「Java题折算」java计算

博主:adminadmin 2022-11-25 04:48:05 53

今天给各位分享Java题折算的知识,其中也会对java计算进行解释,如果能碰巧解决你现在面临的问题,别忘了关注本站,现在开始吧!

本文目录一览:

JAVA编程题目,共四题,做其中一题就够了,

好了,你测试一下咯!

public interface ShapeArea {//定义ShapeArea接口

public double getArea();//double getArea( ):求一个形状的面积

public double getPerimeter();// Double getPerimeter( ):求一个形状的周长。

}

public class MyTriangle implements ShapeArea {

double x,y,z,s;//x,y,z :double型,表示三角形的三条边

public MyTriangle(double x, double y, double z) {//方法:MyTriangle(double x, double y, double z):构造函数,给三条边和s赋值;

this.x = x;

this.y = y;

this.z = z;

this.s = (x+y+z)/2;

}

@Override

public double getArea() {

return Math.sqrt(this.s*(this.s-this.x)*(this.s-this.y)*(this.s-this.z));

}

@Override

public double getPerimeter() {

return (x+y+z);

}

@Override

public String toString() {

System.out.print("此三角形的面积和周长为:");

return this.getArea()+"、"+this.getPerimeter();

}

}

public class Test {//测试类

public static void main(String[] args) {

MyTriangle myTriangle = new MyTriangle(3, 4, 5);

System.out.println(myTriangle);

}

}

求大神帮忙解答JAVA编程题目,谢谢!

import java.util.Scanner;

public class Test1 {

public static void main(String[] args) {

System.out.println("请输入会员积分:");

Scanner sc=new Scanner(System.in);

int a=sc.nextInt();

sc.close();

if(a200){

System.out.println("该会员享受的折扣是:0.9");

}else if(a=2000a=4000){

System.out.println("该会员享受的折扣是:0.8");

}else if(a=4000a=8000){

System.out.println("该会员享受的折扣是:0.7");

}else if(a=8000){

System.out.println("该会员享受的折扣是:0.6");

}

}

}

java编程题

1.package test;

import java.util.Iterator;

import java.util.Map;

import java.util.TreeMap;

public class Test {

public static void main(String[] args) {

String str = "afdjasdg$$jfsjdfjdjjdjdjdjdjdj";

int max=0;

Object chars=null;

Map tree = new TreeMap();

for (int i = 0; i str.length(); i++) {

char ch = str.charAt(i);

if ((ch = 1 ch = 255) ) {

if (!tree.containsKey(ch)) {

tree.put(ch, new Integer(1));

} else {

Integer in = (Integer) tree.get(ch) + 1;

tree.put(ch, in);

}

}

}

Iterator tit = tree.keySet().iterator();

while (tit.hasNext()) {

Object temp = tit.next();

if(max=Integer.parseInt(tree.get(temp)+""))

{

max=Integer.parseInt(tree.get(temp)+"");

chars=temp;

}

}

System.out.print(chars.toString() + "出现" + max + "次");

}

}

只要用assic码做范围就可以了.任何字符都可以过滤.

2.方法很多,hashmap或是arraylist,数组都可以的.就是对应关系而已.

package test;

public class ListTest {

static String[] to_19 = { "zero", "one", "two", "three", "four", "five",

"six", "seven", "eight", "nine", "ten", "eleven", "twelve",

"thirteen", "fourteen", "fifteen", "sixteen", "seventeen",

"eighteen", "nineteen" };

static String[] tens = { "twenty", "thirty", "forty", "fifty", "sixty",

"seventy", "eighty", "ninety" };

static String[] denom = { "", "thousand ", "million", "billion",

"trillion", "quadrillion", "quintillion", "sextillion",

"septillion", "octillion", "nonillion", "decillion", "undecillion",

"duodecillion", "tredecillion", "quattuordecillion",

"sexdecillion", "septendecillion", "octodecillion",

"novemdecillion", "vigintillion" };

public static void main(String[] argv) throws Exception {

long tstValue = 12345;

ListTest itoe = new ListTest();

System.out.println(itoe.english_number(tstValue));

}

private String convert_nn(int val) {

if (val 20) return to_19[val];

int flag = val / 10 - 2; if (val % 10 != 0)

return tens[flag] + "-" + to_19[val % 10];

else return tens[flag];

}

private String convert_nnn(int val) {

String word = "";

int rem = val / 100;

int mod = val % 100;

if (rem 0) {word = to_19[rem] + " hundred ";}

if (mod 0) {word = word + convert_nn(mod);}

return word;

}

public String english_number(long val) {

if (val 100) {System.out.println((int) val);return convert_nn((int) val);}

if (val 1000) {return convert_nnn((int) val); }

for (int v = 0; v denom.length; v++) {

int didx = v - 1;

long dval = new Double(Math.pow(1000, v)).longValue();

if (dval val) {

long mod = new Double(Math.pow(1000, didx)).longValue();

int l = (int) (val / mod);

long r = (long) (val - (l * mod));

String ret = convert_nnn(l) + " " + denom[didx];

if (r 0) {ret = ret + ", " + english_number(r);}

return ret;

}

}

return null;

}

}

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发布于:2022-11-25,除非注明,否则均为首码项目网原创文章,转载请注明出处。