《Thinking in Java》P75
練習10:(5)吸血鬼數字是指位數爲偶數的數字,可以由一對數字相乘而得到,而這對數字各包含乘積的一半位的數字,其中從最初的數字中選取的數字可以任意排序。以兩個0結尾的數字是不允許的,例如,下列數字都是“吸血鬼”數字:
1260 = 21 * 60
1827 = 21 * 87
2187 = 27 * 81
寫一個程序,找出4位數的所有吸血鬼數字(Dan Forhan推薦)。
我的方案1:窮舉四位數得到從左到右的 a1 a2 a3 a4 四個數字,進行排列組合,滿足 n = (a1*10 + a2)*(a3*10 + a4)或類似條件,就打印出來。
好處是時間複雜度爲N,但是四位數窮舉本身比較次數就很高。
import java.util.StringTokenizer;
/**
* Created by sherry on 17/2/14.
*/
public class vampire {
public static void main(String[] args){
for(int n = 1000;n <= 9999;n++)
{
int a1 = n/1000;
int a2 = n/100 - a1*10;
int a3 = n/10 - a2*10 - a1*100;
int a4 = n - a3*10 - a2*100 - a1*1000;
if((a1*10 + a2)*(a3*10 + a4) == n)
{
System.out.print(n + " ");
}
if((a2*10 + a1)*(a3*10 + a4) == n)
{
System.out.print(n + " ");
}
if((a1*10 + a2)*(a4*10 + a3) == n)
{
System.out.print(n + " ");
}
if((a2*10 + a1)*(a4*10 + a3) == n)
{
System.out.print(n + " ");
}
if((a1*10 + a3)*(a2*10 + a4) == n)
{
System.out.print(n + " ");
}
if((a3*10 + a1)*(a2*10 + a4) == n)
{
System.out.print(n + " ");
}
if((a1*10 + a3)*(a4*10 + a2) == n)
{
System.out.print(n + " ");
}
if((a3*10 + a1)*(a4*10 + a2) == n)
{
System.out.print(n + " ");
}
if((a1*10 + a4)*(a2*10 + a3) == n)
{
System.out.print(n + " ");
}
if((a4*10 + a1)*(a2*10 + a3) == n)
{
System.out.print(n + " ");
}
if((a1*10 + a4)*(a3*10 + a2) == n)
{
System.out.print(n + " ");
}
if((a4*10 + a1)*(a3*10 + a2) == n)
{
System.out.print(n + " ");
}
}
}
}
方案2:方案1的簡潔版
public 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 static void main(String[] args) {
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)));
}
}
}
```![這裏寫圖片描述](http://img.blog.csdn.net/20170214214322509?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvdTAxMjk1OTQ5OA==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast)
方案3:另一種思路,通過先窮舉兩位數的方式,再比較四位數結果。時間效率爲N²,但實際比較次數有改善。參考博客http://www.cnblogs.com/vincent-hv/archive/2013/04/19/3030031.html
<div class="se-preview-section-delimiter"></div>
import java.util.StringTokenizer;
import java.util.Arrays;
/**
* Created by sherry on 17/2/14.
*/
public class vampire {
public static void main(String[] args){
int count = 0;
String[] ar_str1, ar_str2;
for(int x = 10;x < 100;x++) {
int from = Math.max(1000 / x, x);
int to = Math.min(10000 / x, 100);
for (int y = from; y < to; y++) {
int integrity = x * y;
//(integrity - x - y) % 9的由來:
//x = 10*a1 + a2, y = 10*a3 + a4
//integrity = 1000*a1 + 100*a2 + 10*a3 + a4
//integrity - x - y = 990*a1 + 99*a2 + 9*a3 = 9*(110*a1 + 11*a2 + a3)
//通過上式保證 x*y = x和y元素的組合四位數(integrity)
if (integrity % 100 == 0 || (integrity - x - y) % 9 != 0) {
continue;
}
ar_str1 = String.valueOf(integrity).split("");
ar_str2 = (String.valueOf(x) + String.valueOf(y)).split("");
Arrays.sort(ar_str1);
Arrays.sort(ar_str2);
if (Arrays.equals(ar_str1, ar_str2)) {// 排序後比較,爲真則找到一組
System.out.println( x + "*" + y + "=" + integrity);
count++;
}
}
}
System.out.print("總數 = " + count + " ");
}
}
“`
運行結果:
問題在於Arrays.sort()之後,無法區別 80*60 = 6880 和 60*80 = 6880,所以只有7行結果。