求出e=1+1/1!+1/2!+1/3!+……+1/n!+……的近似值的java applet程序-JSP教程,Java技巧及代码

//求出e=1+1/1!+1/2!+1/3!+……+1/n!+……的近似值,要求误差小于0.0001import java.applet.*;import java.awt.*;import java.awt.event.*;public class at1_1 extends applet implements actionlistener{  textfield text1; button button1;  public void init() {  text1 = new textfield("0",10);  button1 = new button("清除");  add(text1);  add(button1);  text1.addactionlistener(this);  button1.addactionlistener(this); } public void start(){} public void stop(){} public void destory(){} public void paint(graphics g) {   g.drawstring("在文本区输入数字n后回车",10,100);  g.drawstring("文本区显示1+1/1!+1/2!+1/3!+……+1/n!+……的近似值",10,120); } public void actionperformed(actionevent e) {  if(e.getsource()==text1)  {   double sum=1,a=1;   int i=1;   int n=0;   try   {   n = integer.valueof(text1.gettext()).intvalue();   while(i<=n)   {     a = a*(1.0/i);    sum = sum + a;    i=i+1;   }   sum=sum*10000;   sum=math.round(sum);   sum=sum/10000;   text1.settext(""+sum);     }   catch(numberformatexception event)   {   text1.settext("请输入数字字符");   }  }  else if(e.getsource()==button1)  {   text1.settext("0");  }   }}