lightoj1179 - Josephus Problem&&hdoj2211【约瑟夫环】
2016-01-31 11:02
531 查看
1179 - Josephus Problem
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, Josephus 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 affecting 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 practiced quietly with 41 stones in a dark corner,
or had he calculated mathematically that he should adopt the 31st position in order to survive?
Now you are in a similar situation. There are n persons standing in a circle. The persons are numbered from 1 to n circularly. For example, 1 and n are adjacent and 1 and 2 are
also. The count starts from the first person. Each time you count up to k and the kth person is killed and removed from the circle. Then the count starts from the next person. Finally one person remains. Given n and k you
have to find the position of the last person who remains alive.
Each case contains two positive integers n (1 ≤ n ≤ 105) and k (1 ≤ k < 231).
![]() ![]() | PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 affecting 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 practiced quietly with 41 stones in a dark corner,
or had he calculated mathematically that he should adopt the 31st position in order to survive?
Now you are in a similar situation. There are n persons standing in a circle. The persons are numbered from 1 to n circularly. For example, 1 and n are adjacent and 1 and 2 are
also. The count starts from the first person. Each time you count up to k and the kth person is killed and removed from the circle. Then the count starts from the next person. Finally one person remains. Given n and k you
have to find the position of the last person who remains alive.
Input
Input starts with an integer T (≤ 200), denoting the number of test cases.Each case contains two positive integers n (1 ≤ n ≤ 105) and k (1 ≤ k < 231).
Output
For each case, print the case number and the position of the last remaining person.Sample Input | Output for Sample Input |
6 2 1 2 2 3 1 3 2 3 3 4 6 | Case 1: 2 Case 2: 1 Case 3: 3 Case 4: 3 Case 5: 2 Case 6: 3 |
#include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> #include<queue> #include<list> #include<vector> using namespace std; int main() { int t,i,j,n,k,test=1; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&k); printf("Case %d: ",test++); int last=0; for(i=1;i<=n;++i){ last=(last+k)%i; } printf("%d\n",last+1); } return 0; }
相关文章推荐
- 人生的高境界 - 无所谓
- 个人家园 - 永久免费ASP个人主页空间
- L.WuBi - 发布个最好用的在线五笔编码查询、五笔拆字图解
- 用hta实现制作的无殇 - 快书V1.1打包下载了第1/2页
- Windows SP2 免激活安装版 - DeepinXP PLUS V2 测试版 下载
- 超星图书浏览器(SSReader) v4.00 Bulid 070511 - 数字图书阅览器 下载
- Convenientfox v0.0.1.3 - 基于FireFox的浏览器 下载
- GX::Transcoder v3.20.51.3687 - 音频视频文件转换全能王 下载
- 某市大型汽车网 - vip代码提供下载了
- CSS属性 - white-space 空白属性使用说明
- 测试模式 - XSL教程 - 5
- 选择模式 - XSL教程 - 2
- 匹配模式 - XSL教程 - 4
- XPath入门 - XSL教程 - 3
- 强烈推荐 - Ajax 技术资源中心
- filewipe - 商用级文件擦除工具 提供下载了
- Javascript - HTML的request类
- 关于XSL - XSL教程
- JS创建优美的页面滑动块效果 - Glider.js
- 脚本中出现 window.open() access is denied - 拒绝访问 情况一则及分析