Question

There are 100 doors in a row that are all initially closed.

You make 100 passes by the doors.

The first time through, visit every door and toggle the door (if the door is closed,open it;if it is open,close it).

The second time, only visit every 2nd door (door #2, #4, #6, ...),and toggle it.

The third time, visit every 3rd door (door #3, #6, #9, ...), etc,until you only visit the 100th door.

Task

Answer the question: what state are the doors in after the last pass? Which are open, which are closed?

先把问题缩小: 10道门关着,按以上方式遍历这些门10次。 最后很明显,第一道们肯定是开着的;第二、第三是关着的! X道门的开关次数跟X的约数有关,如果X有偶数个约数则门最终的状态是关着的,否则为开着的。

问题现在变成: 1 到 100 之间公约数个数为奇数的数字是?

[i * i for i in range(1, int(math.sqrt(100)) + 1)] // [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]

Java Solution

public class Doors
{
public static void main(String[] args)
{
for(int i=0;i<10;i++)
System.out.println("Door #"+(i + 1)*(i + 1) +" is open.");
}
}

100 doors的更多相关文章

  1. 100 Door Puzzle

    问题重述: There are 100 doors in a long hallway. They are all closed. The first time you walk by each do ...

  2. 欧拉回路-Door Man 分类: 图论 POJ 2015-08-06 10:07 4人阅读 评论(0) 收藏

    Door Man Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 2476 Accepted: 1001 Description ...

  3. POJ 1300 Door Man - from lanshui_Yang

    Description You are a butler in a large mansion. This mansion has so many rooms that they are merely ...

  4. POJ1300(欧拉回路)

    Door Man Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2139   Accepted: 858 Descripti ...

  5. POJ 1300.Door Man 欧拉通路

    Door Man Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2596   Accepted: 1046 Descript ...

  6. POJ1300Door Man(欧拉回路)

                                                               Door Man Time Limit: 1000MS   Memory Limi ...

  7. POJ1300 Door Man —— 欧拉回路(无向图)

    题目链接:http://poj.org/problem?id=1300 Door Man Time Limit: 1000MS   Memory Limit: 10000K Total Submiss ...

  8. [欧拉回路] poj 1300 Door Man

    题目链接: http://poj.org/problem?id=1300 Door Man Time Limit: 1000MS   Memory Limit: 10000K Total Submis ...

  9. POJ 1300 Door Man(欧拉通路)

    题目描写叙述: 你是一座大庄园的管家. 庄园有非常多房间,编号为 0.1.2.3..... 你的主人是一个心不在 焉的人,常常沿着走廊任意地把房间的门打开.多年来,你掌握了一个诀窍:沿着一个通道,穿 ...

随机推荐

  1. Centos系统使用代理上网时 yum的代理设置

    yum的存在使centos上软件的安装.配置.升级.卸载变得十分的方便,但是当安装centos的机器是通过代理服务器访问外网的话,yum的 使用就变得无从下手了,以下介绍以下怎样为yum配置代理地址及 ...

  2. java的回忆录

    封装的三步骤:(1)加属性(成员变量.全局变量.域field)用private来修饰(2)为对应的属性生成共有的setter.getter方法(3)在对应的setter的方法中可以根据需要加入对应的验 ...

  3. javascript基础知识--函数定义

    函数声明式 function funname( 参数 ){ ...执行的代码 } 声明式的函数并不会马上执行,需要我们调用才会执行:funname(); * 分号是用来分隔可执行JavaScript语 ...

  4. Linux -Yum 命令详解

    yum(全称为 Yellow dog Updater, Modified)是一个在Fedora和RedHat以及SUSE中的Shell前端软件包管理器.基於RPM包管理,能够从指定的服务器自动下载RP ...

  5. popViewControllerAnimated 后,对页面内UITableView 内数据刷新

    popViewControllerAnimated后,这时它不执行viewDidLoad,所以不能及时对viewControler及时刷新,此时对该页面进行操作可以调用viewWillAppear:( ...

  6. [GeekBand] STL与泛型编程(3)

    本篇文章主要介绍泛型算法中的变易.排序.数值算法. 一. 变易算法 所谓变易算法是指那些改变容器中的对象的操作. 1.1 copy组 template <class InputIterator, ...

  7. C#中调用unmanaged DLL

    using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; usin ...

  8. thinkphp3.2引入php 实例化类

    如果你的类库没有采用命名空间的话,需要使用import方法先加载类库文件,然后再进行实例化,例如:我们定义了一个Counter类(位于Com/Sina/Util/Counter.class.php): ...

  9. simple_factory

    #include <stdlib.h> #include <iostream> using namespace std; class Product { public: vir ...

  10. 获取当前<script>节点

    /* get current JavaScript dom object. */ var all_js = document.getElementsByTagName("script&quo ...