uva133-S.B.S.

Posted 2020-07-04 报告振兴哥

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.

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这道题简单，上代码:

 1 // UVa133 The Dole Queue
 2 #include<cstdio>
 3 #define maxn 25
 4 int n, k, m, a[maxn];
 5 int go(int p, int d, int t) {
 6   while(t--) {
 7     do {p=(p+d+n-1)%n+1;} while(a[p] == 0);
 8   }
 9   return p;
10 }
11 
12 int main() {
13   while(scanf("%d%d%d", &n, &k, &m) == 3 && n) {
14     for(int i = 1; i <= n; i++) a[i] = i;
15     int left = n;
16     int p1 = n, p2 = 1;
17     while(left) {
18       p1 = go(p1, 1, k);
19       p2 = go(p2, -1, m);
20       printf("%3d", p1); left--;
21       if(p2 != p1) { printf("%3d", p2); left--; }
22       a[p1] = a[p2] = 0;
23       if(left) printf(",");
24     }
25     printf("\n");
26   }  
27   return 0;
28 }