newton.java

数值算法例子,牛顿插值,拉格朗日,三次样条

/*@author 邱宇峰 ,software project031
 *2004,9,20
 */
import java.awt.*;
import java.awt.event.*;

public class newton extends Frame implements MouseListener,ActionListener{
	static newton frm=new newton();
	static Button b1;
	static int array[][]=new int[30][2];
	static int X=0;
	static int count=0;
	static double p=0;	
	public static void main(String[] args){
		frm.setTitle("牛顿插值法");
		b1=new Button("click to start");
		frm.add(b1,"South");
		frm.setSize(500,500);
 		frm.setLocation(200,200);
		frm.setVisible(true);
		frm.addMouseListener(frm);
		b1.addActionListener(frm);
 		frm.addWindowListener(new WindowAdapter(){public void windowClosing(WindowEvent e){System.exit(0);}
		});
	}

		public void mouseEntered(MouseEvent e){	}
		public void mousePressed(MouseEvent e){	}
		public void mouseExited(MouseEvent e){	}
		public void mouseReleased(MouseEvent e){	}

	/*this mouseevent is to get the basic spots
	 */
	 					
		public void mouseClicked(MouseEvent e){
			Graphics g=getGraphics();
			int x=e.getX();
			int y=e.getY();
			array[count][0]=x;array[count][1]=y;
			g.drawString("。",x,y);
			count++;
		}
	
	/*this actionevent is to draw the line
	 */
	  
		public void actionPerformed(ActionEvent e){
			double arrayD[]=new double[count];
			Graphics g=getGraphics();
			creatA cA=new creatA(count);
			arrayD=cA.creatArrayD(array);
			for(X=array[0][0];X<array[count-1][0];X++){
				for(long h=0;h<5000000;h++);
				double p=arrayD[count-1];
				for(int i=count-2;i>=0;i--){
					p=arrayD[i]+p*((double)X-(double)array[i][0]);
				}
				g.drawString(".",X,(int)p);
			}
	}
}

/*this class is to calculate the constant number of 'a'
 */
 
class creatA{
	static double arrayD[];
	static int count;
	public creatA(int number){
		arrayD=new double[number];
		count=number;
	}
	
	double[] creatArrayD(int location[][]){
		for(int i=0;i<count;i++) arrayD[i]=(double)location[i][1];
		for(int i=1;i<count;i++){
			for(int j=count-1;j>=i;j--){
				arrayD[j]=((double)arrayD[j]-(double)arrayD[j-1])/((double)location[j][0]-(double)location[j-i][0]);
			}
		}
	return (arrayD);	
	}
}