Java 找出四位数的全部吸血鬼数字 基础代码实例
2024-04-11 02:06:15
/**
* 找出四位数的全部吸血鬼数字
* 吸血鬼数字是指位数为偶数的数字，能够由一对数字相乘而得到，而这对数字各包括乘积的一半位数的数字，当中从最初的数字中选取的数字能够随意排序.
* 以两个0结尾的数字是不同意的。
* 比例如以下列数字都是吸血鬼数字
1260=21*60
1827=21*87
2187=27*81
...
* 比較笨的低效率的做法： 遍历全部四位数， 每生成一个四位数的时候，
* 在双重循环遍历两位数，在两位数的内层循环中推断是否与最外层循环的四位数相等。 假设相等把这些数字都存放到数组，进行排序后比較
* 两组数字，假设相等那么输出这个数就是要找的数字；
*/
了解下这个英文參考:吸血鬼数字
An important theoretical result found by Pete Hartley:
If x·y is a vampire number then x·y == x+y (mod 9)
Proof:
Let mod be the binary modulo operator and d(x) the
sum of the decimal digits of x.
It is well-known that 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 are (x mod 9, y mod 9) in {(0,0),
(2,2), (3,6), (5,8), (6,3), (8,5)}
Only these cases (6 out of 81) have to be tested in a vampire search based on testing x·y for different values of x and y.
以下五中方法, 当中还是ThinkinJava给出的參考答案效率最高, 其它高效率做法 , 请网友高手大神补充
import java.util.Arrays;
public class Demo3 { static int a; //千位
static int b; //百位
static int c; //十位
static int d; //个位 static int a1; //十位
static int b1; //个位
static int c1; //十位
static int d1; //个位
static int sum = 0; //总和
static int sum2 = 0; //两数之积 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 VampireNumbers(); //该方法 有反复数字
} 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 * 10) + 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() { // 改进
for (int i = 11; i < 100; 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() { // 官方參考答案 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(num1 + " * " + num2 + " : "
+ product);
}
} } /*
* Output: 15 * 93 : 1395 21 * 60 : 1260 21 * 87 : 1827 27 * 81 :
* 2187 30 * 51 : 1530 35 * 41 : 1435 80 * 86 : 6880
*///:~
} private static void method2() { // 穷举2 //遍历四位数，排除00 从1001開始
for (int i = 1001; i <= 9999; i++) {
//排除00
if (i % 100 != 00) {
for (int k = 0; k < 100; k += 10) {
if (k != 0) {
//10 -99
for (int j2 = 0; j2 <= 9; j2++) {
//生成第一个两位数
for (int j = 0; j < 100; j += 10) {
for (int j3 = 0; j3 <= 9; j3++) {
//生成第二个两位数
//推断两数之积 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() { //穷举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 < 10; j2++) {
c = j2 * 10;
for (int k = 0; k < 10; k++) {
d = k;
sum = a + b + c + d;
//取当中四个数字 中组成两个两位数 ，假设这两个两位数之积 等于 sum ,则输入 这个数
for (int k2 = 1; k2 < 10; k2++) { a1 = k2 * 10;
for (int l = 0; l < 10; l++) {
if (a1 + b1 > 100) {
break;
}
b1 = l;
//得到一个两位数字
for (int l2 = 1; l2 < 10; l2++) {
c1 = l2 * 10;
for (int m = l; m < 10; m++) { if (c1 + d1 > 100) {
break;
}
d1 = m;
//再得到一个两位数字
sum2 = (a1 + b1) * (c1 + d1);
//计算来两个两位数字之积，假设等于sum
if (sum2 == sum) {
//且尾数不能为00
if (c + d != 0) {
// 比較这个几个数字 是否一样
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 j2, int k, int j3) {
int a[] = { i % 10, i / 10 % 10, i / 100 % 10, 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;
}
}
}
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