有2*k个人，围成一圈，问，最小的m（从1到m报数，数到m的人离开队列），使得编号k+1到2*k的人先出队列。

我用单向循环链表来做，不过发现数据大于10的时候速度很慢，我不知道如何优化，

题目原题http://。

我的代码如下：



#include <iostream>
using namespace std;


typedef struct node{
    int number;
    bool exit;
    struct node* next;
}node,*pnode;


int main(){
   
    node head;
    head.number=1;
    head.exit=true;
    head.next=NULL;

   
    node* p=NULL;
    p=&head;
    node* q=NULL;
   
    int k;
    int i;
    int m;
    int count;
   
   
    while(cin>>k){

        if (k==0)return 0;

        for (i=0;i<2*k-1;i++)
        {
            q=new node;
            q->number=i+2;
            q->exit=true;
            q->next=NULL;
            p->next=q;
            p=q;
            
        }
        p->next=&head;

        for (m=1;m<2504884;m++)//0x7fffffff
        {
            for (i=0;i<2*k;i++)
            {
                p->exit=true;
                p=p->next;
            }
            p=&head;
            i=1;
            count=0;
            while(1){
                if (i <= m)
                {
                    if (p->exit==true)
                    {
                        if (i==m)
                        {
                            int fsdfsd=p->number;
                            if (p->number <= k)break;
                            else
                            {
                                count++;
                                p->exit=false;
                                if (count==k)
                                {
                                    goto showm;
                                }
                            }   
                        } //if(i==m)
                        p=p->next;
                        i++;
                    }
                    else
                    {
                        p=p->next;
                    }
                }
                else
                {
                    i=1;
                }   
            }//while(1)
        }//m=0;m<10000;m++
showm:
        cout<<m<<endl;

        p=&head;
        for (i=0;i<2*k;i++)
        {
            p->number=0;
            p->exit=false;
            p=p->next;
        }
        
    }
    return 0;
}