1、3.编写一个应用程序,可以计算result=1!+2!+100!。并输出。public class Demo3 System.out.println(sum(100);+ public static int sum(int n) int sum=0; while(n=0) sum=sum+jiecheng(n); n-; return sum; public static int jiecheng(int n) if(n=1|n=0) else return n*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。public class
2、Demo4 int arr = 9, 5, 8, 6, 4, 2, 3, 18 ; int maxIndex = 0, max = arr0; for (int i = 0; i max) maxIndex = i; 最大的元素为: + arrmaxIndex + 其下标为: + maxIndex);5. 从一堆数组中找出最大的元素和最小的元素代码如下;public class Demo5 int arr = 33, 56, 9, 46, 99, 5656, 5, 6, 4, 9 ; int max, min; max = min = arr0; max = arri; if (arri mi
3、n) min = arri;最大值为: + max + n最小值为: + min);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。public class Demo6 int arr = 11, 23, 66, 77, 5 , 1, 23, 5, 66, 4 , 5, 6, 8, 9 , 1, 3, 4 ; int sum = 0; int maxSum = 0; int indexI = 0; int j; for (j = 0; j maxSum) maxSum = sum; indexI = i; System.out.println(第 + (i+1) +
4、 行的数组之后为: + sum);最大一行为: + (indexI+1);+7. 定义一个计算n!的类Fact,然后在主类Class1中创建一个对象N,求解4!的值。代码如下public class Demo7 Fact N= new Fact(); System.out.println(N.jiecheng(4);+class Fact public int jiecheng(int n) if (n = 0 | n = 1) else return n * jiecheng(n - 1);8.编写一个教师类Teacher,要求:(1) 教师类Teacher属性有:name : String
5、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)phone :long型,代表联系电话(2) 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。intgetAge() :获取年龄作为方法的返回值booleangetSex() : 获取性别作为方法的返回值public String toString() : 以姓名:联系电话的形式作为方法的返回值。public class Demo8 private
6、String name=;/姓名 private int age; /年龄 private boolean sex;/代表性别(其中: private long phone=123456;/联系电话 public String getName() return name; public void setName(String name) this.name = name; public int getAge() return age; public void setAge(int age) this.age = age; public boolean getSex() return sex;
7、public void setSex(boolean sex) this.sex = sex; public long getPhone() return phone; public void setPhone(long phone) this.phone = phone; public String toString() return name+ : +phone; Demo8 d= new Demo8(); System.out.println(d.toString();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pu
8、blic class Demo9 / main方法 public static void main(String a) Complex1 b = new Complex1(2, 5); Complex1 c = new Complex1(3, 4); System.out.println( b+ c = + b.add(c).m+i+b.add(c).n);b- c =i-+/ Complex类class Complex1 public double m; / 实部 public double n; / 虚部 public Complex1(double m, double n) this.m
9、 = m; this.n = n; / add public Complex1 add(Complex1 c) return new Complex1(m + c.m, n + c.n); /jian public Complex1 jian(Complex1 c) return new Complex1(m - c.m, n - c.n);10. 编写一个学生类Student 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(2) 学生类Student方法有:Student(long l ,String
10、a, int 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。获取年龄作为方法的返回值。以姓名、学号的形式作为方法的返回值。public class Demo10 private long id; private String name; public long getId() return id; public void setId(long id) this.id = id; public Demo10(long id, String name, int age) super(); Override public String toString() 、+id; Demo10 d
11、= new Demo10(110, , 23); System.out.println(d.toString();11. 编程从三个整数x,y,z中选出最大者。public class Demo11 int max=max(155, 23, 88); System.out.println(max); System.out.println(+ public static int max(int a, int b, int c) 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s
12、etColor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。package t12;public class vehicle private int speed=110; private String kind; public int getSpeed() return speed; public void setSpeed(int speed) this.speed = speed; public String getKind() return kind; public void setKind(
13、String kind) this.kind = kind;public class Car extends vehicle private int passenger; public int getMaxSpeed() return super.getSpeed(); Car c= new Car(); System.out.println(c.getMaxSpeed()+km/h13. 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Point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().package t1
14、3;public class Point public void print()point 方法public class Point3d extends Point public void print() point3d Point3d p=new Point3d(); p.print(); Point p1= new Point(); p1.print();14. 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,
15、高度分别为4和3的矩形的面积。package t14;public abstract class shape public abstract void getArea();public class Circle extends shape private int r; private final double PI=3.14159; public Circle(int r) this.r = r; public void getArea() double sum=r*r*PI;Circle area is +sum);public class Rectangle extends shape p
16、rivate int a; private int b; public Rectangle(int a, int b) this.a = a; this.b = b; int sum = a * b;Rectangle area is public class Test Rectangle r = new Rectangle(4, 3); r.getArea(); Circle c = new Circle(2); c.getArea();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。packag
17、e t15;public class Account private long CardId; private int balance=1000;/ 余额 private double 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&this.balancem) 余额不足! else if(m=this.balance) this.balance -= m;取款成功! /查询 public void query()余额为:+this.balance); Account a= new Account(); System.out.
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