一道模拟题：改进的Joseph环

题目：改进的Joseph环。一圈人报数，报数上限依次为3,7,11,19，循环进行，直到所有人出列完毕。

思路：双向循环链表模拟。

代码：

#include <cstdio>
#include <cstdlib>

#define N 20

typedef struct node
{
    int id;
    struct node *next;
    struct node *pre;
}Node, *pNode;

//双向循环链表的构建
pNode RingConstruct (int n)
{
    int i;
    pNode head, p, q;
    head = (pNode)malloc(sizeof(Node));
    head->id = 1;
    p = head;
    for (i = 2; i <= n; i++)
    {
        q = (pNode)malloc(sizeof(Node));
        q->id = i;
        p->next = q;
        q->pre = p;
        p = q;
    }
    p->next = head;
    head->pre = p;
    return head;
}

//传入报数的次数序号，返回此次报数的上限值
int boundMachine (int order)
{
    int boundList[4] = {3, 7, 11, 19};
    return boundList[(order - 1)%4];
}

//first为每次报数的第一人，bound为此次报数的上限，返回此次报数的应出列者
pNode count (pNode first, int bound)
{
    while (--bound)
    {
        first = first->next;
    }
    return first;
}

//将currentNode从环中删除，并返回被删除节点的下一节点
pNode removeNode (pNode currentNode)
{
    pNode first = currentNode->next;
    if (first != currentNode)
    {
        currentNode->pre->next = first;
        first->pre = currentNode->pre;
    }
    printf("%d ", currentNode->id);
    free(currentNode);
    return first;
}

int main (int argc, char *argv)
{
    pNode first, toRemove;
    int i;
    first = RingConstruct(N);
    for (i = 1; i <= N; i++)
    {
        toRemove = count(first, boundMachine(i));
        first = removeNode(toRemove);
    }
    return 0;
}

当N为20时，出列顺序是：