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：

1 #include 
     
       2 #include 
      
        3  4 #define N 20 5  6 typedef struct node 7 { 8     int id; 9     struct node *next;10     struct node *pre;11 }Node, *pNode;12 13 //双向循环链表的构建14 pNode RingConstruct (int n)15 {16     int i;17     pNode head, p, q;18     head = (pNode)malloc(sizeof(Node));19     head->id = 1;20     p = head;21     for (i = 2; i <= n; i++)22     {23         q = (pNode)malloc(sizeof(Node));24         q->id = i;25         p->next = q;26         q->pre = p;27         p = q;28     }29     p->next = head;30     head->pre = p;31     return head;32 }33 34 //传入报数的次数序号，返回此次报数的上限值35 int boundMachine (int order)36 {37     int boundList[4] = {
   3, 7, 11, 19};38     return boundList[(order - 1)%4];39 }40 41 //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42 pNode count (pNode first, int bound)43 {44     while (--bound)45     {46         first = first->next;47     }48     return first;49 }50 51 //将currentNode从环中删除，并返回被删除节点的下一节点52 pNode removeNode (pNode currentNode)53 {54     pNode first = currentNode->next;55     if (first != currentNode)56     {57         currentNode->pre->next = first;58         first->pre = currentNode->pre;59     }60     printf("%d ", currentNode->id);61     free(currentNode);62     return first;63 }64 65 int main (int argc, char *argv)66 {67     pNode first, toRemove;68     int i;69     first = RingConstruct(N);70     for (i = 1; i <= N; i++)71     {72         toRemove = count(first, boundMachine(i));73         first = removeNode(toRemove);74     }75     return 0;76 }
      
     

当N为20时，出列顺序是：