UVA 133 环形队列
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 3 0 0 0
Sample output
4
8,
9
5,
3
1,
2
6,
10,
7
where represents a space.
题目链接:http://acm.uva.es/local/online_judge/search_uva.html
题目大意:环形队列问题,有N张牌,k表示从最左开始,往右每次隔k张牌取出一张,到最右边之后再回到最左边;m则是从最右边开始,往左每次隔m张牌取出一张,到最左边之后再回到最右边,直到把牌全部取出
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <iomanip> using namespace std; int queue[25]; int main() { int n,k,m; int front,rear,count1,count2,cc; while(cin>>n>>k>>m) { if(!n&&!k&&!m)//判断输入是否结束 break; for(int i=1; i<=n; i++)//队列数组的初始化 queue[i]=i; front=0; rear=n+1; cc=0; while(cc<n)//控制进度 { count1=0;count2=0; while(1) { if(queue[front]!=0)//读到未抽出的牌并计数 { count1++; } if(front>n)//如果从左往右到最右面返回最左边 front=0; if(count1==k)//按格式输出抽出的牌 { cout<<setw(3)<<queue[front]; cc++; break; } front++; } while(1)//同上,只是这是从右往左开始抽牌 { if(queue[rear]!=0) { count2++; } if(rear<1) rear=n+1; if(count2==m) { if(queue[front]!=queue[rear]) { cout<<setw(3)<<queue[rear]; cc++; } break; } rear--; } queue[front]=0;//将牌抽出后标记为0 queue[rear]=0;//将牌抽出后标记为0 if(cc!=n) cout<<","; } cout<<endl; } return 0; }