（求助）吸血鬼数字 算法问题
public class xixuegui {
public static void main (String[] args) {
int i,j,k,a,b,c,d,n,m;
for(i=1001;i <9999;i++){
a=i/1000;
b=(i-a*1000)/100;
c=(i-a*1000-b*100)/10;
d=i-a*1000-b*100-c*10;
for(j=1;j <10;j++){
for(k=0;k <10;k++){
if(i==(j*1000)+(k*100)) continue;
}
}
if(i==(a*10+b)*(c*10+d)) System.out.println(i+ "is xixuegui ");
if(i==(a*10+b)*(d*10+c)) System.out.println(i+ "is xixuegui ");
if(i==(b*10+a)*(c*10+d)) System.out.println(i+ "is xixuegui ");
if(i==(b*10+a)*(d*10+c)) System.out.println(i+ "is xixuegui ");
if(i==(a*10+c)*(b*10+d)) System.out.println(i+ "is xixuegui ");
if(i==(a*10+c)*(d*10+b)) System.out.println(i+ "is xixuegui ");
if(i==(c*10+a)*(b*10+d)) System.out.println(i+ "is xixuegui ");
if(i==(c*10+a)*(d*10+b)) System.out.println(i+ "is xixuegui ");
if(i==(a*10+d)*(b*10+c)) System.out.println(i+ "is xixuegui ");
if(i==(a*10+d)*(c*10+b)) System.out.println(i+ "is xixuegui ");
if(i==(d*10+a)*(b*10+c)) System.out.println(i+ "is xixuegui ");
if(i==(d*10+a)*(c*10+b)) System.out.println(i+ "is xixuegui ");
if(i==(b*10+c)*(a*10+d)) System.out.println(i+ "is xixuegui ");
if(i==(b*10+c)*(d*10+a)) System.out.println(i+ "is xixuegui ");
if(i==(c*10+b)*(a*10+d)) System.out.println(i+ "is xixuegui ");
if(i==(c*10+b)*(d*10+a)) System.out.println(i+ "is xixuegui ");
}
}
}
求助各位啊~~~这是我刚写的一个输出4位数 "吸血鬼数字 "的程序，虽然可以正确输出，但是我觉着这个做法太麻烦了，想得到一个简化代码的而且高效的方法，而且我这程序实际可以说还没完成，因为输出有重复，而我不希望有重复，虽然我知道一个方法，可是由于代码太多我实在觉得那样写出来有点丢人，所以求助各位帮帮我啦，小弟感激不尽~~~~~~
差点忘了，所谓“吸血鬼数字”就是指位数为偶数的数字(我们算得是4位的)，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字，其中从偶数位数字中选取的数字可以任意排列。以两个0截尾的数字是不允许的，例如：
1260=21*60
1827=21*87
2187=27*81
等等。
大家帮帮忙啊~~~~~小弟感激不尽啊~~~
[解决办法]
java编程思想上的题吧，我也做了，我是先把提取的4个数全排列，然后一个一个测试的。而不是像你这样全列出来，而且4个数字全排列应该有24种，你咋只有16种呢
[解决办法]
public static void main(String[] arg){
String[] ar_str1,ar_str2;
int sum=0;
//双重循环穷举
for(int i=10;i <100;i++){
//j=i+1避免重复
for(int j=i+1;j <100;j++){
int i_val=i*j;
if(i_val <1000||i_val> 9999)
continue; //积小于1000或大于9999排除,继续下一轮环
ar_str1=String.valueOf(i_val).split( " ");
ar_str2=(String.valueOf(i)+String.valueOf(j)).split( " ");
java.util.Arrays.sort(ar_str1);
java.util.Arrays.sort(ar_str2);
if(java.util.Arrays.equals(ar_str1, ar_str2)){
//排序后比较,为真则找到一组
sum++;
System.out.println( "第 "+sum+ "组: "+i+ "* "+j+ "= "+i_val);
}
}
}
System.out.println( "共找到 "+sum+ "组吸血鬼数 ");
}
第1组: 15*93=1395
第2组: 21*60=1260
第3组: 21*87=1827
第4组: 27*81=2187
第5组: 30*51=1530
第6组: 35*41=1435
第7组: 80*86=6880
共找到7组吸血鬼数
[解决办法]
学习
[解决办法]
sureyor 的方法很好，排序后再比较，算法不错。
[解决办法]
public class VampireNumber {
private int isVampire(int a, int b) {
int[] number = new int[4];
number[0] = a / 10;
number[1] = a % 10;
number[2] = b / 10;
number[3] = b % 10;
int c = a * b;
int[] number1 = new int[4];
for (int i = 0, temp = c, div = 1000; i < number1.length; i++, div /= 10) {
number1[i] = temp / div;
temp = temp % div;
}
if (compare(number, number1))
return c;
else
return -1;
}
private boolean compare(int[] number, int[] number1) {
number = sort(number);
number1 = sort(number1);
for (int i = 0; i < number.length; i++)
if (number[i] != number1[i])
return false;
return true;
}
private int[] sort(int[] number) {
for (int i = 0; i < number.length - 1; i++)
for (int j = i; j > = 0; j--) {
if (number[j] > number[j + 1]) {
int temp = number[j];
number[j] = number[j + 1];
number[j + 1] = temp;
}
}
return number;
}
public static void main(String[] args) {
VampireNumber v = new VampireNumber();
for (int a = 10; a < 100; a++) {
for (int b = a; b < 100; b++) {
int c = v.isVampire(a, b);
if (c > 0)
System.out.println( "a = " + a + ", b= " + b + ", c = " + c);
}
}
}
}
a = 15, b= 93, c = 1395
a = 21, b= 60, c = 1260
a = 21, b= 87, c = 1827
a = 27, b= 81, c = 2187
a = 30, b= 51, c = 1530
a = 35, b= 41, c = 1435
a = 80, b= 86, c = 6880
[解决办法]
一个很简单的问题，怎么搞得这么复杂
这只是两个循环就能解决问题了
哪有什么必要什么全排列啊
解决问题算法最重要啊
不要一味只看到题目就动手做
等思路想好了，你就会发现会事半功倍的
public class xixuegui {
public static void main(String[] g) {
int num1, num2, product;
int[] startDigit = new int[4];
int[] productDigit = new int[4];
int count = 0;
int vampCount = 0;
int x, y;
for(num1 = 10; num1 <= 99; num1++)
for(num2 = 10; num2 <= 99; num2++) {
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;
count = 0;
for(x = 0; x < 4; x++)
for(y = 0; y < 4; y++) {
if (productDigit[x] == startDigit[y]){
count++;
productDigit[x] = -1;
startDigit[y] = -2;
if (count == 4) {
vampCount++;
int a = (int)Math.random() * 100;
int b = (int)Math.random() * 100;
int c = (int)Math.random() * 100;
if (vampCount < 10) {
System.out.println( "Vampire number "
+ vampCount + " is " + num1 +
" * " + num2 + " = "+ product);
} else {
System.out.println( "Vampire number "
+ vampCount + " is " + num1 +
" * " + num2 + " = "+ product);
}
}
}
}
}
}
}
[解决办法]
昨天刚看的，还没做那，打算明天做，最近太忙
[解决办法]
我是这样写的
package base;
public class Bloodsucker {
static int rowNum = 1;
public static void main(String[] args) {
for(int i = 20; i <= 99;i++){
for(int y = 50; y <= 99;y++){
if(y < i)
continue;
//get sucked 's number
int check = i * y;
int idown = i%10;
int iup = i/10;
int ydown = y%10;
int yup = y/10;
//get sucker 's number
int checkdown = check%10;
int checkmindown = (check/10)%10;
int checkminup = (check/100)%10;
int checkup = check/1000;
int[] sucked = {idown,iup,ydown,yup};
int[] sucker = {checkdown,checkmindown,checkminup,checkup};
int[] mark ={0,0,0,0};
for(int j = 0;j < 4;j++){
for(int k = 0;k < 4;k++){
if(mark[k] ==0){
if(sucker[j] == sucked[k]){
mark[k] = 1;
break;
}
}
}
}
if(mark[0] == 1 && mark[1] == 1 && mark[2] == 1 && mark[3] == 1)
System.out.println( "row " + rowNum++ + ": the bloodsucker is " + check +
",it sucked " + i + " and " + y + "; ");
}
}
}
}
结果是:
row 1: the bloodsucker is 1260,it sucked 21 and 60;
row 2: the bloodsucker is 1827,it sucked 21 and 87;
row 3: the bloodsucker is 2187,it sucked 27 and 81;
row 4: the bloodsucker is 1530,it sucked 30 and 51;
row 5: the bloodsucker is 6880,it sucked 80 and 86;
[解决办法]
sureyor() 对包里提供的工具很熟悉啊
[解决办法]
//: c03:E11_Vampire.java
// Solution by Dan Forhan
import java.applet.*;
import java.awt.*;
public class Vampire extends Applet {
private int num1, num2, product;
private int[] startDigit = new int[4];
private int[] productDigit = new int[4];
private int count = 0;
private int vampCount = 0;
private int x, y;
public void paint(Graphics g) {
g.drawString( "Vampire Numbers ", 10, 20);
g.drawLine(10, 22, 150, 22);
// Positioning for output to applet:
int column = 10, row = 35;
for(num1 = 10; num1 <= 99; num1++)
for(num2 = 10; num2 <= 99; num2++) {
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;
count = 0;
for(x = 0; x < 4; x++)
for(y = 0; y < 4; y++) {
if (productDigit[x] == startDigit[y]){
count++;
productDigit[x] = -1;
startDigit[y] = -2;
if (count == 4) {
vampCount++;
int a = (int)Math.random() * 100;
int b = (int)Math.random() * 100;
int c = (int)Math.random() * 100;
if (vampCount < 10) {
g.drawString( "Vampire number "
+ vampCount + " is " + num1 +
" * " + num2 + " = "+ product,
column, row);
row += 20;
} else {
g.drawString( "Vampire number "
+ vampCount + " is " + num1 +
" * " + num2 + " = "+ product,
column, row);
row += 20;
}
}
}
}
}
}
} ///:~
这个是Thinking in Java作者给的答案，我在《Thinking in Java, 2nd edition, Annotated Solution Guide Revision 1.0》里找到的。
[解决办法]
up
[解决办法]
都是高手啊。。
[解决办法]
void fun()
{
int b[4]={0,0,0,0};
int c[3]={0,0,0};
int d[2]={0,0};
int e[1]={0};
int f[4]={0,0,0,0};
for(int i=1000;i <9999;i++)
{
//取出千百十个位
b[3]=i/1000;
b[2]=(i/100)%10;
b[1]=i%100/10;
b[0]=i%10;
for(int j=0;j <4;j++)
{
int temp0=(b[2])*(b[0]);
if(temp0%10!=b[j]) continue;
if(temp0%10==b[j])//比较个位
{
c[2]=b[(j+1)%4];
c[1]=b[(j+2)%4];
c[0]=b[(j+3)%4];
f[0]=b[j];
for(int k=0;k <3;k++)
{
int temp1=temp0/10+b[1]*b[2]%10+b[0]*b[3]%10;
if(temp1%10!=c[k])continue;
if(temp1%10==c[k])//比较十位
{
d[0]=c[(k+1)%3];
d[1]=c[(k+2)%3];
f[1]=c[k];
for(int l=0;l <2;l++)
{
int temp2=b[0]*b[3]/10+b[1]*b[2]/10+b[3]*b[1]%10;
if(temp1/10+temp2%10!=d[l])continue;
if(temp1/10+temp2%10==d[l])//比较百位
{
e[0]=d[(l+1)%2];
f[2]=d[l];
if((temp2/10+b[1]*b[3]/10==e[0])&&e[0])//比较千位
{
f[3]=e[0];
for(int m=3;m> =0;m--)
{
cout < <f[m];
}
cout < <endl;
}
}
}
}
}
}
}
}
}
[解决办法]
import java.util.Arrays;
public class Xi {
public static void main(String[] args) {
for(int i=1;i <100;i++){
for(int j=1;j <100;j++){
if(Xi.isXiXueGui(i, j)){
System.out.println(i+ "* "+j+ "= "+i*j);
}
}
}
}
public static boolean isXiXueGui(int x1, int x2) {
String strx1 = Integer.toString(x1);
String strx2 = Integer.toString(x2);
String x1x2 = strx1 + strx2;
String xxg = Integer.toString(x1 * x2);
int x1Len = strx1.length();
int x2Len = strx2.length();
int xxgLen = xxg.length();
if ((strx1.indexOf(strx1.length() - 1) != '0 ' && strx2.indexOf(strx2
.length() - 1) != '0 ')
&& x1Len == x2Len && x1Len + x2Len == xxgLen) {
} else {
return false;
}
char[] x1x2c = x1x2.toCharArray();
Arrays.sort(x1x2c);
char[] xxgc = xxg.toCharArray();
Arrays.sort(xxgc);
x1x2 = new String(x1x2c);
xxg = new String(xxgc);
if (x1x2.equals(xxg)) {
return true;
}
return false;
}
}
[解决办法]
import java.util.Arrays;
public class Test {
public static void main(String[] args) {
int sum = 0;
for(int i = 10; i < 99; i++){
for(int j = i + 1; j < 99; j++){
int product = i * j;
if(product > 9999 || product < 1000 || product % 100 == 0) continue;
String strProduct = " " + product;
String gene = " " + i + j;
String[] arrProduct = strProduct.split( " ");
String[] arrGene = gene.split( " ");
Arrays.sort(arrProduct);
Arrays.sort(arrGene);
if(Arrays.equals(arrProduct, arrGene)){
sum++;
System.out.println( "第 " + sum + "组： " + strProduct);
}
}
}
}
}
[解决办法]
请教，吸血鬼数字有什么实际应用上的意义么
[解决办法]
看到N多好算法。唉，郁闷......自己写的惭愧啊
[解决办法]
呵呵 大家都不错啊 欢迎加入 程序员群 41801837
[解决办法]
重点在于数论中的 （(product - i - j ) %9 == 0 ）
if(product > 9999 || product < 1000 || product % 100 == 0) continue;
if(product > 9999 || product < 1000 || product % 100 == 0 || （(product - i - j )%9 != 0 ）
) continue;
重中之重，影响效率的关键因素