POJ 3750 小孩报数问题(模拟/数据结构—循环链表)

//其实是一道水题，但是可以用它来学习双向循环链表的构建和基本使用方法#include<iostream>#include<cstring>using namespace std;struct List//双向链表结构体{char name[20];struct List *front;//前指针struct List *rear;//后指针};List* create(int n)//构造函数，再过几天尝试将这些函数整合成C++的类模板，应该会更简洁{List *node,*head;head = node = new List;for(int i = 2;i <= n;++i){node->rear = new List;node->rear->front = node;node = node->rear;}node->rear = head;head->front = node;return head;}List* move(List* p,int step)//链表计数移动{for(int i = 1;i < step;++i){p = p->rear;}return p;}List* remove(List *p)//删除节点操作，双向链表的关键所在{p->front->rear = p->rear;p->rear->front = p->front;p = p->rear;return p;}int main(){int n,w,s;//freopen("in.txt","r",stdin);scanf("%d",&n);List *node,*head;head = node = create(n);for(int i = 1;i <= n;++i){scanf("%s",node->name);//读入结点名字node = node->rear;}scanf("%d,%d",&w,&s);node = move(head,w);while(node->rear != node)//循环链表的遍历终止条件{node = move(node,s);printf("%s/n",node->name);node = remove(node);}printf("%s/n",node->name);//剩下最后一个结点时要输出return 0;}