uva133 The Dole Queue ( 约瑟夫环的模拟)

题目链接：

啊哈哈，选我选我

思路是：

相当于模拟约瑟夫环，仅仅只是是从顺逆时针同一时候进行的，然后就是顺逆时针走能够编写一个函数，仅仅只是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 3
0 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("
");
        }
    }
    return 0;
}

Sample output

 4  8,  9  5,  3  1,  2  6,  10,  7