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
先贴上我自己的代码，虽然它比较复杂
1 #include<bits/stdc++.h>
2 using namespace std;
3 int n,k,m;
4 int main()
5 {
6 while(~scanf("%d %d %d",&n,&k,&m),n,k,m)
7 {
8 int a[n+1];
9 for(int i=0;i<=n;i++)
10 a[i]=i;
11 int tot;
12 int t=0;
13 int f=1,s=n;
14 while(t<n){
15 int kk=0,mm=0;
16 while(1)
17 {
18 if(a[f]!=0&&kk<k)
19 kk++;
20 if(kk==k)
21 {
22 t++;
23 printf("%3d",a[f]);
24 tot=f;
25 f=(f+1)%n;
26 if(f==0)f=n;
27 break;
28 }
29 f=(f+1)%n;
30 if(f==0)f=n;
31 }
32 while(1)
33 {
34 if(a[s]!=0&&mm<m)
35 mm++;
36 if(mm==m)
37 {
38 if(s==tot)break;
39 t++;
40 printf("%3d",a[s]);
41 a[s]=0;
42 s=(s+n-1)%n;
43 if(s==0)s=n;
44 break;
45 }
46 s=(s+n-1)%n;
47 if(s==0)s=n;
48 }
49 a[tot]=0;
50 if(t!=n)printf(",");
51 }
52 printf("\n");
53 }
54 return 0;
55 }
有空再补一下大佬的想法，~