SmallTalk — Sat, 13 May 2006 08:07:00 GMT

//Jose.cpp----使用单向循环链表实现�U�瑟夫问�?br />//问题描述�Q�n个�h手拉手排成一个环�Q�从�W�一个�h开始数�Q�每数到�W�m个�h�Q�第m个�h出列�Q�从下一人重新计数重新数�Q�求全部出列后他们的出列��序
#include
struct jose
{
int data;
int no;
struct jose * next;
};

int main()
{
struct jose *head,*p_curent,*p_find;
int n,m;
cout << "Please enter the total of numbers (n):";
cin >> n;
cout << "Please enter the counter number (m):";
cin >> m;

//初始化链�?/span>
head=p_curent=new jose;//标记首表元地址�Q�即头指�?/span>
head->no=1;
cout << "Please enter the first number :";
cin >>head->data;
//形成其余的n-1表元
cout << "Please enter last numbers :"< for (int i=2;i<=n;i++)
{
  p_curent->next=new jose;
  p_curent=p_curent->next;
  cin >> p_curent->data;
  p_curent->no=i;
}//end for
p_curent->next=head;//��指针指向头指针�Q��Ş成环�Q�到�q�完成初始化链表

//开始查询，�W�M个�h出列�Q��ƈ输出
cout << "Now : The numbers of who will quit the cycle in turn are:"< while (n)//全部出列后结束��@�?br /> {
  //掠过m�Q?个表�?br />  for (int j=1;j   p_curent=p_curent->next;//end for
  //扑ֈ��W�M个�h
  p_find=p_curent->next;
  //从表中删除第m个�h�Q��ƈ输出�W�m个�h
  p_curent->next=p_find->next;
  cout << p_find->data<  //释放�W�m个表元占用的�I�间
  delete p_find;
  n--;
}
//end while

return 0;
}

SmallTalk 2006-05-13 16:07 发表评论

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