uva 133 - 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

 1 #include<iostream>  

 2 #include<string.h>  

 3 #include<stdio.h>  

 4 #include<ctype.h>  

 5 #include<algorithm>  

 6 #include<stack>  

 7 #include<queue>  

 8 #include<set>  

 9 #include<math.h>  

10 #include<vector>  

11 #include<map>  

12 #include<deque>  

13 #include<list>  

14 using namespace std;  

15 #define maxn 25

16 int n,k,m,a[maxn];

17 int go(int p,int d,int t)//构建走动函数，跳过数值为0的位置 

18 {

19     while(t--)

20     {

21         do

22         {

23             p=(p+d+n-1)%n+1;

24         }

25         while(a[p]==0);

26     }

27     return p;

28 } 

29 int main()

30 {

31     while(scanf("%d%d%d",&n,&k,&m)==3&&n)

32     {

33         for(int i=1;i<=n;i++)

34         a[i]=i;

35         int left =n;

36         int p1=n,p2=1;//初始化移动位置，逆时针用p1，顺时针用p2

37         while(left)

38         {

39             p1=go(p1,1,k);

40             p2=go(p2,-1,m);

41             printf("%d",p1);

42             left--;

43             if(p1!=p2)

44             {

45                  printf(" %d",p2);

46                  left--;

47             }

48             if(left)

49             printf(",");

50             a[p1]=a[p2]=0;;

51         } 

52         printf("\n");

53     }

54     return 0;

55 }

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