关于24点算法
24点算法是怎么样的，怎么用Java语言实现？ 搜索更多相关的解决方案:
算法
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计算24点算法
基本原理是穷举4个整数所有可能的表达式，然后对表达式求值。
表达式的定义： expression = (expression|number) operator (expression|number)
因为能使用的4种运算符 + - * / 都是2元运算符，所以本文中只考虑2元运算符。2元运算符接收两个参数，输出计算结果，输出的结果参与后续的计算。
由上所述，构造所有可能的表达式的算法如下：
(1) 将4个整数放入数组中
(2) 在数组中取两个数字的排列，共有 P(4,2) 种排列。对每一个排列，
(2.1) 对 + - * / 每一个运算符，
(2.1.1) 根据此排列的两个数字和运算符，计算结果
(2.1.2) 改表数组：将此排列的两个数字从数组中去除掉，将 2.1.1 计算的结果放入数组中
(2.1.3) 对新的数组，重复步骤 2
(2.1.4) 恢复数组：将此排列的两个数字加入数组中，将 2.1.1 计算的结果从数组中去除掉
可见这是一个递归过程。步骤 2 就是递归函数。当数组中只剩下一个数字的时候，这就是表达式的最终结果，此时递归结束。
在程序中，一定要注意递归的现场保护和恢复，也就是递归调用之前与之后，现场状态应该保持一致。在上述算法中，递归现场就是指数组，2.1.2 改变数组以进行下一层递归调用，2.1.3 则恢复数组，以确保当前递归调用获得下一个正确的排列。
括号 () 的作用只是改变运算符的优先级，也就是运算符的计算顺序。所以在以上算法中，无需考虑括号。括号只是在输出时需加以考虑。
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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[i] = Integer.parseInt(args[i]);
}
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[i] + numb[j];
p[temp][1] = numb[i] - numb[j];
p[temp][2] = numb[j] - numb[i];
p[temp][3] = numb[i] * numb[j];
if(numb[j] != 0)
p[temp][4] = numb[i] / (double)numb[j];
else
p[temp][4] = 10000;
if(numb[i] != 0)
p[temp][5] = numb[j] / (double)numb[i];
else
p[temp][5] = 10000;
switch(temp){
case 0:p[temp][6] = numb[2]; p[temp][7] = numb[3];break;
case 1:p[temp][6] = numb[1]; p[temp][7] = numb[3];break;
case 2:p[temp][6] = numb[1]; p[temp][7] = numb[2];break;
case 3:p[temp][6] = numb[0]; p[temp][7] = numb[3];break;
case 4:p[temp][6] = numb[0]; p[temp][7] = numb[2];break;
case 5:p[temp][6] = numb[0]; p[temp][7] = numb[1];
}
}
for(int k = 0,tem = 0; k < 6; k ++)//Get the possible three numbers and store into the array num[36][3] for calculating .
for(int l = 0; l < 6; l ++,tem ++){
num[tem][0] = p[k][l] ;
num[tem][1] = p[k][6] ;
num[tem][2] = p[k][7] ;
for(int t = 2,m = 0, n = 0,te = 0; t >= 0; t --,te ++){//Get two of the three to calculate and then store the new number into the array q;
m = (t + 1)%3;
n = (t + 2)%3;
q[te][6] = num[tem][t];
q[te][0] = num[tem][m] + num[tem][n];
q[te][1] = num[tem][m] - num[tem][n];
q[te][2] = num[tem][n] - num[tem][m];
q[te][3] = num[tem][m] * num[tem][n];
if(num[tem][n] != 0)
q[te][4] = num[tem][m] / (double)num[tem][n];
else
q[te][4] = 10000 ;
if(num[tem][m] != 0)
q[te][5] = num[tem][n] / (double)num[tem][m];
else
q[te][5] = 10000 ;
}
for(int u = 0; u < 3; u ++)
for(int v = 0; v < 6; v ++){
if(u == 2){//We must insure that the old value is in the left ,so the result string can be appended rightly.
total[0] = q[u][6] + q[u][v];
total[1] = q[u][6] - q[u][v];
total[2] = q[u][v] - q[u][6];
total[3] = q[u][v] * q[u][6];
if(q[u][6] != 0)
total[4] = q[u][6] / (double)q[u][v];
else
total[4] = 10000;
if(q[u][v] != 0)
total[5] = q[u][v] / (double)q[u][6];
else
total[5] = 10000;
}
else{
total[0] = q[u][v] + q[u][6];
total[1] = q[u][v] - q[u][6];
total[2] = q[u][6] - q[u][v];
total[3] = q[u][v] * q[u][6];
if(q[u][6] != 0)
total[4] = q[u][v] / (double)q[u][6];
else
total[4] = 10000;
if(q[u][v] != 0)
total[5] = q[u][6] / (double)q[u][v];
else
total[5] = 10000;
}
for(int s = 0 ; s < 6 ; s ++){
if(total[s]>23.9999&&total[s]<24.0001){
//System.out.println("24!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!");
totalSuc ++ ;
//print the expression
char x[] = new char[3];
int n0 = index ;
String expre = "" ;
String expre1 = "" ;
int n1 = index/108;//the first composition and its operator.
int n2 = index%108/6;//the second composition and its operator.
int n3 = index%6;//the last operator.
//now ready to printout the equation.
switch(n1){
case 0:expre += "((a+b)";x[1]=’c’;x[2]=’d’;break;
case 1:expre += "((a-b)";x[1]=’c’;x[2]=’d’;break;
case 2:expre += "((b-a)";x[1]=’c’;x[2]=’d’;break;
case 3:expre += "((a*b)";x[1]=’c’;x[2]=’d’;break;
case 4:expre += "((a/b)";x[1]=’c’;x[2]=’d’;break;
case 5:expre += "((b/a)";x[1]=’c’;x[2]=’d’;break;
case 6:expre += "((a+c)";x[1]=’b’;x[2]=’d’;break;
case 7:expre += "((a-c)";x[1]=’b’;x[2]=’d’;break;
case 8:expre += "((c-a)";x[1]=’b’;x[2]=’d’;break;
case 9:expre += "((a*c)";x[1]=’b’;x[2]=’d’;break;
case 10:expre += "((a/c)";x[1]=’b’;x[2]=’d’;break;
case 11:expre += "((c/a)";x[1]=’b’;x[2]=’d’;break;
case 12:expre += "((a+d)";x[1]=’b’;x[2]=’c’;break;
case 13:expre += "((a-d)";x[1]=’b’;x[2]=’c’;break;
case 14:expre += "((d-a)";x[1]=’b’;x[2]=’c’;break;
case 15:expre += "((a*d)";x[1]=’b’;x[2]=’c’;break;
case 16:expre += "((a/d)";x[1]=’b’;x[2]=’c’;break;
case 17:expre += "((d/a)";x[1]=’b’;x[2]=’c’;break;
case 18:expre += "((b+c)";x[1]=’a’;x[2]=’d’;break;
case 19:expre += "((b-c)";x[1]=’a’;x[2]=’d’;break;
case 20:expre += "((c-b)";x[1]=’a’;x[2]=’d’;break;
case 21:expre += "((b*c)";x[1]=’a’;x[2]=’d’;break;
case 22:expre += "((b/c)";x[1]=’a’;x[2]=’d’;break;
case 23:expre += "((c/b)";x[1]=’a’;x[2]=’d’;break;
case 24:expre += "((b+d)";x[1]=’a’;x[2]=’c’;break;
case 25:expre += "((b-d)";x[1]=’a’;x[2]=’c’;break;
case 26:expre += "((d-b)";x[1]=’a’;x[2]=’c’;break;
case 27:expre += "((b*d)";x[1]=’a’;x[2]=’c’;break;
case 28:expre += "((b/d)";x[1]=’a’;x[2]=’c’;break;
case 29:expre += "((d/b)";x[1]=’a’;x[2]=’c’;break;
case 30:expre += "((c+d)";x[1]=’a’;x[2]=’b’;break;
case 31:expre += "((c-d)";x[1]=’a’;x[2]=’b’;break;
case 32:expre += "((d-c)";x[1]=’a’;x[2]=’b’;break;
case 33:expre += "((c*d)";x[1]=’a’;x[2]=’b’;break;
case 34:expre += "((c/d)";x[1]=’a’;x[2]=’b’;break;
case 35:expre += "((d/c)";x[1]=’a’;x[2]=’b’;
}
switch(n2){
case 0:expre += "+" +x[1] +")";x[1]=x[2];break;//x[0] and x[1].
case 1:expre += "-" +x[1] +")";x[1]=x[2];break;
case 2:expre += "--" +x[1] +")";x[1]=x[2];break;
case 3:expre += "*" +x[1] +")";x[1]=x[2];break;
case 4:expre += "/" +x[1] +")";x[1]=x[2];break;
case 5:expre += "//" +x[1] +")";x[1]=x[2];break;
case 6:expre += "+" +x[2] +")";x[1]=x[1];break;//x[2] and x[0].
case 7:expre += "--" +x[2] +")";x[1]=x[1];break;
case 8:expre += "-" +x[2] +")";x[1]=x[1];break;
case 9:expre += "*" +x[2] +")";x[1]=x[1];break;
case 10:expre += "//" +x[2] +")";x[1]=x[1];break;
case 11:expre += "/" +x[2] +")";x[1]=x[1];break;
case 12:expre1 += x[1] + "+" + x[2] +"))";x[1]=’(’;break;//x[1] and x[2].
case 13:expre1 += x[1] + "-" + x[2] +"))";x[1]=’(’;break;
case 14:expre1 += x[1] + "--" + x[2] +"))";x[1]=’(’;break;
case 15:expre1 += x[1] + "*" + x[2] +"))";x[1]=’(’;break;
case 16:expre1 += x[1] + "/" + x[2] +"))";x[1]=’(’;break;
case 17:expre1 += x[1] + "//" + x[2] +"))";x[1]=’(’;
}
switch(n3){
case 0:expre += "+" +x[1] + expre1;break;
case 1:expre += "-" +x[1] + expre1;break;
case 2:expre += "--" +x[1] + expre1;break;
case 3:expre += "*" +x[1] + expre1;break;
case 4:expre += "/" +x[1] + expre1;break;
case 5:expre += "//" +x[1] + expre1;
}
expre = expre.replace(’a’,(char)(numb[0] + 48));
expre = expre.replace(’b’,(char)(numb[1] + 48));
expre = expre.replace(’c’,(char)(numb[2] + 48));
expre = expre.replace(’d’,(char)(numb[3] + 48));
System.out.println(expre+" ");
}
//System.out.println("total : " + total[s] + " index :" +"\n\n");
index ++ ;
}
}
}System.out.println("The number of total successful expressions is : " + totalSuc);
}
}
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回复 3楼 myhnuhai
"a--b"means"b-a""a//b"means"b/a"
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0
at Test24Point.main(Test24Point.java:34)
会有个这样的错误呢
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