zoukankan      html  css  js  c++  java
  • Java:thinging in java p154 exercise 10

    题目

    A vampire number has an even number of digits and is formed by multiplying a pair of numbers containing half the number of digits of the result. The digits are taken from the original number in any order. Pairs of trailing zeroes are not allowed. Examples include: 1260 = 21 * 60, 1827 = 21 * 87, 2187 = 27 * 81. Write a program that finds all the 4-digit vampire numbers. (Suggested by Dan Forhan.)

     1 public class VampireNumber {
     2 
     3     public static void main(String[] args) {
     4         int a = 0, b = 0, c = 0, d = 0;
     5         int[] m = new int[12];
     6         int[] n = new int[12];
     7 
     8         for (int i = 1000; i < 9999; i++) {
     9             // 分成4个数
    10             a = i / 1000;
    11             b = (i - a * 1000) / 100;
    12             c = (i - a * 1000 - b * 100) / 10;
    13             d = i % 10;
    14 
    15             m[0] = a * 10 + b;
    16             n[0] = c * 10 + d;
    17 
    18             m[1] = a * 10 + b;
    19             n[1] = d * 10 + c;
    20 
    21             m[2] = a * 10 + c;
    22             n[2] = b * 10 + d;
    23 
    24             m[3] = a * 10 + c;
    25             n[3] = d * 10 + b;
    26 
    27             m[4] = a * 10 + d;
    28             n[4] = b * 10 + c;
    29 
    30             m[5] = a * 10 + d;
    31             n[5] = c * 10 + b;
    32 
    33             m[6] = b * 10 + a;
    34             n[6] = c * 10 + d;
    35 
    36             m[7] = b * 10 + a;
    37             n[7] = d * 10 + c;
    38 
    39             m[8] = b * 10 + c;
    40             n[8] = d * 10 + a;
    41 
    42             m[9] = b * 10 + d;
    43             n[9] = c * 10 + a;
    44 
    45             m[10] = c * 10 + a;
    46             n[10] = d * 10 + b;
    47 
    48             m[11] = c * 10 + b;
    49             n[11] = d * 10 + a;
    50             for (int j = 0; j < 12; j++) {
    51                 if (i == m[j] * n[j]) {
    52                     System.out.println(i + " = " + m[j] + " * " + n[j]);
    53                 }
    54             }
    55 
    56         }
    57     }
    58 }

    输出

    1 1260 = 21 * 60
    2 1395 = 15 * 93
    3 1435 = 41 * 35
    4 1530 = 51 * 30
    5 1827 = 87 * 21
    6 2187 = 27 * 81
    7 6880 = 86 * 80
    8 6880 = 80 * 86
  • 相关阅读:
    [转]用汇编实现原子操作
    贪心算法练习集
    链表练习题集
    STL测试3)优先级队列实现二叉堆
    STL测试2)计算器简单实现
    STL测试
    Java学习笔记(三)
    【JAVA】学习笔记(2)
    【Java】学习笔记(1)
    【机器学习】异常检测算法(I)
  • 原文地址:https://www.cnblogs.com/taoxiuxia/p/4434527.html
Copyright © 2011-2022 走看看