数数双链表解法。

#include "stdio.h"#include "malloc.h"typedef struct DulNode{int data;struct DulNode *prior, *next;}NODE;NODE *creat( int n){NODE *head;int i;head = (NODE*)malloc(sizeof(NODE)*n);for(i = 0; i <n; i++){head[i].next = &head[i+1];head[i].prior = &head[i-1];head[i].data = i+1;}head[0].prior = &head[n-1];head[n-1].next = &head[0];return head;}void f(NODE *head, int m){NODE *p = head;int i;for( ; ; ){i = 1;while(i != m){p = p->next;i++;}printf("%d-> ",p->data);p->prior->next = p->next;p->next->prior = p->prior;if(p->next == p || p->prior == p)return ;p = p->next;}}int main(){int n, m;NODE *head;//scanf("%d %d",&n, &m);n = 7, m = 3;printf("1 ~ %d个数，数到%d的退出,依次为/n",n,m);head = creat(n);f(head,m);free(head);printf("/n");return 0;}