模拟约瑟夫环

 Roman Roulette 

The historian Flavius Josephus relates how, in the Romano-Jewish conflict  of 67 A.D., the Romans took the town of Jotapata which he was commanding.   Escaping, Jospehus found himself trapped in a cave with 40 companions.  The  Romans discovered his whereabouts and invited him to surrender, but his  companions refused to allow him to do so.  He therefore suggested that they  kill each other, one by one, the order to be decided by lot.  Tradition has it  that the means for effecting the lot was to stand in a circle, and, beginning  at some point, count round, every third person being killed in turn.  The  sole survivor of this process was Josephus, who then surrendered to the  Romans.  Which begs the question: had Josephus previously practised quietly  with 41 stones in a dark corner, or had he calculated mathematically that he  should adopt the 31st position in order to survive?

Having read an account of this gruesome event you become obsessed with  the fear that you will find yourself in a similar situation at some time in  the future.  In order to prepare yourself for such an eventuality you decide  to write a program to run on your hand-held PC which will determine the  position that the counting process should start in order to ensure that you  will be the sole survivor.

In particular, your program should be able to handle the following variation  of the processes described by Josephus.  n > 0 people are initially  arranged in a circle, facing inwards, and numbered from 1 to n.  The  numbering from 1 to n proceeds consecutively in  a clockwise direction.   Your allocated number is 1.  Starting with person number i, counting  starts in a clockwise direction, until we get to person number k (k > 0),  who is promptly killed.  We then proceed to count a further k people in a  clockwise direction, starting with the person immediately to the left of the  victim.  The person number k so selected has the job of burying the  victim, and then returning to the position in the circle that the victim had  previously occupied.  Counting then proceeds from the person to his  immediate left, with the kth person being killed, and so on, until only one  person remains.

For example, when n = 5, and k = 2, and i = 1, the order of execution is  2, 5, 3, and 1.  The survivor is 4.

Input and Output

Your program must read input lines containing values for n and k (in  that order), and for each input line output the number of the person with  which the counting should begin in order to ensure that you are the sole  survivor.  For example, in the above case the safe starting position is 3.   Input will be terminated by a line containing values of 0 for n and k.

Your program may assume a maximum of 100 people taking part in this  event.

Sample Input

1 1
1 5
0 0

Sample Output

1
1

UVa 130 - Roman Roulette的更多相关文章

  1. Roman Roulette(约瑟夫环模拟)

    Roman Roulette Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)To ...

  2. UVA - 185 Roman Numerals

    题目链接: https://vjudge.net/problem/UVA-185 思路: 剪枝.回溯 注意回溯的时候,是从当前点的下一个开始,而不是从已经遍历的个数点开始!!不然回溯有问题! 思路参考 ...

  3. HOJ题目分类

    各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1 ...

  4. python随机数学习笔记

    #coding:utf-8 import random # random.randint(1,10)产生1,10的随机整数 for i in range(1,5): ranint = random.r ...

  5. ascii码所有字符对照表(包含汉字和外国文字)

    http://www.0xaa55.com/thread-398-1-1.html看到了0xaa55的这个帖子,想起了2年前我在51cto发的一个帖子http://down.51cto.com/dat ...

  6. uva 759 - The Return of the Roman Empire

    #include <cstdio> #include <string> #include <map> using namespace std; ; , , , , ...

  7. UVA 590 二十一 Always on the run

     Always on the run Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit ...

  8. UVa 10048: Audiophobia

    这道题要求我们求出图中的给定的两个节点(一个起点一个终点,但这是无向图)之间所有“路径中最大权值”的最小值,这无疑是动态规划. 我开始时想到根据起点和终点用动态规划直接求结果,但最终由于题中S过大,会 ...

  9. [uva] 10067 - Playing with Wheels

    10067 - Playing with Wheels 题目页:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Ite ...

随机推荐

  1. FTP内容

    1.1.1 作业FTP部署. FTP工具或者浏览器默认使用的都是PASV模式连接FTP服务器 1.先检查有没有安装   rpm -q vsftpd 如果没有安装   yum install vsftp ...

  2. SpringMVC拦截器简单使用

    一.拦截器的配置 1.传统的配置 Xml代码   <bean class="org.springframework.web.servlet.mvc.annotation.Default ...

  3. jQuery EasyUI教程之datagrid应用

    一.利用jQuery EasyUI的DataGrid创建CRUD应用 对网页应用程序来说,正确采集和管理数据通常很有必要,DataGrid的CRUD功能允许我们创建页面来列表显示和编辑数据库记录.本教 ...

  4. ioc和aop的区别?

    ioc:控制反转(Inversion of Control,英文缩写为IoC)把创建对象的权利交给框架,是框架的重要特征,并非面向对象编程的专用术语.它包括依赖注入(Dependency Inject ...

  5. 【Android实战】Android中处理崩溃异常

    public class MainActivity extends ActionBarActivity { public CrashApplication application; @Override ...

  6. tRNA 二级结构预测可视化

    tRNAdb 收录了来自104个物种的623条tRNA 序列,从数据库中下载对应物种的tRNA 序列和二级结构,以人为例 打开下面的链接 http://trna.bioinf.uni-leipzig. ...

  7. jquery中判断选择器,找没找到元素用$().size()==0

    jquery中判断选择器,找没找到元素用$().size()==0

  8. Windows下安装Scrapy

    安装python 根据你的需求下载python安装包,安装python(本文基于python27)https://www.python.org/downloads/ 在 环境变量---"Pa ...

  9. 解决error: Your local changes to the following files would be overwritten by merge

    在项目里我们一般都会把自己第一次提交的配置文件忽略本地跟踪 1 [Sun@webserver2 demo]$ git update-index --assume-unchanged <filen ...

  10. 关于android 调用网页隐藏地址栏

    首先创建项目,在main.xml里 添加好WebView控件R.id为webview1. HelloWebView.java 代码 package liu.ming.com; import andro ...