例4.7弦截法求方程f(x)=x^3-5x^2+16x-80=0的根

 1 import java.util.Scanner;
 2 import java.lang.*;
 3 public class cjava {
 4     public static void main(String[] args) {
 5         double x1,x2,f1,f2,x;
 6         Scanner a=new Scanner(System.in);
 7         do
 8         {
 9             System.out.println("input x1,x2:");
 
10             x1=a.nextDouble();
11             x2=a.nextDouble();
12             f1=f(x1);
13             f2=f(x2);
14         }while(f1*f2>=0);
15         x=root(x1,x2);
16         System.out.println("A root of equation is "+x);
17     }
18     static double f(double x) {
19         double y;
20         y=x*x-5*x+16*x-80;
 
21         return y;
22     }
23     static double xpoint(double x1,double x2)
24     {
25         double y;
26         y=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1));
27         return y;
28     }
29     static double root(double x1,double x2)
30     {
31         double x,y,y1;
32         y1=f(x1);
33         do
 

34         {
35             x=xpoint(x1,x2);
36             y=f(x);
37         if(y*y1>0)
38         {
39             y1=y;
40             x1=x;
41         }
42         else
43             x2=x;
44     }while(Math.sqrt(y)>=0.00001);
45         return x;
46         }
47 }

例4.8遞迴的方法求n！

 1 import java.util.Scanner;
 2 import java.lang.*;
 3 public class cjava {
 4     public static void main(String[] args) {
 5         int n;
 6         long y;
 7         Scanner a=new Scanner(System.in);
 8         System.out.println("please input an integer:");
 9         n=a.nextInt();
10         y=fac(n);
11         System.out.println(n+"!="+y);
12     }
13     static long fac(int n) {
14         long f;
15         if(n<0)
16         {
17             System.out.println("n<0,data error!");
18             f=-1;
19         }
20         else if(n==0||n==1) f=1;
21         else f=fac(n-1)*n;
22         return f;
23     }
24 }

2.遇到的問題：c++中許多的用法java是沒有的

3.明天繼續寫例題