24点算法的Java低级简易实现
package test;/** * @author QuarterLifeForJava */public class Test {public static void main(String[] args) {int x[] = {8,8,3,3};int y[] = new int[x.length];for(int a=0;a<4;a++){for(int b=0;b<4;b++){for(int c=0;c<4;c++){for(int d=0;d<4;d++){if(a!=b&&b!=c&&c!=d&&d!=a&&a!=c&&b!=d){y[0] = x[a];y[1] = x[b];y[2] = x[c];y[3] = x[d];for(int i=0;i<4;i++){circle(y,i);}}}}}}}//边界判断public static boolean boundaryJudgment(double result){if(result>=23.99999999999999&&result<=24.00000000000001){return true;}else{return false;}}//加public static double jia(double x,double y){return x+y;}//减public static double jian(double x,double y){return x-y;}//乘public static double cheng(double x,double y){return x*y;}//除public static double chu(double x,double y){return x/y;}//循环选择public static double change(int num,double x,double y){switch(num){case 0:return jia(x,y);case 1:return jian(x,y);case 2:return cheng(x,y);case 3:return chu(x,y);}return 0;}//循环计算public static void circle(int x[],int choice){double mid = 0;double y = 0;double z = 0;for(int i=0;i<4;i++){mid = change(i,x[0],x[1]);z = mid;for(int j=0;j<4;j++){mid = z;switch(choice){case 0:mid = change(j,mid,x[2]);for(int k=0;k<4;k++){y = mid;mid = change(k,mid,x[3]);if(boundaryJudgment(mid)){System.out.println("可算24点");//想输出结果，请自己写个方法，这里略了System.exit(0);}else{mid = y;}}break;case 1:mid = change(j,mid,x[2]);for(int k=0;k<4;k++){y = mid;mid = change(k,x[3],mid);if(boundaryJudgment(mid)){System.out.println("可算24点");//想输出结果，请自己写个方法，这里略了System.exit(0);}else{mid = y;}}break;case 2:mid = change(j,x[2],mid);for(int k=0;k<4;k++){y = mid;mid = change(k,mid,x[3]);if(boundaryJudgment(mid)){System.out.println("可算24点");//想输出结果，请自己写个方法，这里略了System.exit(0);}else{mid = y;}}break;case 3:mid = change(j,x[2],mid);for(int k=0;k<4;k++){y = mid;mid = change(k,x[3],mid);if(boundaryJudgment(mid)){System.out.println("可算24点");//想输出结果，请自己写个方法，这里略了System.exit(0);}else{mid = y;}}break;}}}}}