java24点算法的简单介绍
今天给各位分享java24点算法的知识，其中也会对进行解释，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：
- 1、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
- 2、java程序设计：算24点
- 3、java算24点：输入4个数算24点，不要代码，求算法
- 4、用JAVA如何算出24点?
- 5、一个java面试题：计算24点游戏 编程
- 6、Java24点
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;
}
}
}
java程序设计：算24点
//这是我自己写的，在VC里可以运行。
#include "stdafx.h"
#includestdio.h
#includestdlib.h
static int NUMBER;
bool Game24(int const nNum, int* arr, int nLen, int nCount, char* pOperator, bool* pFlag){
if(nCount == 1){
if(*arr == nNum){
printf("((%d %c %d) %c %d) %c %d = %d\n",
arr[0],pOperator[0],arr[1],pOperator[1],arr[2],pOperator[2],arr[3],NUMBER);
if(!(*pFlag)) *pFlag = true;
}
return *pFlag;
}
for(int i = 0; i 4; ++i){
switch(i){
case 0:
pOperator[nCount - 2] = '+';
Game24(nNum - arr[nCount - 1], arr,nLen,nCount - 1,pOperator,pFlag);
break;
case 1:
pOperator[nCount - 2] = '-';
Game24(nNum + arr[nCount - 1], arr,nLen,nCount - 1,pOperator,pFlag);
break;
case 2:
pOperator[nCount - 2] = '*';
if( arr[nCount - 1] !(nNum % arr[nCount - 1]))
Game24(nNum / arr[nCount - 1], arr,nLen,nCount - 1,pOperator,pFlag);
break;
case 3:
pOperator[nCount - 2] = '/';
if(arr[nCount - 1])
Game24(nNum * arr[nCount - 1], arr,nLen,nCount - 1,pOperator,pFlag);
break;
}
}
return *pFlag;
}
int fOperating(char pOpe, int x1, int x2){
switch(pOpe){
case '+': return x1 + x2;
case '-': return x1 - x2;
case '*': return x1 * x2;
case '/': return x1 / x2;
}
}
bool fGame24(int const nNum, int* arr,char* pOperator, bool* pFlag){
int nLeft = 0,nRight = 0;
char pOpe[4] = {'+','-','*','/'};
for(int l = 0; l 4; ++l){
if(l == 3 (!arr[1] || arr[0] % arr[1])) continue;
pOperator[0] = pOpe[l];
nLeft = fOperating(pOpe[l],arr[0],arr[1]);
for(int r = 0; r 4; ++r){
if(r == 3 (!arr[3] || arr[2] % arr[3])) continue;
pOperator[2] = pOpe[r];
nRight = fOperating(pOpe[r],arr[2],arr[3]);
for(int m = 0; m 4; ++m){
if(m == 3 (!nRight || nLeft % nRight)) continue;
pOperator[1] = pOpe[m];
if(fOperating(pOpe[m],nLeft,nRight) == nNum){
printf("(%d %c %d) %c (%d %c %d) = %d\n",
arr[0],pOperator[0],arr[1],pOperator[1],arr[2],pOperator[2],arr[3],NUMBER);
if(!(*pFlag)) *pFlag = true;
}
}
}
}
return *pFlag;
}
int main(int argc, char* argv[])
{
puts("start!\nPlease input 4 numbers:");
bool* pFlag = (bool*)malloc(1);
*pFlag = false;
bool flag = 0;
int pNum[4] = {0};
int cNum[4] = {0};
int iNum[4] = {0};
char cOpe[3] = {0};
for(int i = 0; i 4; ++i)
scanf("%d",pNum[i]);
puts("So, what number do you want:");
scanf("%d",NUMBER);
puts("************************************");
for(iNum[0] = 0; iNum[0] 4; ++iNum[0]){
for(iNum[1] = 0; iNum[1] 4; ++iNum[1]){
if(iNum[1] == iNum[0]) continue;
for(iNum[2] = 0; iNum[2] 4; ++iNum[2]){
if(iNum[2] == iNum[0] || iNum[2] == iNum[1]) continue;
for(iNum[3] = 0; iNum[3] 4; ++iNum[3]){
if(iNum[3] == iNum[0] || iNum[3] == iNum[1] || iNum[3] == iNum[2]) continue;
for(int i = 0; i 4; ++i) cNum[i] = pNum[iNum[i]];
if(Game24(NUMBER,cNum,4,4,cOpe,pFlag)) flag = true;
if(fGame24(NUMBER,cNum,cOpe,pFlag)) flag = true;
}
}
}
}
free(pFlag);
if(!flag) printf("No way can be found.\n");
puts("************************************\nEnd!");
system("pause");
return 0;
}
java算24点：输入4个数算24点，不要代码，求算法
例如：4个数A、B、C、D，我们可以用F(A,B,C,D)=24来表示。那么。我们就可以把函数F拆解成F1(B,C,D)=P1(24,A)。（意思是：B，C，D之间的四则运算可以得到A和24之间的四则运算结果）。那么F1又可以继续拆解为C和D之间的四则运算关系得到结果后再和B来一次四则运算结果。这样，就可以得到很简单的一个数组6*6*6=216种结果而已。当然，这是A，B，C，D顺序固定的组合，那么把A，B，C，D换个位置，又一种组合。所以，所有的结果有6*6*6*12种。但，我们还是忽略了一种情况：A和B的值与C和D的值再进行四则运算，那么我们还需要再加一组6*6*6就可以了。
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;
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~
Java24点
粗略看了下，至少发现三个BUG:
1.没有考虑int类型相除时是否整除的问题，如3/4在你的算法中直接以零作为计算结果；
2.如果前两个或前三个数计算结果已经为24，则直接输出，没有等到四个全部参与计算；
3.类似n == 24.0这种浮点数的相等判断在绝大多数计算机语言中都是不正确的。
java24点算法的介绍就聊到这里吧，感谢你花时间阅读本站内容，更多关于、java24点算法的信息别忘了在本站进行查找喔。
发布于：2022-12-24，除非注明，否则均为原创文章，转载请注明出处。