Java編程題解讀

1、1計算 1/1 + 1/2+1/3+ .+1/100 的值。程序題Console £3|_2_. Froblems (® Javadoc ； Declaration I*. Froject Migrationtermiriated Demiol |Java. Application F ； NlyEclipEe FrofesEi ojialbinaryVcom. euil. jai:53197377517639621+CZ04112267徐延遲代碼如下：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");"徐延遲");11計算 1/1 + 1/2+1/3+ .+1/100 的值。#1計算 1/1 + 1/2+1/3+ .+1/100 的值。2.設(shè)計計算階乘的函數(shù)為：m (n) -n!。設(shè)計一個計算

3、階乘的方法和相應(yīng)的類，并利用這個Bemo2 Jiva Appliciti F'MyEclipse Pre類計算4的階乘。#1計算 1/1 + 1/2+1/3+ .+1/100 的值。24+CZ04112267徐延遲代碼如下: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");"徐延遲");3編寫一個應(yīng)用程序，可以計算result=1!+2!+100!。并輸出。旦 Console £3 更 Problems® JavadocBeclarati 電»Java Application) F:MyEelipse Prof-125961703+CZ04112267徐延遲代碼如下: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( "徐延遲");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.將整數(shù)數(shù)組a賦初值（含8個元素），并求出a中的最大元素和最大元素的下標。曰 Consoleerm in ate Bemo4 Jura. Appli cation F：MlyEclip 生電 Profess 最大前元素為其下標為廠7+CZ04112267徐延遲代碼如下：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+) "最大的兀素為：” + arrmaxIndex+ "其下標為：” + maxIndex);System. out .println("+" );System. out .println("CZ04112267");System. out .println("徐延遲"）;5.從一

8、堆數(shù)組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最大值為=5656最小值為:4+4-+CZ04112267徐延遲代碼如下；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("最大值為："+ max + "n最小值為"+" );"CZ04112267");"徐延遲");6計算二維數(shù)組中各行元素之和并查找其值最大的那個行，假定元素值都為正整數(shù)。4曰

10、Console 芙Problems Javadoc倍 Declaration Froject Mligrationft電EnunaiiL邕dD電moE Java Application F" ''iMyEclipse Pro£ess;i onlinaryVcom. sun第1行的數(shù)組之后為：182第2行的數(shù)組之后為：281第3行的數(shù)組之后為：309第4行的數(shù)組之后為：317 最大一行為：4+CZ04112267徐延遲代碼如下: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;"行的數(shù)組之后System. out .println( "第"+ (i+1) +為：&q

12、uot;+ sum);System. out .println("最大一行為:"+ (indexI+1);5#System. out .println(System. out .println();"+""CZ04112267");System. out .println( "徐延遲");#S*_ Frobl ems& Jivadce 風(fēng) Declaration 垢IsmoT Javi Application F：訓(xùn)yEcl ipse Frofessi24+CZ04112267 徐延遲7.定義一個計算n!的

13、類Fact，然后在主類 Classi中創(chuàng)建一個對象 N，求解4!的值。代碼如下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");"徐延遲"); classFact public int jieche

14、ng( int n) if (n = 0 | n = 1) return n; else retur n n * jieche ng(n - 1);8.編寫一個教師類 Teacher要求：(1)教師類Teacher屬性有： name : String類對象，代表姓名 age : int型，代表年齡sex :boolean型，代表性別(其中:true表示男，false表示女)7phone :long型，代表聯(lián)系電話教師類Teacher方法有：Teacher(String n,inta,Booleans,long p):有參構(gòu)造函數(shù)，形象表中的 參數(shù)分別初始化姓名、年齡、性別和聯(lián)系電話。in tg

15、etAge():獲取年齡作為方法的返回值boolea ngetSex():獲取性別作為方法的返回值 public String toStri ng():以姓名：聯(lián)系電話的形式作為方法的返回值?；?Console 滋 此 Frobltms佩 Javadoc屬onProject NligrJav Applicaticti F:?rofessi onalbinaryco徐延遲：123456+CZ04112267徐延遲代碼如下：public class Demo8 private Stringname = "徐延遲”； 姓名private int age ;/ 年齡privateboolea

16、nsex ; / 代表性另廿(其中：true 表示男，false表示女)private long phone =123456; / 聯(lián)系電話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( "徐延遲");9定義一個復(fù)數(shù)類 ComplexNumber，其中定義屬性來表示復(fù)數(shù)的實部和虛部，并實現(xiàn)復(fù)數(shù)的的相加和相減的基本方法曰 Console :直 Frobi am5 Jaradoc冏 D«dar alio讓mi in魚 t 電&Demo9 Java Applicati on F:MyEclips& Frofeb+ c =5.Oi+9.0b- c =5.01-9.0CZ04112267徐延遲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( "徐延遲");/ Complex 類class Complex1 publicdoublem/實部publicdoublen;/虛部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. 編寫一個學(xué)生類 Stude nt要求:(1)學(xué)生類Student屬性有： id : long型代表學(xué)號name : String類對象，代表姓名age : int型，代表年齡學(xué)生類Student方法有：Stude nt(lo ng l ,Stri nga, i nt b)有參構(gòu)造函數(shù)，形參表中的參數(shù)分別初始化學(xué)號、姓名、年齡。in tgetAge():獲取年齡作為方法的返回值。public Stri ng toStri ng() :以姓名、學(xué)號的形式

22、作為方法的返回 曰 CcrtsflleterminatJava. Appliea.tion| F： HyEclipxe徐延遲、+4_4_+(+CZ04112267|徐延遲代碼如下；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,"徐延遲",23);System. out.pri ntln( d.toStri ng();System. out .println(System. out .println(）；"+""CZ04112267");System. out .println( "徐延遲");11. 編程從三個整數(shù)x,y,z中選出最大者。E) Ccnsol# ProblenE ' 1 Javdo'(tarminatedy Dsmall LJwa Applicatio

25、n f TS5+CZO4112267徐延遲代碼如下；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( "徐延遲");public static int max( int a, int b, int c)

26、 return Math. max(Math. max(a, b), c);12. 設(shè)計一個交通工具類 vehicle，其數(shù)據(jù)成員包括速度 speed和種類bind方法包括設(shè)置顏色 setColor和取得顏色getColor，再設(shè)計一個子類 Car，增加屬性passenger表示可容納旅客的 人數(shù)，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 徐延遲代碼如下： 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.定義一個Point類，含有一個print方法，再定義一個類Point3d，繼承自父類并覆蓋父類的同名方法print()。在main方法中分別調(diào)用父類和子類的print().旦備 也 Problems血 Javadocterminated>

29、; Pointed Java Application F:MyEcliSystem. out .println(System. out .println(System. out .println("+" )； "CZ04112267");"徐延遲");Point 類,point3d point方法 +CS04112267徐延遲代碼如下：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("徐延遲");Poin t3

31、d p=new Poi nt3d();p.pr in t(););16 Console £3化 Frobl頤s他 JavadocDecler ati onTestAppli c&ti on F：Frofessi one14.定義一個 shape抽象類，它包含一個抽象方法getArea(),在shape類上派生出 Rectangle和Circle類，兩者都用getArea()方法計算對象的面積。定義一個測試類Test,在該類中計算 半徑為2的圓的面積和寬度，高度分別為4和3的矩形的面積。Rectangle area is 12Circle area is 12.56636+CZ

32、04112267 徐延遲代碼如下：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( "徐延遲");15編寫Account類模擬銀行儲蓄賬戶，一個儲蓄帳號有帳號、余額、年利息等屬性，有存 款、取款、查詢方法。取款方法中，假設(shè)儲蓄帳號不允許透支。Account Java

35、Application Y：底歎，余額為: 取歎:100 取款成功!杳詢：余額為：1900+CZ04112267徐延遲代碼如下；package t15;public class Acco unt privatelongCardIdJprivateintbala nee= 1000;/余額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( " 余額不足！ " ); else if (m<= this . balance )this . balance -= m;System. out .println( " 取款成功！ " );/ 查詢public void query()System. out .println( " 余額為： " +this . balance );public staticvoid main(String args) " 存款，余額為：" 取款： 100n" );Account a=new Account();System. out .println( " +a.deposit(1