JAVA 编程思想 找出所有四位数的吸血鬼数字

/**
 * 找出四位数的所有吸血鬼数字
 * 吸血鬼数字是指位数为偶数的数字，可以由一对数字相乘而得到，而这对数字各包含乘积的一半位数的数字，其中从最初的数字中选取的数字可以任意排序.
 * 以两个0结尾的数字是不允许的。
 *   例如下列数字都是吸血鬼数字
1260=21*60
1827=21*87
 2187=27*81
 …
 * 比较笨的低效率的做法： 遍历所有四位数， 每生成一个四位数的时候，
 *         在双重循环遍历两位数，在两位数的内层循环中判断是否与最外层循环的四位数相等。 如果相等把这些数字都存放到数组，进行排序后比较

 *         两组数字，如果相等那么输出这个数就是要找的数字；

 */

了解下这个英文参考:吸血鬼数字 

An important theoretical result found by Pete Hartley:
          If x·y is a vampire number then x·y == x+y (mod 9)
Proof:
Let mod be the binary modulo operator and d(x) the sum of the decimal digits of x.
It is well-known that d(x) mod 9 = x mod 9, for all x.
Assume x·y is a vampire. Then it contains the same digits as x and y, and in particular d(x·y) = d(x)+d(y). This leads to:
          (x·y) mod 9 = d(x·y) mod 9 = (d(x)+d(y)) mod 9 = (d(x) mod 9 + d(y) mod 9) mod 9
            = (x mod 9 + y mod 9) mod 9 = (x+y) mod 9

The solutions to the congruence are (x mod 9, y mod 9) in {(0,0), (2,2), (3,6), (5,8), (6,3), (8,5)}
Only these cases (6 out of 81) have to be tested in a vampire search based on testing x·y for different values of x and y.

下面五中方法, 其中还是ThinkinJava给出的参考答案效率最高, 其他高效率做法 , 请网友高手大神补充

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1. import java.util.Arrays;  
2. public class Demo3 {  
3.   
4.     static int a;       //千位  
5.     static int b;       //百位  
6.     static int c;       //十位  
7.     static int d;       //个位  
8.   
9.     static int a1;      //十位  
10.     static int b1;      //个位  
11.     static int c1;      //十位  
12.     static int d1;      //个位  
13.     static int sum  = 0; //总和  
14.     static int sum2 = 0; //两数之积  
15.   
16.     public static void main(String[] args) {  
17.   
18.         long startTime = System.nanoTime();  
19.         method1();  
20.         long endTime = System.nanoTime();  
21.         System.out.println(”method1 :” + (endTime - startTime)); //method1 :185671841  
22.         long s = System.nanoTime();  
23.         method2();  
24.         long d = System.nanoTime();  
25.         System.out.println(”method2 :” + (d - s)); //method2 :90556063  
26.         long s3 = System.nanoTime();  
27.         method3();  
28.         long d3 = System.nanoTime();  
29.         System.out.println(”method3 :” + (d3 - s3));//method3 :574735  
30.         long s4 = System.nanoTime();  
31.         method4();  
32.         long d4 = System.nanoTime();  
33.         System.out.println(”method4 :” + (d4 - s4));//method4 :22733469  
34.         long s5 = System.nanoTime();  
35.         method5();  
36.         long d5 = System.nanoTime();  
37.         System.out.println(”method5 :” + (d5 - s5));//method4 :19871660  
38.   
39.     }  
40.   
41.     private static void method5() {  
42.         new VampireNumbers(); //该方法 有重复数字  
43.     }  
44.   
45.     static class VampireNumbers {  
46.         static int a(int i) {  
47.             return i / 1000;  
48.         }  
49.   
50.         static int b(int i) {  
51.             return (i % 1000) / 100;  
52.         }  
53.   
54.         static int c(int i) {  
55.             return ((i % 1000) % 100) / 10;  
56.         }  
57.   
58.         static int d(int i) {  
59.             return ((i % 1000) % 100) % 10;  
60.         }  
61.   
62.         static int com(int i, int j) {  
63.             return (i * 10) + j;  
64.         }  
65.   
66.         static void productTest(int i, int m, int n) {  
67.             if (m * n == i)  
68.                 System.out.println(i + ” = ” + m + “ * ” + n);  
69.         }  
70.   
71.         public VampireNumbers() {  
72.             for (int i = 1001; i < 9999; i++) {  
73.                 productTest(i, com(a(i), b(i)), com(c(i), d(i)));  
74.                 productTest(i, com(a(i), b(i)), com(d(i), c(i)));  
75.                 productTest(i, com(a(i), c(i)), com(b(i), d(i)));  
76.                 productTest(i, com(a(i), c(i)), com(d(i), b(i)));  
77.                 productTest(i, com(a(i), d(i)), com(b(i), c(i)));  
78.                 productTest(i, com(a(i), d(i)), com(c(i), b(i)));  
79.                 productTest(i, com(b(i), a(i)), com(c(i), d(i)));  
80.                 productTest(i, com(b(i), a(i)), com(d(i), c(i)));  
81.                 productTest(i, com(b(i), c(i)), com(d(i), a(i)));  
82.                 productTest(i, com(b(i), d(i)), com(c(i), a(i)));  
83.                 productTest(i, com(c(i), a(i)), com(d(i), b(i)));  
84.                 productTest(i, com(c(i), b(i)), com(d(i), a(i)));  
85.             }  
86.         }  
87.     }  
88.   
89.     private static void method4() { // 改进  
90.         for (int i = 11; i < 100; i++) {  
91.             for (int j = i; j < 100; j++) {  
92.                 int k = i * j;  
93.                 String kStr = Integer.toString(k);  
94.                 String checkStr = Integer.toString(i) + Integer.toString(j);  
95.                 if (kStr.length() != 4)  
96.                     continue;  
97.                 char[] kChar = kStr.toCharArray();  
98.                 char[] checkChar = checkStr.toCharArray();  
99.                 Arrays.sort(kChar);  
100.                 Arrays.sort(checkChar);  
101.                 boolean isVampire = Arrays.equals(kChar, checkChar);  
102.                 if (isVampire) {  
103.                     System.out.println(i + ” * ” + j + “ = ” + k);  
104.                 }  
105.             }  
106.         }  
107.     }  
108.   
109.     private static void method3() { // 官方参考答案  
110.   
111.         int[] startDigit = new int[4];  
112.         int[] productDigit = new int[4];  
113.         for (int num1 = 10; num1 <= 99; num1++)  
114.             for (int num2 = num1; num2 <= 99; num2++) {  
115.                 // Pete Hartley’s theoretical result:  
116.                 // If x·y is a vampire number then  
117.                 // x·y == x+y (mod 9)  
118.                 if ((num1 * num2) % 9 != (num1 + num2) % 9)  
119.                     continue;  
120.                 int product = num1 * num2;  
121.                 startDigit[0] = num1 / 10;  
122.                 startDigit[1] = num1 % 10;  
123.                 startDigit[2] = num2 / 10;  
124.                 startDigit[3] = num2 % 10;  
125.                 productDigit[0] = product / 1000;  
126.                 productDigit[1] = (product % 1000) / 100;  
127.                 productDigit[2] = product % 1000 % 100 / 10;  
128.                 productDigit[3] = product % 1000 % 100 % 10;  
129.                 int count = 0;  
130.                 for (int x = 0; x < 4; x++)  
131.                     for (int y = 0; y < 4; y++) {  
132.                         if (productDigit[x] == startDigit[y]) {  
133.                             count++;  
134.                             productDigit[x] = -1;  
135.                             startDigit[y] = -2;  
136.                             if (count == 4)  
137.                                 System.out.println(num1 + ” * ” + num2 + “ : ”  
138.                                         + product);  
139.                         }  
140.                     }  
141.   
142.             } /* 
143.                * Output: 15 * 93 : 1395 21 * 60 : 1260 21 * 87 : 1827 27 * 81 : 
144.                * 2187 30 * 51 : 1530 35 * 41 : 1435 80 * 86 : 6880 
145.                *///:~  
146.     }  
147.   
148.     private static void method2() { // 穷举2  
149.   
150.         //遍历四位数，排除00 从1001开始  
151.         for (int i = 1001; i <= 9999; i++) {  
152.             //排除00   
153.             if (i % 100 != 00) {  
154.                 for (int k = 0; k < 100; k += 10) {  
155.                     if (k != 0) {  
156.                         //10 -99  
157.                         for (int j2 = 0; j2 <= 9; j2++) {  
158.                             //生成第一个两位数   
159.                             for (int j = 0; j < 100; j += 10) {  
160.                                 for (int j3 = 0; j3 <= 9; j3++) {  
161.                                     //生成第二个两位数   
162.                                     //判断两数之积   
163.   
164.                                     if ((k + j2) * (j + j3) == i) {  
165.                                         if (compare2(i, k / 10, j2, j / 10, j3)) {  
166.                                             System.out  
167.                                                     .println(i + ”=” + (k + j2)  
168.                                                             + ”*” + (j + j3));  
169.                                         }  
170.   
171.                                     }  
172.                                 }  
173.                             }  
174.   
175.                         }  
176.                     }  
177.                 }  
178.             }  
179.   
180.         }  
181.     }  
182.   
183.     public static void method1() { //穷举1  
184.         int x = 0, y = 0;  
185.         for (int i = 1; i <= 9; i++) {  
186.             a = i * 1000;  
187.             for (int j = 0; j <= 9; j++) {  
188.                 b = j * 100;  
189.                 for (int j2 = 0; j2 < 10; j2++) {  
190.                     c = j2 * 10;  
191.                     for (int k = 0; k < 10; k++) {  
192.                         d = k;  
193.                         sum = a + b + c + d;  
194.                         //取其中四个数字 中组成两个两位数 ，如果这两个两位数之积  等于 sum ,则输入 这个数  
195.                         for (int k2 = 1; k2 < 10; k2++) {  
196.   
197.                             a1 = k2 * 10;  
198.                             for (int l = 0; l < 10; l++) {  
199.                                 if (a1 + b1 > 100) {  
200.                                     break;  
201.                                 }  
202.                                 b1 = l;  
203.                                 //得到一个两位数字  
204.                                 for (int l2 = 1; l2 < 10; l2++) {  
205.                                     c1 = l2 * 10;  
206.                                     for (int m = l; m < 10; m++) {  
207.   
208.                                         if (c1 + d1 > 100) {  
209.                                             break;  
210.                                         }  
211.                                         d1 = m;  
212.                                         //再得到一个两位数字  
213.                                         sum2 = (a1 + b1) * (c1 + d1);  
214.                                         //计算来两个两位数字之积，如果等于sum   
215.                                         if (sum2 == sum) {  
216.                                             //且尾数不能为00   
217.                                             if (c + d != 0) {  
218.                                                 // 比较这个几个数字 是否一样  
219.                                                 if (compare(a, b, c, d, a1, b1,  
220.                                                         c1, d1)) {  
221.                                                     System.out.println(sum  
222.                                                             + ”=” + (a1 + b1)  
223.                                                             + ”*” + (c1 + d1));  
224.                                                 }  
225.                                             }  
226.                                         }  
227.                                     }  
228.                                 }  
229.                             }  
230.                         }  
231.                     }  
232.                 }  
233.             }  
234.         }  
235.     }  
236.   
237.     private static boolean compare2(int i, int j, int j2, int k, int j3) {  
238.         int a[] = { i % 10, i / 10 % 10, i / 100 % 10, i / 1000 };  
239.         int b[] = { j, j2, k, j3 };  
240.         Arrays.sort(a);  
241.         Arrays.sort(b);  
242.         if (Arrays.equals(a, b))  
243.             return true;  
244.         else  
245.             return false;  
246.   
247.     }  
248.   
249.     private static boolean compare(int a2, int b2, int c2, int d2, int a12,  
250.                                    int b12, int c12, int d12) {  
251.   
252.         int[] a = new int[4];  
253.         int[] b = new int[4];  
254.   
255.         a[0] = a2 / 1000;  
256.         a[1] = b2 / 100;  
257.         a[2] = c2 / 10;  
258.         a[3] = d2;  
259.   
260.         b[0] = a12 / 10;  
261.         b[1] = b12;  
262.         b[2] = c12 / 10;  
263.         b[3] = d12;  
264.   
265.         Arrays.sort(a);  
266.         Arrays.sort(b);  
267.   
268.         if (Arrays.equals(a, b))  
269.             return true;  
270.         else  
271.             return false;  
272.     }  
273. }  
274. }  

import java.util.Arrays;
public class Demo3 {

    static int a;       //千位
    static int b;       //百位
    static int c;       //十位
    static int d;       //个位

    static int a1;      //十位
    static int b1;      //个位
    static int c1;      //十位
    static int d1;      //个位
    static int sum  = 0; //总和
    static int sum2 = 0; //两数之积

    public static void main(String[] args) {

        long startTime = System.nanoTime();
        method1();
        long endTime = System.nanoTime();
        System.out.println("method1 :" + (endTime - startTime)); //method1 :185671841
        long s = System.nanoTime();
        method2();
        long d = System.nanoTime();
        System.out.println("method2 :" + (d - s)); //method2 :90556063
        long s3 = System.nanoTime();
        method3();
        long d3 = System.nanoTime();
        System.out.println("method3 :" + (d3 - s3));//method3 :574735
        long s4 = System.nanoTime();
        method4();
        long d4 = System.nanoTime();
        System.out.println("method4 :" + (d4 - s4));//method4 :22733469
        long s5 = System.nanoTime();
        method5();
        long d5 = System.nanoTime();
        System.out.println("method5 :" + (d5 - s5));//method4 :19871660

    }

    private static void method5() {
        new VampireNumbers(); //该方法 有重复数字
    }

    static class VampireNumbers {
        static int a(int i) {
            return i / 1000;
        }

        static int b(int i) {
            return (i % 1000) / 100;
        }

        static int c(int i) {
            return ((i % 1000) % 100) / 10;
        }

        static int d(int i) {
            return ((i % 1000) % 100) % 10;
        }

        static int com(int i, int j) {
            return (i * 10) + j;
        }

        static void productTest(int i, int m, int n) {
            if (m * n == i)
                System.out.println(i + " = " + m + " * " + n);
        }

        public VampireNumbers() {
            for (int i = 1001; i < 9999; i++) {
                productTest(i, com(a(i), b(i)), com(c(i), d(i)));
                productTest(i, com(a(i), b(i)), com(d(i), c(i)));
                productTest(i, com(a(i), c(i)), com(b(i), d(i)));
                productTest(i, com(a(i), c(i)), com(d(i), b(i)));
                productTest(i, com(a(i), d(i)), com(b(i), c(i)));
                productTest(i, com(a(i), d(i)), com(c(i), b(i)));
                productTest(i, com(b(i), a(i)), com(c(i), d(i)));
                productTest(i, com(b(i), a(i)), com(d(i), c(i)));
                productTest(i, com(b(i), c(i)), com(d(i), a(i)));
                productTest(i, com(b(i), d(i)), com(c(i), a(i)));
                productTest(i, com(c(i), a(i)), com(d(i), b(i)));
                productTest(i, com(c(i), b(i)), com(d(i), a(i)));
            }
        }
    }

    private static void method4() { // 改进
        for (int i = 11; i < 100; i++) {
            for (int j = i; j < 100; j++) {
                int k = i * j;
                String kStr = Integer.toString(k);
                String checkStr = Integer.toString(i) + Integer.toString(j);
                if (kStr.length() != 4)
                    continue;
                char[] kChar = kStr.toCharArray();
                char[] checkChar = checkStr.toCharArray();
                Arrays.sort(kChar);
                Arrays.sort(checkChar);
                boolean isVampire = Arrays.equals(kChar, checkChar);
                if (isVampire) {
                    System.out.println(i + " * " + j + " = " + k);
                }
            }
        }
    }

    private static void method3() { // 官方参考答案

        int[] startDigit = new int[4];
        int[] productDigit = new int[4];
        for (int num1 = 10; num1 <= 99; num1++)
            for (int num2 = num1; num2 <= 99; num2++) {
                // Pete Hartley's theoretical result:
                // If x·y is a vampire number then
                // x·y == x+y (mod 9)
                if ((num1 * num2) % 9 != (num1 + num2) % 9)
                    continue;
                int product = num1 * num2;
                startDigit[0] = num1 / 10;
                startDigit[1] = num1 % 10;
                startDigit[2] = num2 / 10;
                startDigit[3] = num2 % 10;
                productDigit[0] = product / 1000;
                productDigit[1] = (product % 1000) / 100;
                productDigit[2] = product % 1000 % 100 / 10;
                productDigit[3] = product % 1000 % 100 % 10;
                int count = 0;
                for (int x = 0; x < 4; x++)
                    for (int y = 0; y < 4; y++) {
                        if (productDigit[x] == startDigit[y]) {
                            count++;
                            productDigit[x] = -1;
                            startDigit[y] = -2;
                            if (count == 4)
                                System.out.println(num1 + " * " + num2 + " : "
                                        + product);
                        }
                    }

            } /*
               * Output: 15 * 93 : 1395 21 * 60 : 1260 21 * 87 : 1827 27 * 81 :
               * 2187 30 * 51 : 1530 35 * 41 : 1435 80 * 86 : 6880
               *///:~
    }

    private static void method2() { // 穷举2

        //遍历四位数，排除00 从1001开始
        for (int i = 1001; i <= 9999; i++) {
            //排除00 
            if (i % 100 != 00) {
                for (int k = 0; k < 100; k += 10) {
                    if (k != 0) {
                        //10 -99
                        for (int j2 = 0; j2 <= 9; j2++) {
                            //生成第一个两位数 
                            for (int j = 0; j < 100; j += 10) {
                                for (int j3 = 0; j3 <= 9; j3++) {
                                    //生成第二个两位数 
                                    //判断两数之积 

                                    if ((k + j2) * (j + j3) == i) {
                                        if (compare2(i, k / 10, j2, j / 10, j3)) {
                                            System.out
                                                    .println(i + "=" + (k + j2)
                                                            + "*" + (j + j3));
                                        }

                                    }
                                }
                            }

                        }
                    }
                }
            }

        }
    }

    public static void method1() { //穷举1
        int x = 0, y = 0;
        for (int i = 1; i <= 9; i++) {
            a = i * 1000;
            for (int j = 0; j <= 9; j++) {
                b = j * 100;
                for (int j2 = 0; j2 < 10; j2++) {
                    c = j2 * 10;
                    for (int k = 0; k < 10; k++) {
                        d = k;
                        sum = a + b + c + d;
                        //取其中四个数字 中组成两个两位数 ，如果这两个两位数之积  等于 sum ,则输入 这个数
                        for (int k2 = 1; k2 < 10; k2++) {

                            a1 = k2 * 10;
                            for (int l = 0; l < 10; l++) {
                                if (a1 + b1 > 100) {
                                    break;
                                }
                                b1 = l;
                                //得到一个两位数字
                                for (int l2 = 1; l2 < 10; l2++) {
                                    c1 = l2 * 10;
                                    for (int m = l; m < 10; m++) {

                                        if (c1 + d1 > 100) {
                                            break;
                                        }
                                        d1 = m;
                                        //再得到一个两位数字
                                        sum2 = (a1 + b1) * (c1 + d1);
                                        //计算来两个两位数字之积，如果等于sum 
                                        if (sum2 == sum) {
                                            //且尾数不能为00 
                                            if (c + d != 0) {
                                                // 比较这个几个数字 是否一样
                                                if (compare(a, b, c, d, a1, b1,
                                                        c1, d1)) {
                                                    System.out.println(sum
                                                            + "=" + (a1 + b1)
                                                            + "*" + (c1 + d1));
                                                }
                                            }
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }

    private static boolean compare2(int i, int j, int j2, int k, int j3) {
        int a[] = { i % 10, i / 10 % 10, i / 100 % 10, i / 1000 };
        int b[] = { j, j2, k, j3 };
        Arrays.sort(a);
        Arrays.sort(b);
        if (Arrays.equals(a, b))
            return true;
        else
            return false;

    }

    private static boolean compare(int a2, int b2, int c2, int d2, int a12,
                                   int b12, int c12, int d12) {

        int[] a = new int[4];
        int[] b = new int[4];

        a[0] = a2 / 1000;
        a[1] = b2 / 100;
        a[2] = c2 / 10;
        a[3] = d2;

        b[0] = a12 / 10;
        b[1] = b12;
        b[2] = c12 / 10;
        b[3] = d12;

        Arrays.sort(a);
        Arrays.sort(b);

        if (Arrays.equals(a, b))
            return true;
        else
            return false;
    }
}
}

/**
 * 找出四位数的所有吸血鬼数字
 * 吸血鬼数字是指位数为偶数的数字，可以由一对数字相乘而得到，而这对数字各包含乘积的一半位数的数字，其中从最初的数字中选取的数字可以任意排序.
 * 以两个0结尾的数字是不允许的。
 *   例如下列数字都是吸血鬼数字
1260=21*60
1827=21*87
 2187=27*81
 …
 * 比较笨的低效率的做法： 遍历所有四位数， 每生成一个四位数的时候，
 *         在双重循环遍历两位数，在两位数的内层循环中判断是否与最外层循环的四位数相等。 如果相等把这些数字都存放到数组，进行排序后比较

 *         两组数字，如果相等那么输出这个数就是要找的数字；

 */

了解下这个英文参考:吸血鬼数字 

An important theoretical result found by Pete Hartley:
          If x·y is a vampire number then x·y == x+y (mod 9)
Proof:
Let mod be the binary modulo operator and d(x) the sum of the decimal digits of x.
It is well-known that d(x) mod 9 = x mod 9, for all x.
Assume x·y is a vampire. Then it contains the same digits as x and y, and in particular d(x·y) = d(x)+d(y). This leads to:
          (x·y) mod 9 = d(x·y) mod 9 = (d(x)+d(y)) mod 9 = (d(x) mod 9 + d(y) mod 9) mod 9
            = (x mod 9 + y mod 9) mod 9 = (x+y) mod 9

The solutions to the congruence are (x mod 9, y mod 9) in {(0,0), (2,2), (3,6), (5,8), (6,3), (8,5)}
Only these cases (6 out of 81) have to be tested in a vampire search based on testing x·y for different values of x and y.

下面五中方法, 其中还是ThinkinJava给出的参考答案效率最高, 其他高效率做法 , 请网友高手大神补充

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1. import java.util.Arrays;  
2. public class Demo3 {  
3.   
4.     static int a;       //千位  
5.     static int b;       //百位  
6.     static int c;       //十位  
7.     static int d;       //个位  
8.   
9.     static int a1;      //十位  
10.     static int b1;      //个位  
11.     static int c1;      //十位  
12.     static int d1;      //个位  
13.     static int sum  = 0; //总和  
14.     static int sum2 = 0; //两数之积  
15.   
16.     public static void main(String[] args) {  
17.   
18.         long startTime = System.nanoTime();  
19.         method1();  
20.         long endTime = System.nanoTime();  
21.         System.out.println(”method1 :” + (endTime - startTime)); //method1 :185671841  
22.         long s = System.nanoTime();  
23.         method2();  
24.         long d = System.nanoTime();  
25.         System.out.println(”method2 :” + (d - s)); //method2 :90556063  
26.         long s3 = System.nanoTime();  
27.         method3();  
28.         long d3 = System.nanoTime();  
29.         System.out.println(”method3 :” + (d3 - s3));//method3 :574735  
30.         long s4 = System.nanoTime();  
31.         method4();  
32.         long d4 = System.nanoTime();  
33.         System.out.println(”method4 :” + (d4 - s4));//method4 :22733469  
34.         long s5 = System.nanoTime();  
35.         method5();  
36.         long d5 = System.nanoTime();  
37.         System.out.println(”method5 :” + (d5 - s5));//method4 :19871660  
38.   
39.     }  
40.   
41.     private static void method5() {  
42.         new VampireNumbers(); //该方法 有重复数字  
43.     }  
44.   
45.     static class VampireNumbers {  
46.         static int a(int i) {  
47.             return i / 1000;  
48.         }  
49.   
50.         static int b(int i) {  
51.             return (i % 1000) / 100;  
52.         }  
53.   
54.         static int c(int i) {  
55.             return ((i % 1000) % 100) / 10;  
56.         }  
57.   
58.         static int d(int i) {  
59.             return ((i % 1000) % 100) % 10;  
60.         }  
61.   
62.         static int com(int i, int j) {  
63.             return (i * 10) + j;  
64.         }  
65.   
66.         static void productTest(int i, int m, int n) {  
67.             if (m * n == i)  
68.                 System.out.println(i + ” = ” + m + “ * ” + n);  
69.         }  
70.   
71.         public VampireNumbers() {  
72.             for (int i = 1001; i < 9999; i++) {  
73.                 productTest(i, com(a(i), b(i)), com(c(i), d(i)));  
74.                 productTest(i, com(a(i), b(i)), com(d(i), c(i)));  
75.                 productTest(i, com(a(i), c(i)), com(b(i), d(i)));  
76.                 productTest(i, com(a(i), c(i)), com(d(i), b(i)));  
77.                 productTest(i, com(a(i), d(i)), com(b(i), c(i)));  
78.                 productTest(i, com(a(i), d(i)), com(c(i), b(i)));  
79.                 productTest(i, com(b(i), a(i)), com(c(i), d(i)));  
80.                 productTest(i, com(b(i), a(i)), com(d(i), c(i)));  
81.                 productTest(i, com(b(i), c(i)), com(d(i), a(i)));  
82.                 productTest(i, com(b(i), d(i)), com(c(i), a(i)));  
83.                 productTest(i, com(c(i), a(i)), com(d(i), b(i)));  
84.                 productTest(i, com(c(i), b(i)), com(d(i), a(i)));  
85.             }  
86.         }  
87.     }  
88.   
89.     private static void method4() { // 改进  
90.         for (int i = 11; i < 100; i++) {  
91.             for (int j = i; j < 100; j++) {  
92.                 int k = i * j;  
93.                 String kStr = Integer.toString(k);  
94.                 String checkStr = Integer.toString(i) + Integer.toString(j);  
95.                 if (kStr.length() != 4)  
96.                     continue;  
97.                 char[] kChar = kStr.toCharArray();  
98.                 char[] checkChar = checkStr.toCharArray();  
99.                 Arrays.sort(kChar);  
100.                 Arrays.sort(checkChar);  
101.                 boolean isVampire = Arrays.equals(kChar, checkChar);  
102.                 if (isVampire) {  
103.                     System.out.println(i + ” * ” + j + “ = ” + k);  
104.                 }  
105.             }  
106.         }  
107.     }  
108.   
109.     private static void method3() { // 官方参考答案  
110.   
111.         int[] startDigit = new int[4];  
112.         int[] productDigit = new int[4];  
113.         for (int num1 = 10; num1 <= 99; num1++)  
114.             for (int num2 = num1; num2 <= 99; num2++) {  
115.                 // Pete Hartley’s theoretical result:  
116.                 // If x·y is a vampire number then  
117.                 // x·y == x+y (mod 9)  
118.                 if ((num1 * num2) % 9 != (num1 + num2) % 9)  
119.                     continue;  
120.                 int product = num1 * num2;  
121.                 startDigit[0] = num1 / 10;  
122.                 startDigit[1] = num1 % 10;  
123.                 startDigit[2] = num2 / 10;  
124.                 startDigit[3] = num2 % 10;  
125.                 productDigit[0] = product / 1000;  
126.                 productDigit[1] = (product % 1000) / 100;  
127.                 productDigit[2] = product % 1000 % 100 / 10;  
128.                 productDigit[3] = product % 1000 % 100 % 10;  
129.                 int count = 0;  
130.                 for (int x = 0; x < 4; x++)  
131.                     for (int y = 0; y < 4; y++) {  
132.                         if (productDigit[x] == startDigit[y]) {  
133.                             count++;  
134.                             productDigit[x] = -1;  
135.                             startDigit[y] = -2;  
136.                             if (count == 4)  
137.                                 System.out.println(num1 + ” * ” + num2 + “ : ”  
138.                                         + product);  
139.                         }  
140.                     }  
141.   
142.             } /* 
143.                * Output: 15 * 93 : 1395 21 * 60 : 1260 21 * 87 : 1827 27 * 81 : 
144.                * 2187 30 * 51 : 1530 35 * 41 : 1435 80 * 86 : 6880 
145.                *///:~  
146.     }  
147.   
148.     private static void method2() { // 穷举2  
149.   
150.         //遍历四位数，排除00 从1001开始  
151.         for (int i = 1001; i <= 9999; i++) {  
152.             //排除00   
153.             if (i % 100 != 00) {  
154.                 for (int k = 0; k < 100; k += 10) {  
155.                     if (k != 0) {  
156.                         //10 -99  
157.                         for (int j2 = 0; j2 <= 9; j2++) {  
158.                             //生成第一个两位数   
159.                             for (int j = 0; j < 100; j += 10) {  
160.                                 for (int j3 = 0; j3 <= 9; j3++) {  
161.                                     //生成第二个两位数   
162.                                     //判断两数之积   
163.   
164.                                     if ((k + j2) * (j + j3) == i) {  
165.                                         if (compare2(i, k / 10, j2, j / 10, j3)) {  
166.                                             System.out  
167.                                                     .println(i + ”=” + (k + j2)  
168.                                                             + ”*” + (j + j3));  
169.                                         }  
170.   
171.                                     }  
172.                                 }  
173.                             }  
174.   
175.                         }  
176.                     }  
177.                 }  
178.             }  
179.   
180.         }  
181.     }  
182.   
183.     public static void method1() { //穷举1  
184.         int x = 0, y = 0;  
185.         for (int i = 1; i <= 9; i++) {  
186.             a = i * 1000;  
187.             for (int j = 0; j <= 9; j++) {  
188.                 b = j * 100;  
189.                 for (int j2 = 0; j2 < 10; j2++) {  
190.                     c = j2 * 10;  
191.                     for (int k = 0; k < 10; k++) {  
192.                         d = k;  
193.                         sum = a + b + c + d;  
194.                         //取其中四个数字 中组成两个两位数 ，如果这两个两位数之积  等于 sum ,则输入 这个数  
195.                         for (int k2 = 1; k2 < 10; k2++) {  
196.   
197.                             a1 = k2 * 10;  
198.                             for (int l = 0; l < 10; l++) {  
199.                                 if (a1 + b1 > 100) {  
200.                                     break;  
201.                                 }  
202.                                 b1 = l;  
203.                                 //得到一个两位数字  
204.                                 for (int l2 = 1; l2 < 10; l2++) {  
205.                                     c1 = l2 * 10;  
206.                                     for (int m = l; m < 10; m++) {  
207.   
208.                                         if (c1 + d1 > 100) {  
209.                                             break;  
210.                                         }  
211.                                         d1 = m;  
212.                                         //再得到一个两位数字  
213.                                         sum2 = (a1 + b1) * (c1 + d1);  
214.                                         //计算来两个两位数字之积，如果等于sum   
215.                                         if (sum2 == sum) {  
216.                                             //且尾数不能为00   
217.                                             if (c + d != 0) {  
218.                                                 // 比较这个几个数字 是否一样  
219.                                                 if (compare(a, b, c, d, a1, b1,  
220.                                                         c1, d1)) {  
221.                                                     System.out.println(sum  
222.                                                             + ”=” + (a1 + b1)  
223.                                                             + ”*” + (c1 + d1));  
224.                                                 }  
225.                                             }  
226.                                         }  
227.                                     }  
228.                                 }  
229.                             }  
230.                         }  
231.                     }  
232.                 }  
233.             }  
234.         }  
235.     }  
236.   
237.     private static boolean compare2(int i, int j, int j2, int k, int j3) {  
238.         int a[] = { i % 10, i / 10 % 10, i / 100 % 10, i / 1000 };  
239.         int b[] = { j, j2, k, j3 };  
240.         Arrays.sort(a);  
241.         Arrays.sort(b);  
242.         if (Arrays.equals(a, b))  
243.             return true;  
244.         else  
245.             return false;  
246.   
247.     }  
248.   
249.     private static boolean compare(int a2, int b2, int c2, int d2, int a12,  
250.                                    int b12, int c12, int d12) {  
251.   
252.         int[] a = new int[4];  
253.         int[] b = new int[4];  
254.   
255.         a[0] = a2 / 1000;  
256.         a[1] = b2 / 100;  
257.         a[2] = c2 / 10;  
258.         a[3] = d2;  
259.   
260.         b[0] = a12 / 10;  
261.         b[1] = b12;  
262.         b[2] = c12 / 10;  
263.         b[3] = d12;  
264.   
265.         Arrays.sort(a);  
266.         Arrays.sort(b);  
267.   
268.         if (Arrays.equals(a, b))  
269.             return true;  
270.         else  
271.             return false;  
272.     }  
273. }  
274. }  

import java.util.Arrays;
public class Demo3 {

    static int a;       //千位
    static int b;       //百位
    static int c;       //十位
    static int d;       //个位

    static int a1;      //十位
    static int b1;      //个位
    static int c1;      //十位
    static int d1;      //个位
    static int sum  = 0; //总和
    static int sum2 = 0; //两数之积

    public static void main(String[] args) {

        long startTime = System.nanoTime();
        method1();
        long endTime = System.nanoTime();
        System.out.println("method1 :" + (endTime - startTime)); //method1 :185671841
        long s = System.nanoTime();
        method2();
        long d = System.nanoTime();
        System.out.println("method2 :" + (d - s)); //method2 :90556063
        long s3 = System.nanoTime();
        method3();
        long d3 = System.nanoTime();
        System.out.println("method3 :" + (d3 - s3));//method3 :574735
        long s4 = System.nanoTime();
        method4();
        long d4 = System.nanoTime();
        System.out.println("method4 :" + (d4 - s4));//method4 :22733469
        long s5 = System.nanoTime();
        method5();
        long d5 = System.nanoTime();
        System.out.println("method5 :" + (d5 - s5));//method4 :19871660

    }

    private static void method5() {
        new VampireNumbers(); //该方法 有重复数字
    }

    static class VampireNumbers {
        static int a(int i) {
            return i / 1000;
        }

        static int b(int i) {
            return (i % 1000) / 100;
        }

        static int c(int i) {
            return ((i % 1000) % 100) / 10;
        }

        static int d(int i) {
            return ((i % 1000) % 100) % 10;
        }

        static int com(int i, int j) {
            return (i * 10) + j;
        }

        static void productTest(int i, int m, int n) {
            if (m * n == i)
                System.out.println(i + " = " + m + " * " + n);
        }

        public VampireNumbers() {
            for (int i = 1001; i < 9999; i++) {
                productTest(i, com(a(i), b(i)), com(c(i), d(i)));
                productTest(i, com(a(i), b(i)), com(d(i), c(i)));
                productTest(i, com(a(i), c(i)), com(b(i), d(i)));
                productTest(i, com(a(i), c(i)), com(d(i), b(i)));
                productTest(i, com(a(i), d(i)), com(b(i), c(i)));
                productTest(i, com(a(i), d(i)), com(c(i), b(i)));
                productTest(i, com(b(i), a(i)), com(c(i), d(i)));
                productTest(i, com(b(i), a(i)), com(d(i), c(i)));
                productTest(i, com(b(i), c(i)), com(d(i), a(i)));
                productTest(i, com(b(i), d(i)), com(c(i), a(i)));
                productTest(i, com(c(i), a(i)), com(d(i), b(i)));
                productTest(i, com(c(i), b(i)), com(d(i), a(i)));
            }
        }
    }

    private static void method4() { // 改进
        for (int i = 11; i < 100; i++) {
            for (int j = i; j < 100; j++) {
                int k = i * j;
                String kStr = Integer.toString(k);
                String checkStr = Integer.toString(i) + Integer.toString(j);
                if (kStr.length() != 4)
                    continue;
                char[] kChar = kStr.toCharArray();
                char[] checkChar = checkStr.toCharArray();
                Arrays.sort(kChar);
                Arrays.sort(checkChar);
                boolean isVampire = Arrays.equals(kChar, checkChar);
                if (isVampire) {
                    System.out.println(i + " * " + j + " = " + k);
                }
            }
        }
    }

    private static void method3() { // 官方参考答案

        int[] startDigit = new int[4];
        int[] productDigit = new int[4];
        for (int num1 = 10; num1 <= 99; num1++)
            for (int num2 = num1; num2 <= 99; num2++) {
                // Pete Hartley's theoretical result:
                // If x·y is a vampire number then
                // x·y == x+y (mod 9)
                if ((num1 * num2) % 9 != (num1 + num2) % 9)
                    continue;
                int product = num1 * num2;
                startDigit[0] = num1 / 10;
                startDigit[1] = num1 % 10;
                startDigit[2] = num2 / 10;
                startDigit[3] = num2 % 10;
                productDigit[0] = product / 1000;
                productDigit[1] = (product % 1000) / 100;
                productDigit[2] = product % 1000 % 100 / 10;
                productDigit[3] = product % 1000 % 100 % 10;
                int count = 0;
                for (int x = 0; x < 4; x++)
                    for (int y = 0; y < 4; y++) {
                        if (productDigit[x] == startDigit[y]) {
                            count++;
                            productDigit[x] = -1;
                            startDigit[y] = -2;
                            if (count == 4)
                                System.out.println(num1 + " * " + num2 + " : "
                                        + product);
                        }
                    }

            } /*
               * Output: 15 * 93 : 1395 21 * 60 : 1260 21 * 87 : 1827 27 * 81 :
               * 2187 30 * 51 : 1530 35 * 41 : 1435 80 * 86 : 6880
               *///:~
    }

    private static void method2() { // 穷举2

        //遍历四位数，排除00 从1001开始
        for (int i = 1001; i <= 9999; i++) {
            //排除00 
            if (i % 100 != 00) {
                for (int k = 0; k < 100; k += 10) {
                    if (k != 0) {
                        //10 -99
                        for (int j2 = 0; j2 <= 9; j2++) {
                            //生成第一个两位数 
                            for (int j = 0; j < 100; j += 10) {
                                for (int j3 = 0; j3 <= 9; j3++) {
                                    //生成第二个两位数 
                                    //判断两数之积 

                                    if ((k + j2) * (j + j3) == i) {
                                        if (compare2(i, k / 10, j2, j / 10, j3)) {
                                            System.out
                                                    .println(i + "=" + (k + j2)
                                                            + "*" + (j + j3));
                                        }

                                    }
                                }
                            }

                        }
                    }
                }
            }

        }
    }

    public static void method1() { //穷举1
        int x = 0, y = 0;
        for (int i = 1; i <= 9; i++) {
            a = i * 1000;
            for (int j = 0; j <= 9; j++) {
                b = j * 100;
                for (int j2 = 0; j2 < 10; j2++) {
                    c = j2 * 10;
                    for (int k = 0; k < 10; k++) {
                        d = k;
                        sum = a + b + c + d;
                        //取其中四个数字 中组成两个两位数 ，如果这两个两位数之积  等于 sum ,则输入 这个数
                        for (int k2 = 1; k2 < 10; k2++) {

                            a1 = k2 * 10;
                            for (int l = 0; l < 10; l++) {
                                if (a1 + b1 > 100) {
                                    break;
                                }
                                b1 = l;
                                //得到一个两位数字
                                for (int l2 = 1; l2 < 10; l2++) {
                                    c1 = l2 * 10;
                                    for (int m = l; m < 10; m++) {

                                        if (c1 + d1 > 100) {
                                            break;
                                        }
                                        d1 = m;
                                        //再得到一个两位数字
                                        sum2 = (a1 + b1) * (c1 + d1);
                                        //计算来两个两位数字之积，如果等于sum 
                                        if (sum2 == sum) {
                                            //且尾数不能为00 
                                            if (c + d != 0) {
                                                // 比较这个几个数字 是否一样
                                                if (compare(a, b, c, d, a1, b1,
                                                        c1, d1)) {
                                                    System.out.println(sum
                                                            + "=" + (a1 + b1)
                                                            + "*" + (c1 + d1));
                                                }
                                            }
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }

    private static boolean compare2(int i, int j, int j2, int k, int j3) {
        int a[] = { i % 10, i / 10 % 10, i / 100 % 10, i / 1000 };
        int b[] = { j, j2, k, j3 };
        Arrays.sort(a);
        Arrays.sort(b);
        if (Arrays.equals(a, b))
            return true;
        else
            return false;

    }

    private static boolean compare(int a2, int b2, int c2, int d2, int a12,
                                   int b12, int c12, int d12) {

        int[] a = new int[4];
        int[] b = new int[4];

        a[0] = a2 / 1000;
        a[1] = b2 / 100;
        a[2] = c2 / 10;
        a[3] = d2;

        b[0] = a12 / 10;
        b[1] = b12;
        b[2] = c12 / 10;
        b[3] = d12;

        Arrays.sort(a);
        Arrays.sort(b);

        if (Arrays.equals(a, b))
            return true;
        else
            return false;
    }
}
}