问题：一群人站成一个圆圈，从一个人开始报数，1， 2 ，。。。m，报到m的拉出去砍了，求被砍的顺序和最后一个活下来的。

利用单向循环链表实现

C++代码如下：（参考书籍：数据结构与算法实验指导书）

#include <iostream>
using namespace std;struct NodeType
{int num;char name[20];NodeType* next;
};
class Jose
{
private:NodeType* p_head;
public:Jose(){p_head = new NodeType;  //带空头的链表p_head->next = p_head;  //空的循环链表}~Jose(){}void creat();void print();
};
void Jose::creat()
{int i = 0, n;NodeType *newp, *tempNode;tempNode = p_head;cout << "\n enter total nums of people: ";cin >> n;for(i = 0; i < n; ++i){newp = new NodeType;newp->num = i+1;cout << "\n enter name: ";cin >> newp->name;newp->next = p_head;    //此处p_head为尾部哨兵tempNode->next = newp;   //不断地往尾部（尾部哨兵之前）插入节点tempNode = newp;}tempNode->next = p_head->next;  //断开空头哨兵delete p_head;                  //释放哨兵节点p_head = tempNode->next;        //头结点指向第一个节点
}
void Jose::print()
{int m,i;NodeType *del = p_head, *tempNode;cout << "\n enter value m(m>=2):";cin >> m;cout << "\n start:" << endl;while(del->next != del)         //链表节点个数不为1{for(i = 1; i < m; ++i)        //del往后移动m位{tempNode = del;del = del->next;}cout << del->num << " " << del->name << endl;tempNode->next = del->next;     //断开del节点delete del;                     //释放del节点del = tempNode->next;}cout << del->num << " " << del->name << endl;delete del;             //链表只剩一个节点直接删除
}int main()
{Jose gameList;gameList.creat();gameList.print();cout << "press any key to exit!" ;cin.get();return 0;
}

Valgrind检查结果：

假设有5个人，报数3的出列，手动模拟如下