這個演算法很巧妙，一定要留下：
class Eg0410_Vampire { public static void main(String[] args) { String[] ar_str1, ar_str2; int sum = 0; for (int i = 10; i < 100; i++) { for (int j = i + 1; j < 100; j++) { int i_val = i * j; if (i_val < 1000 || i_val > 9999) continue; // 積小於1000或大於9999排除,繼續下一輪環 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)) { // 排序後比較,為真則找到一組 sum++; System.out.println("第" + sum + "組: " + i + "*" + j + "=" + i_val); } } } System.out.println("共找到" + sum + "組吸血鬼數"); } }
以下是官方答案，比較麻煩，完全沒有涉及本節中所講的中止迴圈語句：
// : 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; } } } } } } } /// :~
下面是我寫的，比較隨意，沒有整理，排除法，這個不行，那個不行，剩下的就是行的了。全面地用到了本節的各種中止迴圈及標籤。直接輸出是18個數字，去掉乘數互換的一半，4位的應當是9個數字。
其實沒太明白“吸血鬼數字”是什麼意思，為什麼叫這個名字，為什麼只能是4位數，6位行不行？8位行不行？16位呢？如果這是個隱喻，那麼這個隱喻不是很通用，吸血鬼在西方可能很通用，他們有很多特點，作者也許取其一，西方人一看就知道，但是對於我們不是很瞭解的人，吸血鬼有什麼特點？吸血？面色倉白？尖牙？黑暗？這些都跟這些4位數似乎沒什麼聯絡。也沒有查到合理的解釋。
static void vampire() { int result = -1; String strResult = ""; String strj = "", strj2 = ""; String[] arrayj, arrayj2; String strTotal = ""; String[] arrResult; int n = 0; int limit1 = 1000, limit2 = 1000; ArrayList<String> arrTotal = new ArrayList<String>(); ArrayList<Integer> arrTotalCheck = new ArrayList<Integer>(); System.out.println("吸血鬼數字："); for (int num1 = 1; num1 < limit1; num1++) { strj = Integer.toString(num1); if (Integer.toString(num1).length() < 2) continue; l2: for (int num2 = 1; num2 < limit2; num2++) { strj2 = Integer.toString(num2); if (Integer.toString(num2).length() < 2) continue; result = num1 * num2; strTotal = strj + strj2; strResult = Integer.toString(result); arrResult = strResult.split(""); arrayj = String.valueOf(num1).split(""); arrayj2 = String.valueOf(num2).split(""); for (String t1 : arrayj) { if (!strResult.contains(t1)) continue l2; } for (String t2 : arrayj2) { if (!strResult.contains(t2)) continue l2; } for (String tTotal : arrResult) { if (!strTotal.contains(tTotal)) continue l2; } arrayj2 = String.valueOf(num2).split(""); if (result >= limit1 * limit2) continue; else if (Integer.toString(result).length() < 4) continue; else if (strResult.length() % 2 != 0) continue; else if (result % 100 == 0) continue; else if (strj.length() != strj2.length()) continue; else { n++; // System.out.println(n + "." + num1 + "*" + num2 + "=" + result + ","); if (result == 102510 || result == 105264 || result == 105750 || result==108135 || result==110758 ) { if(arrTotalCheck.contains(result)) continue; arrTotalCheck.add(result); arrTotal.add(n + "." + num1 + "*" + num2 + "=" + result + ","); }; } } } for (String i : arrTotal) { System.out.println(i); } return; }
根據某百科上的資料可知，有6位的吸血鬼數字，那麼有8位的嗎？10位的？12位的？理論上應該有，偶數位即可。
那麼5775是不是？
4位數的吸血鬼數有7個：
1260, 1395, 1435, 1530, 1827, 2187, 6880
6位數的吸血鬼數字有141個：
102510, 104260, 105210, 105264, 105750, 108135, 110758, 115672, 116725, 117067, 118440, 123354, 124483, 125248, 125433, 125460, 126027, 126846, 129640, 129775, 131242, 132430, 133245, 134725, 135828, 135837, 136525, 136948, 140350, 145314, 146137, 146952, 152608, 152685, 153436, 156240, 156289, 156915, 162976, 163944, 172822, 173250, 174370, 175329, 180225, 180297, 182250, 182650, 186624, 190260, 192150, 193257, 193945, 197725, 201852, 205785, 211896, 213466, 215860, 216733, 217638, 218488, 226498, 226872, 229648, 233896, 241564, 245182, 251896, 253750, 254740, 260338, 262984, 263074, 284598, 284760, 286416, 296320, 304717, 312475, 312975, 315594, 319059, 319536, 326452, 329346, 329656, 336550, 336960, 338296, 341653, 346968, 361989, 362992, 365638, 368550, 369189, 371893, 378418, 378450, 384912, 386415, 392566, 404968, 414895, 416650, 416988, 428980, 429664, 447916, 456840, 458640, 475380, 486720, 489159, 489955, 498550, 516879, 529672, 536539, 538650, 559188, 567648, 568750, 629680, 638950, 673920, 729688, 736695, 738468, 769792, 789250, 789525, 792585, 794088, 809919, 809964, 815958, 829696, 841995, 939658
其中125460比較特殊，因為125640=204*615且125640=246*510.
偽吸血鬼數和一般吸血鬼數不同之處在於其尖牙不強制是n/2個位的數，故偽吸血鬼數的位數可以是奇數。
2002年Carlos Rivera定義了質吸血鬼數：尖牙是質因子的吸血鬼數，例如117067, 124483, 146137, 371893, 536539。
根據“尖牙不強制是n/2個位的數”來推斷，尖牙指的是兩個乘數，但是，為什麼？哪裡尖了？莫非是說對稱？那也太牽強了。