Uva133-优快云博客

博客围绕 The Dole Queue 问题展开，该问题是新政党为缩减救济队列制定的策略。需输入 N、k 和 m 三个数，程序要确定申请人接受再培训的顺序。还给出了输入输出示例，作者贴出自己代码，称后续补充大佬思路。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 }