UVa133.The Dole Queue

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=69

 

13874119

133

The Dole Queue

Accepted

C++

0.009

2014-07-13 02:44:49

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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

4 8, 9 5, 3 1, 2 6, 10, 7

where represents a space.

 

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解题思路：一道非常类似约瑟夫问题的题目。http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1197
　　　　 （关于约瑟夫问题，可以翻阅《具体数学》第一章引例。）

　　　　　直接数组模拟就好，没特殊机巧。白书上标程的写法更加精炼，要多学学优化自身的代码！

 

 1 #include <iostream>

 2 #include <cstring>

 3 #include <cstdio>

 4 

 5 using namespace std;

 6 

 7 const int maxn = 25;

 8 int n, m, k, peo[maxn];

 9 

10 int solve_z(int cur_first, int step_time) {

11     int ans = cur_first;

12     while (step_time --) {

13         ans ++;

14         if (ans > n) ans = 1;

15         while (peo[ans] == -1) {

16             ans ++;

17             if (ans > n) ans = 1;

18         }

19     }

20     return ans;

21 }

22 

23 int solve_f(int cur_first, int step_time) {

24     int ans = cur_first;

25     while (step_time --) {

26         ans --;

27         if (ans == 0)  ans = n;

28         while (peo[ans] == -1) {

29             ans --;

30             if (!ans) ans = n;

31         }

32     }

33     return ans;

34 }

35 

36 int main() {

37     while (cin >> n >> k >> m) {

38         if (n + k + m == 0) break;

39         for (int i =  0;  i <= n; i ++) {

40             peo[i] = i;

41         }

42 

43         int left_peo = n;

44         int cur1 = n, cur2 = 1;

45         while (left_peo != 0) {

46             cur1 = solve_z(cur1, k);

47             //cout << cur1 << endl;

48             cur2 = solve_f(cur2, m);

49             //cout << cur2 << endl;

50             //system("pause");

51             printf("%3d", cur1); left_peo --;

52             if (cur2 != cur1) {

53                 printf("%3d", cur2);

54                 left_peo --;

55             }

56             //cout << ",";

57             peo[cur1] = peo[cur2] = -1;

58             if(left_peo) cout << ",";

59         }

60         cout << endl;

61     }

62 

63     return 0;

64 }

 

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