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

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 30 0 0

Sample output

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

where represents a space.

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 }