Java finds all four-digit vampire digital Base code examples

/**
* Find out four digits of all vampire numbers
* Vampire numbers are numbers with an even number of digits that can be multiplied by a pair of numbers, each of which includes the number of half-digits of the product, in which the numbers selected from the original numbers can be sorted arbitrarily.
* With two 0 The ending number is not agreeable.
* Proportions such as the following numbers are vampire numbers
1260=21*60
1827=21*87
2187=27*81
...
* Rather stupid inefficient approach: traverse all four digits, each generating a four-digit number,
* Traverse two digits in a double loop to infer whether the four-bit number is equal to the outermost loop in the inner loop of the two-digit number. Assuming equality to store these numbers in an array, the comparison

* Two sets of numbers, assuming equal then the output of this number is the number to find;

*/

Read the following English: vampire numbers

An important theoretical result found by Pete Hartley:
If x y is a vampire number thenx y = = x+y (mod 9)
Proof:
LetMoDBe the binary modulo operator andd (x)The sum of the decimal digits of X.
It is well-known this 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 is(x mod 9, y mod 9)inch{(0,0), (2,2), (3,6), (5,8), (6,3), (8,5)}
Only these cases (6 out of Bayi) has to be tested in a vampire search based on testing x y for different values of x an D y.

The following V method, which is still thinkinjava given the most efficient answer, other high-efficiency practices, please netizen master God to add

Import Java.util.arrays;public class Demo3 {static int A;       thousand static int b;       hundred static int C;       10-bit static int D;      single-digit static int A1;      10-bit static int B1;      single-digit static int C1;      10-bit static int d1; digit static int sum = 0; sum static int sum2 = 0;        Two numbers of the product 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 Vampire Numbers ();        The method has repeated numbers} 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 *) + 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 () {//improved for (int i = one; i <; 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 () {//Official Test answer int[] startdigit = new Int[4];        int[] Productdigit = new Int[4]; for (int num1 = ten; num1 <=; num1++) for (int num2 = NUM1; num2 <=; num2++) {//Pet                E 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; + +) 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 * Bayi: * 2187 * 51:1530 * 41:1435 * 86:6880 *///:~} private S              tatic void Method2 () {//Poor lift 2//traverse four digits, exclude 00 from 1001 for (int i = 1001; I <= 9999; i++) {//exclude 00                        if (i% = xx) {for (int k = 0; k <; k + =) {if (k! = 0) { 10-99 for (int j2 = 0; J2 <= 9; j2++) {//Generate The first two-digit for (int j = 0; J <; J + =) {for (int j3 = 0; J3 <= 9;                                     j3++) {//Generate a second double digit//infer the product of two numbers 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 () {//poor lift 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 < j2++) {c = J2 * 10;                        for (int k = 0; k < k++) {d = k;                        sum = a + B + C + D;                            Take the four number of the two two-digit number, assuming that the two two-digit product equals sum, enter this number for (int k2 = 1; k2 < k2++) {                            A1 = K2 * 10; for (int l = 0; l < l++) {if (A1 +B1 >) {break;                                } B1 = l; Get a two-bit number for (int l2 = 1; L2 < l2++) {C1 = L2                                    * 10;                                            for (int m = l; m < m++) {if (c1 + D1 > 100) {                                        Break                                        } d1 = m;                                        Then get a two digit sum2 = (A1 + b1) * (c1 + d1);                                            Calculates the product of two two-bit numbers, assuming equal to sum if (sum2 = = sum) {                                                And the mantissa cannot be an if (c + d! = 0) {                       Is it the same as a few numbers?                         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 J        2, int k, int j3) {int a[] = {i%, i/10%, i/100%, 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; }}}