uva 133 The Dole Queue

题意：http://www.nocow.cn/index.php/UVA_133（中文翻译）

解题思路：双向循环链表，模拟操作即可

解题代码：

// File Name: uva133.cpp
// Author: darkdream
// Created Time: 2013年04月27日 星期六 09时50分42秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>

using namespace std;
struct node
{
    int date;
    struct node *next ,*last;
};
int main(){
    int n,k,m;
    while(scanf("%d %d %d",&n,&k,&m) != EOF)
    {
        //k = k%(n+1); m = m%(n+1);
        if(n == 0)
            break;
        struct node *head,*p,*q;
        struct node a[30];
        for(int i = 1; i<= n;i ++)
        {  a[i].date = i;
            a[i].next = &a[i+1];
            a[i].last = &a[i-1];
        }
        a[1].last = &a[n];
        a[n].next = &a[1];
        p = &a[1];
        q = &a[n];
        while(1)
        {   
            
            if(n == 1)
            { 
                printf("%3d\n",p->date);
                break;
            }
           for(int i= 1; i< (k%n==0?n:k%n); i ++)
            {
                p = p->next;
            }
            for(int i = 1; i< (m%n==0?n:m%n) ; i++)
            {
                q = q->last;
            }
            if(p->next == q && p->last == q && p != q)
            {
               printf("%3d%3d\n",p->date,q->date);
               break;
            }
            if(p == q)
            {
               
                printf("%3d,",p->date);
                p->next->last = p->last;
                p->last->next = p->next;
                p = p->next;
                q = q->last;
                n--;
               
            }
            else{
              printf("%3d%3d,",p->date,q->date);
              p->last->next = p->next;
              p->next->last = p->last;
              if(p->next == q)
                  p = q->next;
              else
                p= p->next;
              q->last->next = q->next;
              q->next->last = q->last;
              q = q->last;
              n = n-2;
                
            }
              

        }


    }
    return 0;
}