【TIJ4】第四章全部习题

第四章 没啥好说的......

4.1

 package ex0401;
 //[4.1]写一个程序打印从1到100的值

 public class PrintOneToHundred
 {
     public static void main(String[] args)
     {
         for(int i = 1; i <= 100; ++i)
             System.out.print(i + " ");
     }
 }

4.1

4.2-4.3

 package ex040203;
 //[4.2]写一个程序,产生25个int类型的随机数.对于每一个随机值,使用if-else语句来讲其分类为大于\小于,或等于紧随它产生的值
 //[4.3]修改练习2,把代码用一个while无限循环包括起来.然后运行它直至键盘中断其运行(通常是通过按Ctrl+C).

 import java.util.Random;

 public class GetRandomInteger
 {
     public static void main(String[] args)
     {
         int i = 0;
         Random rand = new Random();
         int curr = rand.nextInt();
         int prev = 0;
         //[4.2]
         while(++i < 25)
         //[4.3]
         //while(true)
         {
             prev = rand.nextInt();
             System.out.print(curr + " ");
             if(curr > prev)
                 System.out.print("greater than ");
             else if(curr < prev)
                 System.out.print("less than ");
             else
                 System.out.print("equals ");
             curr = prev;
         }
         System.out.print(prev);
     }
 }

4.2-4.3

4.4

 package ex0404;
 //[4.4]写一个程序,使用两个嵌套的for循环和取余操作符来探测和打印素数

 public class GetPrime
 {
     public static void main(String[] args)
     {
         System.out.println(2);
         for(int number = 3; number < 10000; number += 2)
         {
             int index;
             for(index = 3; index < number ; index += 2)
             {
                 if(number % index == 0)
                     break;
             }
             if(index * index > number)
                 System.out.println(number);
         }
     }
 }

4.4

4.5

 package ex0405;
 //重复练习[3.10],不要用Integer.toBinaryString()方法,而是用三元操作符合按位操作符来显示二进制的1和0

 public class AnotherBitOperation
 {
     public static void main(String[] args)
     {
         int first = 0x2AA;
         int second =0x133;

         for(int i = 0; ((first >> i) & 0xFFFF) != 0; ++i)
             System.out.print((first >> i ) & 1);
         System.out.println("");
         for(int i = 0; ((second >> i) & 0xFFFF) != 0; ++i)
             System.out.print((second >> i ) & 1);
         System.out.println("");
         for(int i = 0; (((first & second) >> i) & 0xFFFF) != 0; ++i)
             System.out.print(((first & second) >> i ) & 1);
         System.out.println("");
         for(int i = 0; (((first | second) >> i) & 0xFFFF) != 0; ++i)
             System.out.print(((first | second) >> i ) & 1);
         System.out.println("");
         for(int i = 0; (((first ^ second) >> i) & 0xFFFF) != 0; ++i)
             System.out.print(((first ^ second) >> i ) & 1);
         System.out.println("");
     }
 }

4.5

4.7

 package ex0407;
 //[4.7]修改本章练习1,通过使用break关键词,使得程序在打印到99时退出.然后尝试用return达到相同目的

 public class TestBreakAndReturn
 {
     public static void main(String[] args)
     {
         for(int i = 1; i <= 100; ++i)
         {
             if(i == 100)
                 return;
                 //break;
             System.out.print(i + " ");
         }
     }
 }

4.7

4.8

 package ex0408;
 //[4.8]写一个switch开关语句,为每个case打印一个消息.然后把这个switch放进for循环来测试每个case.
 //先让每个case后面都有break,测试一下会怎么样;然后把break删除,看看会怎样.

 public class TestSwitch
 {
     public static void main(String[] args)
     {
         for(int i = 0; i < 3; ++i)
         {
             switch(i)
             {
             case 0:
                 System.out.println("Cat");//break;
             case 1:
                 System.out.println("Dog");//break;
             case 2:
                 System.out.println("Bird");//break;
             default:
                 break;
             }
         }
     }
 }

4.8

4.9

 package ex0409;
 //[4.9]一个斐波那契数列是由数字从第三个数字其都是前两个数字的和.
 //创建一个方法,接收一个整数参数,并显示从第一个元素开始,总共有该参数指定的个数所构成的所有斐波那契数字.

 class Fibonacci
 {
     private static final int FIRST = 1;
     private static final int SECOND = 1;

     public static void printFibonacci(int length)
     {
         if(length <= 0)
             System.out.println("error!");
         switch(length)
         {
         case 2:
             System.out.println(FIRST);
         case 1:
             System.out.println(SECOND);break;
         default:
         {
             int prevprev = FIRST;
             int prev = SECOND;
             int curr;
             System.out.println(FIRST);
             System.out.println(SECOND);
             for(int i = 3; i <= length; ++i)
             {
                 curr = prevprev + prev;
                 System.out.println(curr);
                 prevprev = prev;
                 prev = curr;
             }
         }
         }
     }
 }
 public class ShowFibonacci
 {
     public static void main(String[] args)
     {
         Fibonacci.printFibonacci(10);
     }
 }

4.9