Uva

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.

循环时候的增长步长方式注意

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <algorithm>

using namespace std;

int n, k, m;
int a[25];

int main()
{
	while (scanf("%d%d%d", &n, &k, &m) && n) {
		// 记得清空，虽然这道题目不清空没有什么影响
		memset(a, 0, sizeof(a));
		// 不清空不会影响结果，是因为这里将需要用到的都重新赋值了
		for (int i = 1; i <= n; i++) {
			a[i] = i;
		}
		// 剩下的人数
		int leftPeople = n;
		int pa = n, pb = 1;
		while (leftPeople) {
			// 用两个新的变量，不要改变k和m
			int ka = k;
			int mb = m;
			while (ka--) {
				pa = (pa + n) % n + 1;
				while (a[pa] == 0) {
					pa = (pa + n) % n + 1;
				}
			}
			while (mb--) {
				pb = (pb - 2 + n) % n + 1;
				while (a[pb] == 0) {
					pb = (pb - 2 + n) % n + 1;
				}
			}
			printf("%3d", pa);
			leftPeople--;
			if (pa != pb) {
				printf("%3d", pb);
				leftPeople--;
			}
			a[pa] = a[pb] = 0;
			if (leftPeople) {
				printf(",");
			}
		}
		printf("
");
	}

	return 0;
}