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/*我编的小程序，逻辑有点混乱，注释也少得可怜，但应该能列出任何能用四则运算得出24的四个1～10的数的等于24的等式。 大家若有空，请指点。*/ 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); } } 返回类别: 教程 上一教程: 浅谈Java Virtual Machine 下一教程: 构建高性能J2EE应用的10个技巧 您可以阅读与"24点算法的java代码"相关的教程: · JAVA 的MD5加密算法源代码 · JAVA 的MD5加密算法源代码 · 运用加密技术保护JAVA源代码 · JAVA代码编写的一般性指导 · JAVA代码编写的30条建议 |
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