uva-133 - The Dole Queue
2015-08-01 08:46
477 查看
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 threenumbers 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-clockwiseofficial first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma).
Sample input
10 4 3 0 0 0
Sample output
4
8,
9
5,
3
1,
2
6,
10,
7
where
represents a space.
代码:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | #include <cstdio> #define maxn 25 using namespace std; int n, k, m, a[maxn]; int go(int p, int d, int t) { while( t-- ) { do{ p = (p+d+n-1)%n+1; }while(a[p] == 0); } return p; } int main() { while(scanf("%d%d%d",&n,&k,&m) == 3 && n) { for(int i = 1; i <= n; i++) a[i] = i; int remain = n; int p1 = n, p2 = 1; while(remain) { p1 = go(p1, 1, k); p2 = go(p2, -1, m); printf("%3d",p1);remain--; if(p2 != p1){ printf("%3d",p2); remain--;} a[p1] = a[p2] = 0; if(remain) printf(","); } printf("\n"); } } |
相关文章推荐
- iOS UI03_UIViewController视图控制器
- iOS UI03_LTView
- 写在Demo战斗系统之前,先用原型工具做套UI第四篇-回合制战斗UI界面制作
- 【LeetCode-面试算法经典-Java实现】【063-Unique Paths II(唯一路径问题II)】
- 【LeetCode-面试算法经典-Java实现】【062-Unique Paths(唯一路径)】
- UI常用字体定义和继承的实例,ResearchKitCode
- iOS 错误:… is being deallocated while key value observing are still registered with it
- Java_Web_request.setAttribute("result",username);
- 【Android UI】ListView系列一(基础篇)
- 深入源码解析Android中的Handler,Message,MessageQueue,Looper
- android 各种UI控件的特殊使用方式(不定期更新)
- UIButton简单总结
- hdu 5288 OO’s Sequence
- Packet for query is too large
- mysql异常Packet for query is too large(mysql写入数据过大)
- HDU - 4396 More lumber is required (BFS 最短路)
- Lombok(1.14.8) - @NoArgsConstructor, @RequiredArgsConstructor & @AllArgsConstructor
- 杭电1242 Rescue
- apue 第14章 高级I/O
- Android BuildConfig使用