packagetest4;importjava.util.Arrays;/*** 从TIJ中第4章的练习10看到“吸血鬼数字”,以下几种方法实现以及执行时间对比

* 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*/

public classTest1 {public static voidmain(String[] args) {long start =System.nanoTime();

fun1();long end =System.nanoTime();

System.out.println("方法1所用时间：" + (end - start)+"\n");

start=System.nanoTime();

fun2();

end=System.nanoTime();

System.out.println("方法2所用时间：" + (end - start)+"\n");

start=System.nanoTime();

fun3();

end=System.nanoTime();

System.out.println("方法3所用时间：" + (end - start)+"\n");

start=System.nanoTime();

fun4();

end=System.nanoTime();

System.out.println("方法4所用时间：" + (end - start)+"\n");

}private static voidfun1() {//参考答案

int sum = 0;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("第" + sum + "组: " + num1 + " * " + num2 + " : " +product);

sum++;

}

}

}

}

System.out.println("方法1共找到" + sum + "组吸血鬼数");

}private static voidfun2() {

String[] ar_str1, ar_str2;int sum = 0;intfrom;intto;inti_val;for (int i = 10; i < 100; i++) {

from= Math.max(1000 / i, i + 1);

to= Math.min(10000 / i, 100);//2个数的乘积是4位数(大于等于1000,小于10000),i确定时,另一个数范围随之确定

for (int j = 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 的理解://假设val = 1000a + 100b + 10c + d, 因为满足val = x * y, 则有x =//10a + b, y = 10c + d//可得val - x - y = 990a + 99b + 9c = 9 * (110a + 11b + c),//所以val - x - y能被9整除。//满足该条件的数字必定能被9整除,可以直接过滤其他数字。

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("第" + sum + "组: " + i + "*" + j + "=" +i_val);

}

}

}

System.out.println("方法2共找到" + sum + "组吸血鬼数");

}private static voidfun3() {int sum = 0;for (int i = 11; i < 100; i++) {for (int j = i; j < 100; j++) {int k = i *j;//有另一种变为字符串来操作,比较发现下面的这种方法耗时更少

int[] a = { k / 1000, k / 100 % 10, k / 10 % 100 % 10, k % 1000 % 100 % 10};int[] b = { i % 10, i / 10, j % 10, j / 10};

Arrays.sort(a);

Arrays.sort(b);if(Arrays.equals(a, b)) {

sum++;

System.out.println("第" + sum + "组: " + i + " * " + j + " = " +k);

}

}

}

System.out.println("方法3共找到" + sum + "组吸血鬼数");

}private static voidfun4() {//逆向思维

String[] targetNum = null;

String[] gunNum= null;int sum = 0;for (int i = 10; i < 100; i++) {for (int j = i + 1; j < 100; j++) {//没有哪个两位数满足ab*ab=abab,所以这里j从i+1开始就可以了

int i_target = i *j;if (i_target < 1000 || i_target > 9999)continue; //积不是4位数则跳过

targetNum = 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("第" + sum + "组: " + i_target + "=" + i + "*" +j);

}

}

}

System.out.println("方法4找到" + sum + "个吸血鬼数字。");

}

}