[JAVA程式語言] 範例1-3 範例1-4 通過下面4點的Lagrange內插法求P(1.5)之值?

[JAVA程式語言] 範例1-4通過下面4點的Lagrange內插法求P(1.5)之值?
[JAVA程式語言] 範例1-3通過下面4點的Lagrange內插法求P(1.5)之值?

(1 , 0.0)  (2 , 0.693)  (3 , 1.099)  (4 , 1.386)
x               f(x)
==================
1.0              0.0
2.0              0.693
3.0              1.099
4.0              1.386
===================

/* ex1-4.java: 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])
 */
import java.util.Scanner;

public class Main {
    public static void main(String []args) {
        Scanner scanner = new Scanner(System.in);
        float[]  x, f ;
        float xa , l ,ff;
        int i,k,n;
        x = new float[10]; // 利用new指令產生物件
        f = new float[10]; // 利用new指令產生物件

        n=scanner.nextInt();
        xa=scanner.nextFloat();
        System.out.printf("n=%2d\txa=%2.2f",n,xa);
        System.out.println();
        for(k=0;k<=n;k++)
        {
            //Scanner scanner = new Scanner(System.in);
            x[k]=scanner.nextFloat();
            f[k]=scanner.nextFloat();
            System.out.printf("x[k]=%2.2f, f[k]=%2.2f",x[k],f[k]);
            System.out.println();
        }
     
        ff=0;
        for(k=0;k<=n;k++){
           l=1;
           i=0;
           do{
        if(i !=k) {
                l=l*(xa-x[i])/(x[k]-x[i]);
        }
        i=i+1;
        System.out.printf("l=%f n=%2d i=%2d",l, n, i);
                System.out.println();
           } while (i<=n);
        ff=ff+l*f[k];
        System.out.printf("l=%f k=%2d The value of p(%f)=%f",l, k,xa ,ff);
        System.out.println();
        }
        System.out.printf("The value of p(%f)=%f",xa ,ff);
    }
}

STDIN資料

3    1.5
1.0  0.0
2.0  0.693
3.0  1.099
4.0  1.386

輸出畫面

$javac Main.java
$java -Xmx128M -Xms16M Main
n= 3 xa=1.50
x[k]=1.00, f[k]=0.00
x[k]=2.00, f[k]=0.69
x[k]=3.00, f[k]=1.10
x[k]=4.00, f[k]=1.39
l=1.000000 n= 3 i= 1
l=0.500000 n= 3 i= 2
l=0.375000 n= 3 i= 3
l=0.312500 n= 3 i= 4
l=0.312500 k= 0 The value of p(1.500000)=0.000000
l=0.500000 n= 3 i= 1
l=0.500000 n= 3 i= 2
l=0.750000 n= 3 i= 3
l=0.937500 n= 3 i= 4
l=0.937500 k= 1 The value of p(1.500000)=0.649688
l=0.250000 n= 3 i= 1
l=-0.125000 n= 3 i= 2
l=-0.125000 n= 3 i= 3
l=-0.312500 n= 3 i= 4
l=-0.312500 k= 2 The value of p(1.500000)=0.306250
l=0.166667 n= 3 i= 1
l=-0.041667 n= 3 i= 2
l=0.062500 n= 3 i= 3
l=0.062500 n= 3 i= 4
l=0.062500 k= 3 The value of p(1.500000)=0.392875
The value of p(1.500000)=0.392875