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 |
| 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.
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 }