Description

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 Format

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 Format

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).

10 4 3
0 0 0​

4  8,  9  5,  3  1,  2  6, 10,  7​

Hint

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n开始顺时针数。在每一轮中，官员A数k个就停下来，官员B数m个就停下来（注意有可能两个官员停在同一个人上）。接下来被官员选中的人（1个或者2个）离开队伍。 输入n，k，m输出每轮里被选中的人的编号（如果有两个人，先输出被A选中的）。例如，n=10，k=4，m=3，输出为 4 8, 9 5, 3 1, 2 6, 10, 7。（每一轮以逗号隔开）注意：输出的每个数应当恰好占3列 (printf("%3d",num);)。

Source

数组与字符串