约瑟夫问题

问题：n个人围成一个环，从第i个开始，由1至interval不断报数，凡报到interval的出列，直到环空为止。

#include <iostream>
#include <string>
using namespace std;
//结构体
struct jonsephus
{
    int code;//数字
    jonsephus *next;//next指针
};
void jonsInt(jonsephus *&head,int n)
{
    jonsephus *g;//g为跟踪指针
    g = head = new jonsephus;//新结点
    for (int j = 1; j <= n; j++)
    {
        g->code = j;
        if (j == n) g->next = head;//最后一个结点的next指针指向头指针 形成环状
        else g->next = new jonsephus;//新结点插入链表
        g = g->next;//跟踪指针指向新结点
    }
}
void showJ(jonsephus *head)
{
    cout << "Now include:" << endl;
    jonsephus *pp = head;//跟踪指针pp
    do
    {
        cout << pp->code << '	';
        pp = pp->next;//移动跟踪指针位置
    }
    while (head != pp);//遍历 当head指针等于pp 表示指针移到了头指针位置
    cout << endl;
}
void standUp(jonsephus *&head,int n,int i,int interval)
{
    jonsephus *q = NULL,*p = NULL;//定义两个跟踪指针
    int c;//当前计数
    jonsInt(head, n);
    showJ(head);//检查初始化是否成功
    p = q = head;//先设定pq均为头指针
    while (p->next != head)//遍历得到指向头指针的指针作为p
    {
        p = p->next;
    }
    for (c = 1;c < i; c++)//循环移位等到从第i位开始数数
    {
        p = p->next;
        q = p->next;
    }
    while (p != q)//若果p指针的next指向它自身 表明剩下最后一个对象
    {
        for ( c = 1; c < interval; c++)//每次循环interval-1次挪动指针即可，再下面代码会挪动一次指针
        {
            p = p->next;
            q = p->next;
        }
        cout << q->code << " is stand up!" << endl;//输出的是q指针的值 下面要释放q指针指向的内存空间
        p->next = q->next;//p指针的next指针指向q指针的next指针 这三句代码相当于挪动了一次指针
        delete q;//释放q指针指向的内存空间
        q = p->next;//q指针指向p的next指针
    }
    cout << q->code << " is stand up!" << endl;//输出最后一个
}
void main()
{
    int n, i, interval;
    cout << "How many persons?" << endl;
    cin >> n;//输入多少人
    cout << "Start with which person?" << endl;
    cin >> i;//输入哪个人开始报数
    cout << "Which number will stand up?" << endl;
    cin >> interval;//输入数到第几则出列
    jonsephus *head = NULL;//定义头指针
    standUp(head, n, i, interval);//调用函数
    getchar();
    getchar();
}