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 counterclockwise 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 Writeaprogramthatwillsuccessivelyreadin(inthatorder)thethreenumbers(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 counterclockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma). Note: The symbol ? in the Sample Output below represents a space. Sample Input 10 4 3 0 0 0 Sample Output ??4??8,??9??5,??3??1,??2??6,?10,??7
题意：n(n<20)个人站成一圈，逆时针编号为1~n。有两个官员，A从1开始逆时针数，B从顺时针开始数。在每一轮中，A数K个就停下来，B数m个就停下来（可能停在同一个人身上）。接下来被官员选中的人（1个或2个）离开队伍。先输出被A选中的！
注意：p=(p+d+n-1)%n+1;可以自己模拟几遍；
do{ }while在此题很应景；
归纳出相似代码的特点并用函数简化，例如:https://vjudge.net/problem/18191/origin(The Blocks Problem UVA - 101)
1 #include<iostream>
2 #include<algorithm>
3 #include<cstdio>
4 #include<cstring>
5 using namespace std;
6
7 int num[30];
8 int n,x,y;
9
10 int go(int p,int d,int t){
11 while(t--){
12 do{
13 p=(p+d+n-1)%n+1;
14 }while(num[p]==0);
15 }
16 return p;
17 }
18
19 int main()
20 {
21 while(~scanf("%d%d%d",&n,&x,&y)){
22 if(n==0||x==0||y==0) break;
23
24 for(int i=1;i<=n;i++) num[i]=i;
25
26 int left=n;
27 int p=n,q=1;
28 while(left){
29 p=go(p,1, x);
30 q=go(q,-1,y);
31 printf("%3d",p);
32 left--;
33 if(p!=q){
34 printf("%3d",q);
35 left--;
36 }
37 num[p]=0,num[q]=0;
38 if(left) printf(",");
39 }
40 cout<<endl;
41 }
42 return 0;
43 }