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24点速算java代码的简单介绍
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java算24点代码：输入4个数算24点，能够在命令提示符下就可以运行。100多
import java.util.Scanner;
/** 给定4个数字计算24 */
public class Core {
private double expressionResult = 24;
// private int maxLine=10;
private boolean error = true;
private double numbers[] = new double[4];
public Object resultReturn;
/**
* 该对象拥有3个私有变量 expressionResult,所需结果 maxLine,输出结果每页行数 error,是否出错
* numbers[4],记录用来运算的4个数
*
* 其次,该对象拥有以下方法供外部调用 setNumbers(double[] 运算的数) 输入用来运算的数,4个时才能计算,无返回
* setMaxLine(int 行数) 输入每页的行数,无返回 getMaxLine() 返回每页的行数,类型为int
* setExpressionResult(double 所需结果) 输入所需结果,无返回 getExpressionResult()
* 返回所需结果,类型为double getExpression() 返回可得出所需结果的表达式,类型为字符串数组
*
* 最后,私有方法均为计算与表达式转换部分
*/
// 测试使用
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int[] arr = new int[4];
System.out.print("输入第一个数：");
arr[0] = scanner.nextInt();
System.out.print("输入第二个数：");
arr[1] = scanner.nextInt();
System.out.print("输入第三个数：");
arr[2] = scanner.nextInt();
System.out.print("输入第四个数：");
arr[3] = scanner.nextInt();
Core s = new Core();
s.setNumbers(arr);
String[] output = s.getExpression();
for (int i = 0; i output.length; i++) {
System.out.println(output[i]);
}
}
/** 设定被计算的四个数，由于是数组，所以具有容错功能(不为4个数) */
public void setNumbers(double[] n) {
if (n.length == 4) {
error = false;
numbers = n;
} else
error = true;
}
public void setNumbers(int[] n) {
if (n.length == 4) {
error = false;
for (int i = 0; i 4; i++) {
numbers[i] = n[i];
}
} else
error = true;
}
/** 设定每页显示的行数 */
// public void setMaxLine(int n) {
// if (n0) {
// maxLine=n;
// }
// }
// /** 返回每页显示的行数 */
// public int getMaxLine() {
// return maxLine;
// }
/** 设定需要得到的结果 */
public void setExpressionResult(double n) {
expressionResult = n;
}
/** 返回所需结果 */
public double expressionResult() {
return expressionResult;
}
/** 返回符合条件的表达式 */
public String[] getExpression() {
if (!error) {
String[] expression = calculate(numbers);
return expression;
} else
return new String[] { "出错了,输入有误" };
}
/** cal24()，输出结果为24的表达式 */
private String[] calculate(double[] n) {
if (n.length != 4)
return new String[] { "Error" };
double[] n1 = new double[3];
double[] n2 = new double[2];
String[] resultString = new String[1024]; // 最多1000组解,暂时未溢出
int count = 0;
boolean isRepeat = false;
for (int t1 = 0; t1 6; t1++) {
for (int c1 = 0; c1 6; c1++) {
for (int t2 = 0; t2 3; t2++) {
for (int c2 = 0; c2 6; c2++) {
for (int c3 = 0; c3 6; c3++) {
if ((c1 / 3 == c2 / 3 (c1 % 3) * (c2 % 3) != 0)
|| (c2 / 3 == c3 / 3 (c2 % 3) * (c3 % 3) != 0)
|| (c1 / 3 == c3 / 3
(c1 % 3) * (c3 % 3) != 0 t2 == 2)) {
// 去除连减连除的解,因为x/(y/z)=x*z/y
continue;
}
n1 = cal1(n, t1, c1);
n2 = cal2(n1, t2, c2);
double result = cal(n2[0], n2[1], c3);
if ((result - expressionResult) 0.00000001
(expressionResult - result) 0.00000001) {
resultString[count] = calString(n, t1, c1, t2,
c2, c3)
+ "=" + (int) expressionResult;
for (int i = 0; i count; i++) {
isRepeat = false;
if (resultString[i]
.equals(resultString[count])) { // 去除完全重复的解
isRepeat = true;
break; // 提前退出循环
}
}
if (c1 == c2 c2 == c3 c1 % 3 == 0
t1 + t2 != 0) { // 连加连乘
isRepeat = true;
}
if (!isRepeat) {
count++;
}
}
}
}
}
}
}
if (count == 0)
return new String[] { "该组数无解" };
String[] resultReturn = new String[count];
System.arraycopy(resultString, 0, resultReturn, 0, count);
return resultReturn;
}
/** cal1()，将4个数计算一次后返回3个数 */
private double[] cal1(double[] n, int t, int c) { // t为原来的t1，c为原来的c1
double[] m = new double[3];
switch (t) {
case 0:
m[1] = n[2];
m[2] = n[3];
m[0] = cal(n[0], n[1], c);
break;
case 1:
m[1] = n[1];
m[2] = n[3];
m[0] = cal(n[0], n[2], c);
break;
case 2:
m[1] = n[1];
m[2] = n[2];
m[0] = cal(n[0], n[3], c);
break;
case 3:
m[1] = n[0];
m[2] = n[3];
m[0] = cal(n[1], n[2], c);
break;
case 4:
m[1] = n[0];
m[2] = n[2];
m[0] = cal(n[1], n[3], c);
break;
default:
m[1] = n[0];
m[2] = n[1];
m[0] = cal(n[2], n[3], c);
}
return m;
}
/** cal2()，将3个数计算一次后返回2个数 */
private double[] cal2(double[] n, int t, int c) { // t为原来的t2，c为原来的c2
double[] m = new double[2];
switch (t) {
case 0:
m[1] = n[2];
m[0] = cal(n[0], n[1], c);
break;
case 1:
m[1] = n[1];
m[0] = cal(n[0], n[2], c);
break;
default:
m[1] = n[0];
m[0] = cal(n[1], n[2], c);
}
return m;
}
/** cal()，将2个数计算后返回结果 */
private double cal(double n1, double n2, int c) { // n1,n2为运算数，c为运算类型
switch (c) {
case 0:
return n1 + n2;
case 1:
return n1 - n2;
case 2:
return n2 - n1;
case 3:
return n1 * n2;
case 4:
if (n2 == 0)
return 9999; // 使计算结果必不为24
else
return n1 / n2;
default:
if (n1 == 0)
return 9999; // 同上
else
return n2 / n1;
}
}
/** calString()，输出表达式 */
private String calString(double[] n, int t1, int c1, int t2, int c2, int c3) {
String[] nString = new String[4];
switch (t1) {
case 0:
nString[0] = calString2("" + (int) n[0], "" + (int) n[1], c1);
nString[1] = "" + (int) n[2];
nString[2] = "" + (int) n[3];
break;
case 1:
nString[0] = calString2("" + (int) n[0], "" + (int) n[2], c1);
nString[1] = "" + (int) n[1];
nString[2] = "" + (int) n[3];
break;
case 2:
nString[0] = calString2("" + (int) n[0], "" + (int) n[3], c1);
nString[1] = "" + (int) n[1];
nString[2] = "" + (int) n[2];
break;
case 3:
nString[0] = calString2("" + (int) n[1], "" + (int) n[2], c1);
nString[1] = "" + (int) n[0];
nString[2] = "" + (int) n[3];
break;
case 4:
nString[0] = calString2("" + (int) n[1], "" + (int) n[3], c1);
nString[1] = "" + (int) n[0];
nString[2] = "" + (int) n[2];
break;
default:
nString[0] = calString2("" + (int) n[2], "" + (int) n[3], c1);
nString[1] = "" + (int) n[0];
nString[2] = "" + (int) n[1];
}
if ((c2 / 3 c1 / 3 (t2 != 2 || c2 / 3 == c3 / 3))
|| ((c3 / 3 c1 / 3 + c2 / 3) t2 == 2)
|| (c3 == 1 c1 / 3 == 0)) // 特定情况下加上一个括号*****************************
nString[0] = '(' + nString[0] + ')';
switch (t2) {
case 0:
nString[0] = calString2(nString[0], "" + nString[1], c2);
nString[1] = nString[2];
break;
case 1:
nString[0] = calString2(nString[0], nString[2], c2);
break;
default:
nString[3] = nString[0];
nString[0] = calString2(nString[1], nString[2], c2);
nString[1] = nString[3];
}
if (c3 / 3 c2 / 3 || (c3 == 2 nString[0].indexOf('+') = 0)) // 特定情况下加上一个括号*****************************
nString[0] = '(' + nString[0] + ')';
return calString2(nString[0], nString[1], c3);
}
/** calString()，根据符号输出一部运算表达式 */
private String calString2(String n1, String n2, int c) {
switch (c) {
case 0:
return n1 + '+' + n2;
case 1:
return n1 + '-' + n2;
case 2:
return n2 + '-' + n1;
case 3:
return n1 + '*' + n2;
case 4:
return n1 + '/' + n2;
default:
return n2 + '/' + n1;
}
}
}
相关问答
Q1: 一个java面试题：计算24点游戏 编程
import java.util.Random;
public class test2
{
public static void main(String[] args)
{
Random random = new Random();
int a[] = new int[4];
for(int i=0; i4; i++)
{
a[i] = random.nextInt(13)+1;
}
for(int j=0; j4; j++)
{
System.out.println("第" + j +"个数：" + a[j]);
}
Calcula(a);
}
public static void Calcula(int[] a)
{
int add, sub, multi, div;
add = 0;
sub = 0;
multi = 0;
div = 0;
for(int i=0; i4; i++)
{
add = add + a[i];
sub = sub - a[i];
multi = multi * a[i];
div = div/a[i];
}
if(add==24)
{
System.out.println(a[0] + "+" + a[1] + "+" + a[2] + "+" + a[3]
+ "=" + add);
}
else if(sub==24)
{
System.out.println(a[0] + "-" + a[1] + "-" + a[2] + "-" + a[3]
+ "=" + sub);
}
else if(multi==24)
{
System.out.println(a[0] + "*" + a[1] + "*" + a[2] + "*" + a[3]
+ "=" + multi);
}
else if(div==24)
{
System.out.println(a[0] + "÷" + a[1] + "÷" + a[2] + "÷" + a[3]
+ "=" + div);
}
else
{
System.out.println("对不起，没有实现24点的数");
}
}
}
已编译通过~
Q2: 用JAVA如何算出24点?
24点的源代码,因该可以计算出4则运算24 public class Test24Point{ public static void main(String[] args){ int index = 0 ; int temp = 0 ; int totalSUC = 0 ; int numb[] = new int[4];//the first four numbers double num[][] = new double[36][3];//three numbers after calculating double total[] = new double[6];//the number after three steps of calculating double p[][] = new double[6][8]; double q[][] = new double[3][7]; //System.out.println(2465%108); //System.out.println(2465/108); System.out.println("\"a--b\"means\"b-a\""); System.out.println("\"a//b\"means\"b/a\"\n"); /* for(int h = 0; h = 9; h ++)//Get the first four numbers for calculating and store into the array numb[4]; for(int i = 0; i = 9; i ++) for(int j = 0; j = 9; j ++) for(int k = 0; k = 9; k ++){ numb[0] = h ; numb[1] = i ; numb[2] = j ; numb[3] = k ; }*/ for(int i = 0 ; i 4 ; i ++){ numb = Integer.parseInt(args); } for(int i = 0; i 3; i ++)//Get two of the four to calculate and then store the new number into the array p; for(int j = i + 1; j 4 ; j ++,temp ++){ p[temp][0] = numb + numb[j]; p[temp][1] = numb - numb[j]; p[temp][2] = numb[j] - numb; p[temp][3] = numb * numb[j]; if(numb[j] != 0) p[temp][4] = numb / (double)numb[j]; else p[temp][4] = 10000; if(numb != 0) p[temp][5] = numb[j] / (double)numb; else p[temp][5] = 10000;
Q3: 算24点 java代码
C的代码要吗？我对java不是很熟，我试着用java写下吧。给我点时间！
package test.cardgame;
public class BinaryTreeNode
{
private BinaryTreeNode leftSon=null;
private BinaryTreeNode rightSon=null;
private BinaryTreeNode parent=null;
private double data=0;
private int sign=-1;
public int getSign()
{
return sign;
}
public void setSign(int sign)
{
this.sign = sign;
}
public BinaryTreeNode(BinaryTreeNode parent,BinaryTreeNode leftSon,BinaryTreeNode rightSon)
{
this.parent=parent;
this.leftSon=leftSon;
this.rightSon=rightSon;
}
public BinaryTreeNode()
{
}
public BinaryTreeNode getLeftSon()
{
return leftSon;
}
public void setLeftSon(BinaryTreeNode leftSon)
{
this.leftSon = leftSon;
leftSon.setParent(this);
}
public BinaryTreeNode getParent()
{
return parent;
}
public void setParent(BinaryTreeNode parent)
{
this.parent = parent;
}
public BinaryTreeNode getRightSon()
{
return rightSon;
}
public void setRightSon(BinaryTreeNode rightSon)
{
this.rightSon = rightSon;
rightSon.setParent(this);
}
public boolean isLeaf()
{
return (this.leftSon==nullthis.rightSon==null);
}
public boolean isRoot()
{
return this.parent==null;
}
public double getData()
{
return data;
}
public void setData(double data)
{
this.data = data;
}
}
package test.cardgame;
import java.util.ArrayList;
public class CardGame
{
private ArrayListString expressions=new ArrayListString();
public void solute(ArrayListBinaryTreeNode nodes,double target)
{
//whether the root data equals target
if (nodes.size()==1)
{
if (nodes.get(0).getData()==target)
{
String expression=printBinaryTree(nodes.get(0));
addExpression(expression);
return;
}
}
for (int i=0;inodes.size();i++)
{
for (int j=0;jnodes.size();j++)
{
if (i==j)
{
continue;
}
for (int k=0;k4;k++)
{
BinaryTreeNode node=new BinaryTreeNode();
BinaryTreeNode leftSon=nodes.get(i);
BinaryTreeNode rightSon=nodes.get(j);
if (k==0)
{
node.setData(leftSon.getData()+rightSon.getData());
}
else if (k==1)
{
node.setData(leftSon.getData()-rightSon.getData());
}
else if (k==2)
{
node.setData(leftSon.getData()*rightSon.getData());
}
else if (k==3)
{
if (rightSon.getData()==0)
{
continue;
}
node.setData(leftSon.getData()/rightSon.getData());
}
node.setLeftSon(leftSon);
node.setRightSon(rightSon);
node.setSign(k);
ArrayListBinaryTreeNode clonedArrayList=cloneArrayList(nodes);
//remove nodes from the tree
clonedArrayList.remove(leftSon);
clonedArrayList.remove(rightSon);
clonedArrayList.add(node);
solute(clonedArrayList,target);
}
}
}
}
public void printResult()
{
for (int i=0;iexpressions.size();i++)
{
System.out.println("Solution "+i+": "+expressions.get(i));
}
}
private void addExpression(String expression)
{
if (expressions.contains(expression))
{
return;
}
expressions.add(expression);
}
private ArrayListBinaryTreeNode cloneArrayList(ArrayListBinaryTreeNode source)
{
ArrayListBinaryTreeNode result=new ArrayListBinaryTreeNode();
for (int i=0;isource.size();i++)
{
result.add(source.get(i));
}
return result;
}
private String printBinaryTree(BinaryTreeNode resultRoot)
{
if (resultRoot.isLeaf())
{
return doubleToString(resultRoot.getData());
}
else
{
String expression="(";
expression+=printBinaryTree(resultRoot.getLeftSon());
int sign=resultRoot.getSign();
if (sign==0)
{
expression+="+";
}
else if (sign==1)
{
expression+="-";
}
else if (sign==2)
{
expression+="*";
}
else if (sign==3)
{
expression+="/";
}
expression+=printBinaryTree(resultRoot.getRightSon());
expression+=")";
return expression;
}
}
private String doubleToString(double value)
{
int intValue=(int)value;
if (value==intValue)
{
return String.valueOf(intValue);
}
else
{
return String.valueOf(value);
}
}
public BinaryTreeNode buildBinaryTreeNode(double value)
{
BinaryTreeNode node=new BinaryTreeNode();
node.setData(value);
return node;
}
public static void main(String[] args)
{
CardGame cardGame=new CardGame();
ArrayListBinaryTreeNode nodes=new ArrayListBinaryTreeNode();
nodes.add(cardGame.buildBinaryTreeNode(4));
nodes.add(cardGame.buildBinaryTreeNode(6));
nodes.add(cardGame.buildBinaryTreeNode(1));
nodes.add(cardGame.buildBinaryTreeNode(1));
cardGame.solute(nodes, 24);
cardGame.printResult();
}
}
Q4: 急求 JAVA的 算24点的代码！进来看！有特别要求！
package ceshi;
/** 给定4个数字计算24 */
public class Core {
private double expressionResult = 24;
// private int maxLine=10;
private boolean error = true;
private double numbers[] = new double[4];
public Object resultReturn;
/**
* 该对象拥有3个私有变量 expressionResult,所需结果 maxLine,输出结果每页行数 error,是否出错
* numbers[4],记录用来运算的4个数
*
* 其次,该对象拥有以下方法供外部调用 setNumbers(double[] 运算的数) 输入用来运算的数,4个时才能计算,无返回
* setMaxLine(int 行数) 输入每页的行数,无返回 getMaxLine() 返回每页的行数,类型为int
* setExpressionResult(double 所需结果) 输入所需结果,无返回 getExpressionResult()
* 返回所需结果,类型为double getExpression() 返回可得出所需结果的表达式,类型为字符串数组
*
* 最后,私有方法均为计算与表达式转换部分
*/
// 测试使用
public static void main(String[] args) {
Core s = new Core();
s.setNumbers(new int[] { 3, 4, 8, 6 });
String[] output = s.getExpression();
for (int i = 0; i output.length; i++) {
System.out.println(output[i]);
}
}
/** 设定被计算的四个数，由于是数组，所以具有容错功能(不为4个数) */
public void setNumbers(double[] n) {
if (n.length == 4) {
error = false;
numbers = n;
} else
error = true;
}
public void setNumbers(int[] n) {
if (n.length == 4) {
error = false;
for (int i = 0; i 4; i++) {
numbers[i] = n[i];
}
} else
error = true;
}
/** 设定每页显示的行数 */
// public void setMaxLine(int n) {
// if (n0) {
// maxLine=n;
// }
// }
// /** 返回每页显示的行数 */
// public int getMaxLine() {
// return maxLine;
// }
/** 设定需要得到的结果 */
public void setExpressionResult(double n) {
expressionResult = n;
}
/** 返回所需结果 */
public double expressionResult() {
return expressionResult;
}
/** 返回符合条件的表达式 */
public String[] getExpression() {
if (!error) {
String[] expression = calculate(numbers);
return expression;
} else
return new String[] { "出错了,输入有误" };
}
/** cal24()，输出结果为24的表达式 */
private String[] calculate(double[] n) {
if (n.length != 4)
return new String[] { "Error" };
double[] n1 = new double[3];
double[] n2 = new double[2];
String[] resultString = new String[1024]; // 最多1000组解,暂时未溢出
int count = 0;
boolean isRepeat = false;
for (int t1 = 0; t1 6; t1++) {
for (int c1 = 0; c1 6; c1++) {
for (int t2 = 0; t2 3; t2++) {
for (int c2 = 0; c2 6; c2++) {
for (int c3 = 0; c3 6; c3++) {
if ((c1 / 3 == c2 / 3 (c1 % 3) * (c2 % 3) != 0)
|| (c2 / 3 == c3 / 3 (c2 % 3) * (c3 % 3) != 0)
|| (c1 / 3 == c3 / 3
(c1 % 3) * (c3 % 3) != 0 t2 == 2)) {
// 去除连减连除的解,因为x/(y/z)=x*z/y
continue;
}
n1 = cal1(n, t1, c1);
n2 = cal2(n1, t2, c2);
double result = cal(n2[0], n2[1], c3);
if ((result - expressionResult) 0.00000001
(expressionResult - result) 0.00000001) {
resultString[count] = calString(n, t1, c1, t2,
c2, c3)
+ "=" + (int) expressionResult;
for (int i = 0; i count; i++) {
isRepeat = false;
if (resultString[i]
.equals(resultString[count])) { // 去除完全重复的解
isRepeat = true;
break; // 提前退出循环
}
}
if (c1 == c2 c2 == c3 c1 % 3 == 0
t1 + t2 != 0) { // 连加连乘
isRepeat = true;
}
if (!isRepeat) {
count++;
}
}
}
}
}
}
}
if (count == 0)
return new String[] { "该组数无解" };
String[] resultReturn = new String[count];
System.arraycopy(resultString, 0, resultReturn, 0, count);
return resultReturn;
}
/** cal1()，将4个数计算一次后返回3个数 */
private double[] cal1(double[] n, int t, int c) { // t为原来的t1，c为原来的c1
double[] m = new double[3];
switch (t) {
case 0:
m[1] = n[2];
m[2] = n[3];
m[0] = cal(n[0], n[1], c);
break;
case 1:
m[1] = n[1];
m[2] = n[3];
m[0] = cal(n[0], n[2], c);
break;
case 2:
m[1] = n[1];
m[2] = n[2];
m[0] = cal(n[0], n[3], c);
break;
case 3:
m[1] = n[0];
m[2] = n[3];
m[0] = cal(n[1], n[2], c);
break;
case 4:
m[1] = n[0];
m[2] = n[2];
m[0] = cal(n[1], n[3], c);
break;
default:
m[1] = n[0];
m[2] = n[1];
m[0] = cal(n[2], n[3], c);
}
return m;
}
/** cal2()，将3个数计算一次后返回2个数 */
private double[] cal2(double[] n, int t, int c) { // t为原来的t2，c为原来的c2
double[] m = new double[2];
switch (t) {
case 0:
m[1] = n[2];
m[0] = cal(n[0], n[1], c);
break;
case 1:
m[1] = n[1];
m[0] = cal(n[0], n[2], c);
break;
default:
m[1] = n[0];
m[0] = cal(n[1], n[2], c);
}
return m;
}
/** cal()，将2个数计算后返回结果 */
private double cal(double n1, double n2, int c) { // n1,n2为运算数，c为运算类型
switch (c) {
case 0:
return n1 + n2;
case 1:
return n1 - n2;
case 2:
return n2 - n1;
case 3:
return n1 * n2;
case 4:
if (n2 == 0)
return 9999; // 使计算结果必不为24
else
return n1 / n2;
default:
if (n1 == 0)
return 9999; // 同上
else
return n2 / n1;
}
}
/** calString()，输出表达式 */
private String calString(double[] n, int t1, int c1, int t2, int c2, int c3) {
String[] nString = new String[4];
switch (t1) {
case 0:
nString[0] = calString2("" + (int) n[0], "" + (int) n[1], c1);
nString[1] = "" + (int) n[2];
nString[2] = "" + (int) n[3];
break;
case 1:
nString[0] = calString2("" + (int) n[0], "" + (int) n[2], c1);
nString[1] = "" + (int) n[1];
nString[2] = "" + (int) n[3];
break;
case 2:
nString[0] = calString2("" + (int) n[0], "" + (int) n[3], c1);
nString[1] = "" + (int) n[1];
nString[2] = "" + (int) n[2];
break;
case 3:
nString[0] = calString2("" + (int) n[1], "" + (int) n[2], c1);
nString[1] = "" + (int) n[0];
nString[2] = "" + (int) n[3];
break;
case 4:
nString[0] = calString2("" + (int) n[1], "" + (int) n[3], c1);
nString[1] = "" + (int) n[0];
nString[2] = "" + (int) n[2];
break;
default:
nString[0] = calString2("" + (int) n[2], "" + (int) n[3], c1);
nString[1] = "" + (int) n[0];
nString[2] = "" + (int) n[1];
}
if ((c2 / 3 c1 / 3 (t2 != 2 || c2 / 3 == c3 / 3))
|| ((c3 / 3 c1 / 3 + c2 / 3) t2 == 2)
|| (c3 == 1 c1 / 3 == 0)) // 特定情况下加上一个括号*****************************
nString[0] = '(' + nString[0] + ')';
switch (t2) {
case 0:
nString[0] = calString2(nString[0], "" + nString[1], c2);
nString[1] = nString[2];
break;
case 1:
nString[0] = calString2(nString[0], nString[2], c2);
break;
default:
nString[3] = nString[0];
nString[0] = calString2(nString[1], nString[2], c2);
nString[1] = nString[3];
}
if (c3 / 3 c2 / 3 || (c3 == 2 nString[0].indexOf('+') = 0)) // 特定情况下加上一个括号*****************************
nString[0] = '(' + nString[0] + ')';
return calString2(nString[0], nString[1], c3);
}
/** calString()，根据符号输出一部运算表达式 */
private String calString2(String n1, String n2, int c) {
switch (c) {
case 0:
return n1 + '+' + n2;
case 1:
return n1 + '-' + n2;
case 2:
return n2 + '-' + n1;
case 3:
return n1 + '*' + n2;
case 4:
return n1 + '/' + n2;
default:
return n2 + '/' + n1;
}
}
}
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