uva133--The Dole Queue

看到这个题，跟约瑟夫环感觉差不多，但是这个数据范围比较小，可以模拟实现，我用的是循环链表实现的过程。中间用到了标记，因为对于样例10 4 3来说，就会到后边只剩下7 10这两个节点形成的环，但是now_n这时候指向已删除的2节点，now_f指向已删除的6节点，然而，这就需要让now_n继续向后指，也指向6节点，这样就需要标记是否还在环内。

#include<stdio.h>#include<iostream>using namespace std;struct node {    int a;    bool flg;       //mark the node is the part of the ring or not    node *next;    node *fron;};node *head, *endd;void bui_cir_link ( int nn ) {      //form a ring with two-way linked list    node *tmp, *temp;    head = new ( node );    head->a = 1;    head->flg = 1;    head->fron = NULL;    head->next = NULL;    tmp = head;    for ( int i = 2; i <= nn; i++ )    {        temp = new node;        tmp->next = temp;        temp->fron = tmp;        temp->a = i;        temp->flg = 1;        tmp = temp;    }    endd = tmp;    endd->next = head;    head->fron = endd;}void operate ( int nn, int kk, int mm ) {    int n = nn, num = 0;    node *now_f, *now_n;    now_n = head;    now_f = endd;    while ( nn-- ) {        if ( nn != n - 1 )      //specially operate the star              now_n = now_n->next;        for ( int i = 1; i < kk; i++ ) {            now_n = now_n->next;        }        if ( nn != n - 1 )            now_f = now_f->fron;        for ( int i = 1; i < mm; i++ ) {            now_f = now_f->fron;        }        now_n->next->fron = now_n->fron;    //delete         now_n->fron->next = now_n->next;        num++;                              //records the number of deleted node        now_n->flg = 0;                     //mark        printf("%3d",now_n->a);        if ( num == n ) {            break;        }        if ( now_n != now_f ) {            now_f->fron->next = now_f->next;            now_f->next->fron = now_f->fron;            num++;            now_f->flg = 0;            printf("%3d",now_f->a);            if ( num == n ) {                break;            }        }//        else//           now_f= now_n;                    //delete the same node        for ( int i = 0;; i++ ) {           //finding the closest deleted node with ring            if ( !now_n->next->flg ) {                now_n = now_n->next;            }            else                break;        }        for ( int i = 0;; i++ ) {            if ( !now_f->fron->flg ) {                now_f = now_f->fron;            }            else                break;        }        cout << ",";    }}int main() {    int n, m, k;    while ( scanf ( "%d%d%d", &n, &k, &m ) != EOF && n && m && k ) {        bui_cir_link ( n );        operate ( n, k, m );        cout << endl;    }    return 0;}