Java编程题解读

1、1计算 1/1 + 1/2+1/3+ .+1/100 的值。程序题Console £3|_2_. Froblems (® Javadoc ； Declaration I*. Froject Migrationtermiriated Demiol |Java. Application F ； NlyEclipEe FrofesEi ojialbinaryVcom. euil. jai:53197377517639621+CZ04112267徐延迟代码如下：main( Stri ng args) public class Demol public static void dou

2、ble sum = 0; double n 二 1; while (n <= 100) sum += 1 / n;n+;System. out .println(sum);System. out .println(System. out .println(System. out .println("+" );"CZ04112267");"徐延迟");11计算 1/1 + 1/2+1/3+ .+1/100 的值。#1计算 1/1 + 1/2+1/3+ .+1/100 的值。2.设计计算阶乘的函数为：m (n) -n!。设计一个计算

3、阶乘的方法和相应的类，并利用这个Bemo2 Jiva Appliciti F'MyEclipse Pre类计算4的阶乘。#1计算 1/1 + 1/2+1/3+ .+1/100 的值。24+CZ04112267徐延迟代码如下:int n) public class Demo2 public static int jieChe ng( if (n = 1 | n = 0)#return n; elsereturn n * jieChe ng (n - 1);public static void main( Stri ng args) System. out .println(System.

4、 out .println(System. out .println(System. out .println(jieChe ng(4);"+" ); "CZ04112267");"徐延迟");3编写一个应用程序，可以计算result=1!+2!+100!。并输出。旦 Console £3 更 Problems® JavadocBeclarati 电»Java Application) F:MyEelipse Prof-125961703+CZ04112267徐延迟代码如下:public class De

5、mo3 public static void main( Stri ng args) System. out .println(System. out .println(System. out .println(sum(100);"+" )；"CZ04112267");System. out .println( "徐延迟");public static int sum( int n)int sum=0;while (n >=0)sum=sum+ jieche ng(n);n-；return sum;public static i

6、nt jieche ng(int n)if (n=1|n=0)return n; else returnn* jiecheng(n-1);4.将整数数组a赋初值（含8个元素），并求出a中的最大元素和最大元素的下标。曰 Consoleerm in ate Bemo4 Jura. Appli cation F：MlyEclip 生电 Profess 最大前元素为其下标为厂7+CZ04112267徐延迟代码如下：public class Demo4 publicstaticvoidmain( Stri ng args) int arr = 9, 5, 8, 6, 4, 2, 3, 18 ;intma

7、x In dex = 0, max = arr0;for (int i = 0; i < arr. if (arri > max) maxln dex = i;System. out .println(len gth ; i+) "最大的兀素为：” + arrmaxIndex+ "其下标为：” + maxIndex);System. out .println("+" );System. out .println("CZ04112267");System. out .println("徐延迟"）;5.从一

8、堆数组中找出最大的元素和最小的元素旦 CoiLEole 茨憶Problems Javadoctermiinat ed)* BemoS Java Appli cati on F: VMyEclips最大值为=5656最小值为:4+4-+CZ04112267徐延迟代码如下；public class Demo5 public static void main( Stri ng args) int arr = 33,56,9, 46,99,5656,5, 6, 4, 9 ;len gth ; i+) int max, min; max = min = arr0;for (int i = 0; i &l

9、t; arr. if (arri > max) max = arri;if (arri < mi n) min = arri;System. out .println( "+ min);System. out .println(System. out .println(System. out .println("最大值为："+ max + "n最小值为"+" );"CZ04112267");"徐延迟");6计算二维数组中各行元素之和并查找其值最大的那个行，假定元素值都为正整数。4曰

10、Console 芙Problems Javadoc倍 Declaration Froject Mligrationft电EnunaiiL邕dD电moE Java Application F" ''iMyEclipse Pro£ess;i onlinaryVcom. sun第1行的数组之后为：182第2行的数组之后为：281第3行的数组之后为：309第4行的数组之后为：317 最大一行为：4+CZ04112267徐延迟代码如下:public class Demo6 public static void main( Stri ng args) int arr =

11、 11,23, 66, 77, 5 , 1,23, 5, 66, 4 , 5, 6, 8, 9 , 1, 3, 4 ;intsum = 0;intmaxSum =0;intin dexI = 0;for(int i =(0; i < arr.len gth ; i+) int j;for (j = 0; j < arri.len gth; j+) sum += arrij;if (sum > maxSum) maxSum = sum;in dexl = i;"行的数组之后System. out .println( "第"+ (i+1) +为：&q

12、uot;+ sum);System. out .println("最大一行为:"+ (indexI+1);5#System. out .println(System. out .println();"+""CZ04112267");System. out .println( "徐延迟");#S*_ Frobl ems& Jivadce 風 Declaration 垢IsmoT Javi Application F：训yEcl ipse Frofessi24+CZ04112267 徐延迟7.定义一个计算n!的

13、类Fact，然后在主类 Classi中创建一个对象 N，求解4!的值。代码如下public class Demo7 public static void main( Stri ng args) Fact N= new Fact();System. out .println(N.jiecheng(4);System. out .println(System. out .println(System. out .println("+" );"CZ04112267");"徐延迟"); classFact public int jieche

14、ng( int n) if (n = 0 | n = 1) return n; else retur n n * jieche ng(n - 1);8.编写一个教师类 Teacher要求：(1)教师类Teacher属性有： name : String类对象，代表姓名 age : int型，代表年龄sex :boolean型，代表性别(其中:true表示男，false表示女)7phone :long型，代表联系电话教师类Teacher方法有：Teacher(String n,inta,Booleans,long p):有参构造函数，形象表中的 参数分别初始化姓名、年龄、性别和联系电话。in tg

15、etAge():获取年龄作为方法的返回值boolea ngetSex():获取性别作为方法的返回值 public String toStri ng():以姓名：联系电话的形式作为方法的返回值。凰 Console 滋 此 Frobltms佩 Javadoc屬onProject NligrJav Applicaticti F:?rofessi onalbinaryco徐延迟：123456+CZ04112267徐延迟代码如下：public class Demo8 private Stringname = "徐延迟”； 姓名private int age ;/ 年龄privateboolea

16、nsex ; / 代表性另廿(其中：true 表示男，false表示女)private long phone =123456; / 联系电话publicStri ng getName() return n ame;public void setName(Stri ng n ame) this . name = name;public int getAge() return age ;publicvoid setAge(int age) this . age = age;public boolea n getSex() return sex ;publicvoidsetSex( boolea n

17、sex) this.sex=sex;publiclonggetPho ne() returnphone ;publicvoidsetPh one( longphone)this . phone = phone; publicString toString()retur nn ame+" :" +ph onepublic static void main( Stri ng args) Demo8 d= new Demo8();System. out .println(d.toString();System. out .println(System. out .println(

18、);"+""CZ04112267");System. out .println( "徐延迟");9定义一个复数类 ComplexNumber，其中定义属性来表示复数的实部和虚部，并实现复数的的相加和相减的基本方法曰 Console :直 Frobi am5 Jaradoc冏 D«dar alio让mi in鱼 t 电&Demo9 Java Applicati on F:MyEclips& Frofeb+ c =5.Oi+9.0b- c =5.01-9.0CZ04112267徐延迟public class De

19、mo9 / mai n 方法public static void main( Stri ng a) Complexl b =new Complex1(2, 5);Complexl c =new Complex1(3, 4);"b+ c ="+n);"b- c ="+n);"+" "CZ04112267"););System. out .println( b.add(c).m+"i+" +b.add(c).System. out .println( b.add(c).m+"i-"

20、;+b.add(c).System. out .println(System. out .println(System. out .println( "徐延迟");/ Complex 类class Complex1 publicdoublem/实部publicdoublen;/虚部public Complex1( double m, double n) this . m = m;this . n 二 n;/ addpublic Complex1 add(Complex1 c) retur nnew Complex1( m+ c.m, n + c. n);/jia npubl

21、icComplex1 jia n( Complex1 c) retur nnew Complex1( m- c.m, n - c. n);10. 编写一个学生类 Stude nt要求:(1)学生类Student属性有： id : long型代表学号name : String类对象，代表姓名age : int型，代表年龄学生类Student方法有：Stude nt(lo ng l ,Stri nga, i nt b)有参构造函数，形参表中的参数分别初始化学号、姓名、年龄。in tgetAge():获取年龄作为方法的返回值。public Stri ng toStri ng() :以姓名、学号的形式

22、作为方法的返回 曰 CcrtsflleterminatJava. Appliea.tion| F： HyEclipxe徐延迟、+4_4_+(+CZ04112267|徐延迟代码如下；public class DemolO privatelong id ;privateStr ing n ame;privateintage ;publiclong getId() return id ;publicvoid setId(long id) this . id = id;publicStri ng getName() returnn ame;public void setName(Stri ng n a

23、me) this . name = name;public int getAge() return age ;publicvoid setAge(int age) this . age = age;public Demo10( long id, String name,int age) super ();this . id = id;this . name = name;this . age = age; OverridepublicStri ng toStri ng() returnname+"、" +idpublicstaticvoid main( Stri ng ar

24、gs) Demo10 d=new Demo10(110,"徐延迟",23);System. out.pri ntln( d.toStri ng();System. out .println(System. out .println(）；"+""CZ04112267");System. out .println( "徐延迟");11. 编程从三个整数x,y,z中选出最大者。E) Ccnsol# ProblenE ' 1 Javdo'(tarminatedy Dsmall LJwa Applicatio

25、n f TS5+CZO4112267徐延迟代码如下；public class Demo11 public static void main( Stri ng args) int max二 max(155, 23, 88);System. out .println(max);System. out .println(System. out .println("+""CZ04112267"););System. out .println( "徐延迟");public static int max( int a, int b, int c)

26、 return Math. max(Math. max(a, b), c);12. 设计一个交通工具类 vehicle，其数据成员包括速度 speed和种类bind方法包括设置颜色 setColor和取得颜色getColor，再设计一个子类 Car，增加属性passenger表示可容纳旅客的 人数，添加方法取得最大速度 getMaxSpeed。貝 ConsoleFr JavadocDeclarati on14.terminated. Car JAppli cati on Fz MyEclipse Professi onalHOkm/h +CZ04112267 徐延迟代码如下： package

27、t12;public class vehicle private int speed =110; privateStringkind ;public int getSpeed() returnspeed ;public void setSpeed( int speed) this . speed = speed;publicStri ng getKi nd() return kind ;public void setK in d(Stri ng kind) this . kind = kind; package t12;public class Carexte ndsvehicleprivat

28、e intpasse nger;public int getMaxSpeed() returnsuper .getSpeed();public static void main( Stri ng args) "km/h");Car c= new Car();System. out .println(c.getMaxSpeed()+13.定义一个Point类，含有一个print方法，再定义一个类Point3d，继承自父类并覆盖父类的同名方法print()。在main方法中分别调用父类和子类的print().旦备 也 Problems血 Javadocterminated>

29、; Pointed Java Application F:MyEcliSystem. out .println(System. out .println(System. out .println("+" )； "CZ04112267");"徐延迟");Point 类,point3d point方法 +CS04112267徐延迟代码如下：package t13;public class Point public void prin t()System. out .println("point方法”); package t13;

30、public classPoin t3dexte nds PointOverridepublic void prin t() System. out .println("point3d");public staticvoid main( Stri ng args) 15Poi nt p1 =new Poin t();p1.pri nt();System.out.pr in tl n("+System.out.pr in tl n("CZ04112267");System.out.pr intln("徐延迟");Poin t3

31、d p=new Poi nt3d();p.pr in t(););16 Console £3化 Frobl颐s他 JavadocDecler ati onTestAppli c&ti on F：Frofessi one14.定义一个 shape抽象类，它包含一个抽象方法getArea(),在shape类上派生出 Rectangle和Circle类，两者都用getArea()方法计算对象的面积。定义一个测试类Test,在该类中计算 半径为2的圆的面积和宽度，高度分别为4和3的矩形的面积。Rectangle area is 12Circle area is 12.56636+CZ

32、04112267 徐延迟代码如下：package t14;public abstract class shape public abstract void getArea();package t14;public class Circle exte nds shape private int r;private finaldoublePI =3.14159;publicCircle(int r) this . r = r;Overridepublic void getArea() double sum= r * r * PI ;System. out .println("Circle

33、 area is+sum);package t14;public class Recta ngle privateinta;privateintb;publicRecta ngle(intthis.a =a;this.b =b;Overrideexte nds shape a, int b) 18public void getArea() int sum = a * b; System. out .println("Recta ngle area is+ sum);package t14;public class Test public static voidmain( Stri n

34、g args) Recta ngle r = r.getArea();Circle c = c.getArea();new Rectangle(4, 3);new Circle(2);System. out .println(System. out .println("+" );"CZ04112267");19System. out .println( "徐延迟");15编写Account类模拟银行储蓄账户，一个储蓄帐号有帐号、余额、年利息等属性，有存 款、取款、查询方法。取款方法中，假设储蓄帐号不允许透支。Account Java

35、Application Y：底歎，余额为: 取歎:100 取款成功!杳询：余额为：1900+CZ04112267徐延迟代码如下；package t15;public class Acco unt privatelongCardIdJprivateintbala nee= 1000;/余额privatedouble accrual ; /年利息/ 存款public int deposit( int m) this . balance += m;return balance ;/ 取款public void qukuan( int m) if ( this . balance <= 0&am

36、p;& this . balance <m) System. out .println( " 余额不足！ " ); else if (m<= this . balance )this . balance -= m;System. out .println( " 取款成功！ " );/ 查询public void query()System. out .println( " 余额为： " +this . balance );public staticvoid main(String args) " 存款，余额为：" 取款： 100n" );Account a=new Account();System. out .println( " +a.deposit(1