《Thinking in Java》习题——吸血鬼数字

 1 public class Test2 {
 2     /*
 3      * 将4位数拆分成4个数
 4      * */
 5   public int [] array(int num){
 6       int [] a = new int [4];
 7       int i=0;
 8       while(num!=0){
 9         a[i++]=num%10;
10         num=num/10;
11      }
12     return a;
13   } 
14   /*
15    * 将4给数组合成两个数，相乘与4位数比较。相等就是吸血鬼数字
16    * 
17    * */
18     public boolean equal(int num,int [] a){
19         for(int i=0;i<4;i++){
20             for(int j=0;j<4;j++){
21                 if(i==j){
22                     continue;
23                 }else{
24                     if(6-i-j==5){//i,j=0或1
25                         if((a[i]*10+a[j])*(a[3]*10+a[2])==num||(a[i]+a[j]*10)*(a[3]*10+a[2])==num){
26                             return true;
27                         }
28                     }
29                     else if(6-i-j==4){//i,j=0或2  
30                         if((a[i]*10+a[j])*(a[3]*10+a[1])==num||(a[i]+a[j]*10)*(a[3]*10+a[1])==num){
31                             return true;
32                     }
33                      }
34                     else if(6-i-j==3&&(i==3||j==3)){ //i,j或0,3 
35                         if((a[i]*10+a[j])*(a[2]*10+a[1])==num||(a[i]+a[j]*10)*(a[2]*10+a[1])==num){
36                             return true;
37                      }
38                         }
39                         else if(6-i-j==3&&(i==1||j==1)){//i,j=1或2
40                          if((a[i]*10+a[j])*(a[3]*10+a[0])==num||(a[i]+a[j]*10)*(a[3]*10+a[0])==num){
41                                 return true;
42                      }
43                         } 
44                         else if(6-i-j==2){//i,j=1或3 
45                          if((a[i]*10+a[j])*(a[2]*10+a[0])==num||(a[i]+a[j]*10)*(a[2]*10+a[0])==num){
46                                 return true;
47                      }
48                         }
49                         else if(6-i-j==1){//i,j=2或3 
50                          if((a[i]*10+a[j])*(a[1]*10+a[0])==num||(a[i]+a[j]*10)*(a[1]*10+a[0])==num){
51                                 return true;
52                        }
53                       }
54                      }
55                    }
56                 }
57              return false;
58             }
59       public static void main(String[] args) {
60           Test2 t2 = new Test2();
61          for(int i =1001;(i<=9999);i++){
62              if(t2.equal(i, t2.array(i))){
63                  System.out.println(i);
64              }
65          }
66     }
67         
68    }

 1 public class XiXueGui {   
 2     /**  
 3      * 吸血鬼数字算法  
 4      * 如：12*60=1260  
 5      * YangL.  
 6      */   
 7     public static void main(String[] args) {   
 8         String[] ar_str1 = null, ar_str2;   
 9         int sum = 0;   
10         int count=0;
11         for (int i = 10; i < 100; i++) {   
12             for (int j = i + 1; j < 100; j++) {   
13                 int i_val = i * j;   
14                 if (i_val < 1000 || i_val > 9999)   
15                     continue; // 积小于1000或大于9999排除,继续下一轮环   
16                 count++;
17                 ar_str1 = String.valueOf(i_val).split("");   
18                 ar_str2 = (String.valueOf(i) + String.valueOf(j)).split("");   
19                 java.util.Arrays.sort(ar_str1);   
20                 java.util.Arrays.sort(ar_str2);   
21                 if (java.util.Arrays.equals(ar_str1, ar_str2)) {   
22                     // 排序后比较,为真则找到一组   
23                     sum++;   
24                     System.out.println("第" + sum + "组: " + i + "*" + j + "="   
25                             + i_val);   
26                 }   
27             }   
28         }   
29         System.out.println("共找到" + sum + "组吸血鬼数"+"\ncount"+count);
30        
31     }   
32 } 

 1 //: control/E10_Vampire.java
 2 /****************** Exercise 10 *********************
 3 * A vampire number has an even number of digits and
 4 * is formed by multiplying a pair of numbers containing
 5 * half the number of digits of the result. The digits
 6 * are taken from the original number in any order.
 7 * Pairs of trailing zeroes are not allowed. Examples
 8 * include:
 9 * 1260 = 21 * 60
10 * 1827 = 21 * 87
11 * 2187 = 27 * 81
12 * Write a program that finds all the 4-digit vampire
13 * numbers. (Suggested by Dan Forhan.)
14 ****************************************************/
15 public class E10_Vampire {
16 public static void main(String[] args) {
17     int[] startDigit = new int[4];
18     int[] productDigit = new int[4];
19     for(int num1 = 10; num1 <= 99; num1++)
20         for(int num2 = num1; num2 <= 99; num2++) {
21             // Pete Hartley's theoretical result:
22             // If x·y is a vampire number then
23             // x·y == x+y (mod 9)
24             if((num1 * num2) % 9 != (num1 + num2) % 9)
25                 continue;
26     int product = num1 * num2;
27     startDigit[0] = num1 / 10;
28     startDigit[1] = num1 % 10;
29     startDigit[2] = num2 / 10;
30     startDigit[3] = num2 % 10;
31     productDigit[0] = product / 1000;
32     productDigit[1] = (product % 1000) / 100;
33     productDigit[2] = product % 1000 % 100 / 10;
34     productDigit[3] = product % 1000 % 100 % 10;
35     int count = 0;
36     for(int x = 0; x < 4; x++)
37       for(int y = 0; y < 4; y++) {
38           if(productDigit[x] == startDigit[y]) {
39               count++;
40               productDigit[x] = -1;
41               startDigit[y] = -2;
42               if(count == 4)
43                   System.out.println(num1 + " * " + num2
44                           + " : " + product);
45                 }
46             }
47         }
48     }
49 }