UVa 133 - The Dole Queue-优快云博客

UVa 133 - The Dole Queue

1 题目
2 思路
3 代码
4 参考

1题目

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The Dole Queue

In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour Rhinoceros Party has decided on the following strategy. Every day all dole applicants will be placed in a large circle, facing inwards. Someone is arbitrarily chosen as number 1, and the rest are numbered counter-clockwise up to N (who will be standing on 1's left). Starting from 1 and moving counter-clockwise, one labour official counts off k applicants, while another official starts from N and moves clockwise, counting m applicants. The two who are chosen are then sent off for retraining; if both officials pick the same person she (he) is sent off to become a politician. Each official then starts counting again at the next available person and the process continues until no-one is left. Note that the two victims (sorry, trainees) leave the ring simultaneously, so it is possible for one official to count a person already selected by the other official.

Input

Write a program that will successively read in (in that order) the three numbers (N, k and m; k, m > 0, 0 < N < 20) and determine the order in which the applicants are sent off for retraining. Each set of three numbers will be on a separate line and the end of data will be signalled by three zeroes (0 0 0).

Output

For each triplet, output a single line of numbers specifying the order in which people are chosen. Each number should be in a field of 3 characters. For pairs of numbers list the person chosen by the counter-clockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma).

Sample input

10 4 3
0 0 0

Sample output

48,95,31,26,10,7

whererepresents a space.

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2思路

这是一个双向约瑟夫环问题，使用循环双向链表实现。题目本身不难，关键在于理解题目的意思，并考虑到一些特殊状态的处理。注意是两个officer先都确定下人，然后这两个人再出去，不是先出去一个，再出去一个。另外，需要考虑两人选择相同人时的情况；还要考虑被选的两人位置相邻时的情况，这和一般情况下对链表的操作是不一样的。

3代码

/*
 * Problem: UVa 133 The Dole Queue
 * Lang: ANSI C
 * Time: 0.009s
 * Author: minix
 */
#include <stdio.h>

#define N 20

typedef struct _node {
  struct _node *pre;
  struct _node *next; 
  int seq;
}Node;

Node g_node[N];

int main() {
  int n, k, m, i;
  Node *pa = NULL;
  Node *pb = NULL;
  while (scanf ("%d%d%d", &n,&k,&m) != EOF) {
    if (n==0 && k==0 && m==0) break;
    /* init double & circle queue */
    for (i=1; i<=n; i++) {
      g_node[i].seq = i;
      g_node[i].pre = &g_node[i-1];
      g_node[i].next = &g_node[i+1];
    }
    g_node[1].pre = &g_node[n];
    g_node[n].next = &g_node[1];

    /* kill */
    pa = &g_node[1];
    pb = &g_node[n];

    while (n != 0) {
      for (i=1; i<=(k-1); i++)
        pa = pa->next;
      for (i=1; i<=(m-1); i++)
        pb = pb->pre;

      if (pa == pb) {
        printf ("%3d", pa->seq);
        pa->pre->next = pa->next;
        pa->next->pre = pa->pre;
        pa = pa->next;
        pb = pb->pre;
        n -= 1;
      } else {
        printf ("%3d%3d", pa->seq, pb->seq);
        n -= 2;
        if (pa->next == pb) {
          pa->pre->next = pb->next;
          pb->next->pre = pa->pre;
          pa = pb->next;
          pb = pa->pre;
        } else {
          pa->pre->next = pa->next;
          pa->next->pre = pa->pre;
          pb->pre->next = pb->next;
          pb->next->pre = pb->pre;
          pa = pa->next;
          pb = pb->pre;
        }
      }

      if (n != 0) printf (",");
    }
    printf ("\n");
  }
  return 0;
}