c语言题目：拉格朗日插值法

今天编程学习网小编为大家讲解一道试题：c语言题目：拉格朗日插值法，希望对大家有所帮助！一起来看看解答思路吧：

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);
	

		}
	

以上就是“c语言题目：拉格朗日插值法”的详细内容，想要了解更多C语言教程欢迎持续关注编程学习网