拉格朗日插值法

示例源代码如下 test.c : 

#include <stdio.h>
 
int main()
{
    static float Lx[10],Ly[10];
    int n,i,j;
    float x,y=0,p;

    printf("enter n=");
    scanf("%d",&n);
 
    printf("enter xi\n");
    for(i=1;i<=n;i++)
        scanf("%f",&Lx[i]);
 
    printf("enter yi\n");
    for(i=1;i<=n;i++)
        scanf("%f",&Ly[i]);
 
    printf("enter x=");
    scanf("%f",&x);
    
    for( i=1;i<=n;i++ )
    {
        p=1;
        for( j=1;j<=n;j++ )
        {
            if(i!=j)
                p=p*(x-Lx[j])/(Lx[i]-Lx[j]);
        }

        y=y+p*Ly[i];
    }
 
    printf("y=%f\n",y);
    return 0;
}

Makefile:

test:
	gcc test.c -o test
clean:
	rm -rf test

编译并运行过程

# make
# ./test
enter n=4
enter xi
1
2
3
4
enter yi
2.3
4.3
5.5
7.8
enter x=2.5
y=4.881250