\n",xx3,yy3);}