UVa 133 The Dole Queue

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.

题目还是比较简单的，理解起来可能有点困难，就是双向约瑟夫环问题，直接数组模拟+多次循环。

删除节点很方便，令该点等于0即可。

代码如下：

#include <stdio.h>

int n,m,k;
int a[30];

int main(void)
{
    int i,j;
    int tip1,tip2;
    int clockwise(int tip1);
    int anticlockwise(int tip2);
    
    while(scanf("%d%d%d",&n,&k,&m)&&(n||k||m))
    {
     for(i=1;i<=n;i++)
      a[i]=i;
     int flag=0;
     tip1=0;tip2=n+1;
     while(flag<n)
     {
      tip1=clockwise(tip1+1);
      tip2=anticlockwise(tip2-1);
      printf("%3d",tip1);
      if(a[tip1]==a[tip2])
       flag++;
      else 
      {
       flag+=2;
       printf("%3d",tip2);
      }
      if(flag-n)
       printf(",");
      else
       printf("\n");
      a[tip1]=0;a[tip2]=0;
     }
    }
    return 0;
}

int clockwise(int tip1)
{
    int i,j,flag;
    for(i=tip1,flag=0;flag!=k;i++)
    {
     if(i>n)
      for(j=1;;j++)
      {
       if(j>n) j=1;
       if(a[j]) 
       {i=j;break;}      
      }
     if(a[i]) 
      flag++;
    }
    return i-1;
}
int anticlockwise(int tip2)
{
     int i,j,flag;
     for(i=tip2,flag=0;flag!=m;i--)
     {
      if(i<1) 
       for(j=n;;j--)
       {
        if(j<1) j=n;
        if(a[j]) 
        {i=j;break;}      
       }
      if(a[i]) 
       flag++;
     }
     return i+1;
}