uva133 The Dole Queue ( 约瑟夫环的模拟)
来源:互联网 发布:编程达人全套视频 编辑:程序博客网 时间:2024/05/23 01:18
题目链接:
思路是:
相当于模拟约瑟夫环,只不过是从顺逆时针同时进行的,然后就是顺逆时针走可以编写一个函数,只不过是走的方向的标志变量相反。。还有就是为了(pos+flag+n-1)%n+1的妙用。。。
题目:
The Dole Queue
In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour Rhinoceros Party has decided on the following strategy. Every day all dole applicants will be placed in a large circle, facing inwards. Someone is arbitrarily chosen as number 1, and the rest are numbered counter-clockwise up to N (who will be standing on 1's left). Starting from 1 and moving counter-clockwise, one labour official counts off k applicants, while another official starts from N and moves clockwise, counting m applicants. The two who are chosen are then sent off for retraining; if both officials pick the same person she (he) is sent off to become a politician. Each official then starts counting again at the next available person and the process continues until no-one is left. Note that the two victims (sorry, trainees) leave the ring simultaneously, so it is possible for one official to count a person already selected by the other official.
Input
Write a program that will successively read in (in that order) the three numbers (N, k and m; k, m > 0, 0 < N < 20) and determine the order in which the applicants are sent off for retraining. Each set of three numbers will be on a separate line and the end of data will be signalled by three zeroes (0 0 0).
Output
For each triplet, output a single line of numbers specifying the order in which people are chosen. Each number should be in a field of 3 characters. For pairs of numbers list the person chosen by the counter-clockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma).
Sample input
10 4 30 0 0
Sample output
4 8, 9 5, 3 1, 2 6, 10, 7
where represents a space.
代码为:
#include<cstdio>#include<cstring>const int maxn=25+10;bool vis[maxn];int n,k,m;int Move(int pos,int flag,int step){ for(int i=1;i<=step;i++) { do { pos=(pos+flag+n-1)%n+1; }while(vis[pos]); } return pos;}int main(){ int left,p1,p2; while(~scanf("%d%d%d",&n,&k,&m)) { if(n==0&&k==0&&m==0) return 0; memset(vis,false,sizeof(vis)); p1=n; p2=1; left=n; while(left) { p1=Move(p1,1,k); p2=Move(p2,-1,m); printf("%3d",p1); left--; if(p1!=p2) { printf("%3d",p2); left--; } vis[p1]=vis[p2]=1; if(left) printf(","); else printf("\n"); } } return 0;}
Sample output
4 8, 9 5, 3 1, 2 6, 10, 7
- uva133 The Dole Queue ( 约瑟夫环的模拟)
- UVA133 The Dole Queue (模拟)
- The Dole Queue 约瑟夫环 模拟
- uva133 The Dole Queue
- UVA133:The Dole Queue
- UVA133- The Dole Queue
- The Dole Queue UVA133
- UVA133 The Dole Queue
- uva133-The Dole Queue
- UVA133 The Dole Queue
- UVA133 - The Dole Queue
- uva133--The Dole Queue
- UVA133-The Dole Queue
- UVa133 - The Dole Queue
- UVA133 The Dole Queue
- Uva133 The Dole Queue
- UVa133 - The Dole Queue
- The Dole Queue uva133
- 人脸识别《一》opencv人脸识别之预处理
- 计时器--精确到10毫秒(精确度可以自行设定)
- jobdu1475 非常可乐
- “Spark上流式机器学习算法实现”中期检查报告
- 使用百度地图API根据坐标显示地图
- uva133 The Dole Queue ( 约瑟夫环的模拟)
- dedecms 标题和关键词的调用
- Popular Cows+求强联通量简单题+tarjan算法+POJ
- 参与unity非游戏行业开发者大会小结
- 给 C# 开发者的代码审查清单
- hdoj 2277 Change the ball 【找规律】
- CSS 参考手册
- 利用xib自定义view,在uicontrollerview中使用
- Linux命令行之逗趣无极限