Java_ Vampire Digital Multi-method implementation

 Packagetest4;Importjava.util.Arrays;/*** from the Tij in the 4th chapter of exercise 10 See "Vampire numbers", the following methods of implementation and execution time comparison * Find out four digits of all vampire numbers * Vampire numbers are digits with even numbers, which can be multiplied by a pair of numbers, which each contain a number of half-digits of the product, * The numbers selected from the initial numbers can be sorted arbitrarily.  * Numbers ending with two 0 are not allowed. * For example the following numbers are vampire numbers * 1260=21*60 * 1827=21*87 * 2187=27*81*/ Public classTest1 { Public Static voidMain (string[] args) {LongStart =System.nanotime ();        Fun1 (); LongEnd =System.nanotime (); System.out.println ("Method 1 Time:" + (End-start) + "\ n"); Start=System.nanotime ();        Fun2 (); End=System.nanotime (); System.out.println ("Method 2 Time:" + (End-start) + "\ n"); Start=System.nanotime ();        Fun3 (); End=System.nanotime (); System.out.println ("Method 3 Time:" + (End-start) + "\ n"); Start=System.nanotime ();        Fun4 (); End=System.nanotime (); System.out.println ("Method 4 Time:" + (End-start) + "\ n"); }    Private Static voidfun1 () {//Reference Answer        intsum = 0; int[] Startdigit =New int[4]; int[] Productdigit =New int[4];  for(intNUM1 = 10; NUM1 <= 99; num1++)             for(intnum2 = 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; intProduct = 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; intCount = 0;  for(intx = 0; x < 4; X + +)                     for(inty = 0; Y < 4; y++) {                        if(Productdigit[x] = =Startdigit[y]) {Count++; PRODUCTDIGIT[X]=-1; Startdigit[y]=-2; if(Count = = 4) {System.out.println ("First" + Sum + "group:" + NUM1 + "*" + num2 + ":" +product); Sum++; } }}} System.out.println ("Method 1 found" + sum + "Group of Vampires"); }    Private Static voidfun2 () {string[] ar_str1, AR_STR2; intsum = 0; intFrom ; intto ; intI_val;  for(inti = 10; I < 100; i++) { from= Math.max (1000/i, i + 1); to= Math.min (10000/i, 100); //the product of 2 numbers is a 4-digit number (greater than or equal to 1000, less than 10000), and when I determines, the other number range is determined             for(intj = from; J < to; J + +) {I_val= i *J; if(i_val% 100 = = 0 | | (I_val-i-J)% 9! = 0) {                    //(I_val-i-j)% 9! = 0 Understanding://Suppose val = 1000a + 100b + 10c + D, because the val = x * y is satisfied, there is x =//10a + b, y = 10c + D//available Val-x-y = 990a + 99b + 9c = 9 * (110a + 11b + c),//so val-x-y can be divisible by 9. //The number that satisfies the condition must be divisible by 9, and other numbers can be filtered directly.                     Continue; } ar_str1= String.valueof (I_val). Split (""); AR_STR2= (String.valueof (i) + string.valueof (j)). Split ("");                Arrays.sort (AR_STR1);                Arrays.sort (AR_STR2); if(Arrays.equals (AR_STR1, ar_str2)) {sum++; System.out.println ("First" + Sum + "group:" + i + "*" + j + "=" +i_val); }}} System.out.println ("Method 2 Found" + Sum + "Group of Vampires"); }    Private Static voidFun3 () {intsum = 0;  for(inti = 11; I < 100; i++) {             for(intj = i; J < 100; J + +) {                intK = i *J; //there is another way to manipulate the string, and the comparison finds that the following method takes less time                int[] A = {k/1000, k/100%, k/10%, K% 1000 100 10 }; int[] b = {i%, I/10, J, J/10 };                Arrays.sort (a);                Arrays.sort (b); if(Arrays.equals (A, b)) {Sum++; System.out.println ("First" + Sum + "group:" + i + "*" + j + "=" +k); }}} System.out.println ("Method 3 found" + Sum + "Group of Vampires"); }    Private Static voidFun4 () {//Reverse Thinkingstring[] Targetnum =NULL; String[] Gunnum=NULL; intsum = 0;  for(inti = 10; I < 100; i++) {             for(intj = i + 1; J < 100; J + +) {                //no two-digit number satisfies ab*ab=abab, so here J starts with I+1.                intI_target = i *J; if(I_target < | | i_target > 9999)                    Continue;//the product is not a 4-digit skipTargetnum = string.valueof (I_target). Split (""); Gunnum= (String.valueof (i) + string.valueof (j)). Split ("");                Arrays.sort (Targetnum);                Arrays.sort (Gunnum); if(Arrays.equals (Targetnum, gunnum)) {sum++; System.out.println ("First" + Sum + "group:" + i_target + "=" + i + "*" +j); }}} System.out.println ("Method 4 finds" + sum + "vampire numbers. "); }}

Execution Result:

Group No. 0:15 * 93:1395Group 1th:21 * 60:1260Group 2nd:21 * 87:1827Group 3rd:27 * 81:2,187Group 4th:30 * 51:1530Group 5th:35 * 41:1435Group 6th:80 * 86:6,880Method 1 found 7 groups of vampire number method 1 used time:4880538Group 1th:15*93=1395Group 2nd:21*60=1260Group 3rd:21*87=1827Group 4th:27*81=2187Group 5th:30*51=1530Group 6th:35*41=1435Group 7th:80*86=6880Method 2 found 7 groups of vampire number Method 2 used time:43971275Group 1th:15 * 93 = 1395Group 2nd:21 * 60 = 1260Group 3rd:21 * 87 = 1827Group 4th:27 * 81 = 2187Group 5th:30 * 51 = 1530Group 6th:35 * 41 = 1435Group 7th:80 * 86 = 6880Method 3 found 7 groups of Vampire number method 3 used time:19352070Group 1th:1395=15*93Group 2nd:1260=21*60Group 3rd:1827=21*87Group 4th:2187=27*81Group 5th:1530=30*51Group 6th:1435=35*41Group 7th:6880=80*86Method 4 found 7 vampire numbers. Method 4 Time:125098134

Java_ Vampire Digital Multi-method implementation