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.

就是一堆人站成一个环，一个人从第一个人顺时针数，一个人从最后一个人逆时针数，数到同一个人就出来一个，不同的人就出来两个，要求输出过程。和约瑟夫环是一样的，我手动写了两个数组标记模拟，先找出选中的两个人，判断是否一个人，一直循环到没有人为止。

#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int leap1[30], leap2[30];
int main()
{
	int i, j, m, n, ans, t, k, temp1, temp2, num, pre1, pre2;
	while (scanf("%d%d%d", &n, &k, &m) != EOF)
	{
		if (n == 0 && k == 0 && m == 0)break;
		memset(leap1, 0, sizeof(leap1));
		memset(leap2, 0, sizeof(leap2));
		num = n; pre1 = pre2 = 0;
		while (num)
		{
			ans = 0; i = 0;
			for (i = 0; ans < k; i++)
			{
				if (!leap1[(pre1 + i) % n])
				{
					ans++;
				}
				if (ans == k)
				{
					temp1 = (pre1 + i) % n;
					pre1 = temp1;
				}
			}
			ans = 0; i = 0;
			for (i = 0; ans < m; i++)
			{
				if (!leap2[(pre2 + i) % n])
				{
					ans++;
				}
				if (ans == m)
				{
					temp2 = n - (pre2 + i) % n;
					pre2 = (pre2 + i) % n;
				}
			}
			temp1++;
			if (temp1 == temp2)
				num--;
			else num = num - 2;
			leap1[temp1 - 1] = 1;
			leap2[n - temp1] = 1;
			leap2[n - temp2] = 1;
			leap1[temp2 - 1] = 1;
			if (temp2 == -1)
			{
				if (temp1 < 10)
					cout << "  " << temp1;
				else
					cout << " " << temp1;
			}
			else if (temp1 == temp2)
			{
				if (temp1 < 10)
					cout << "  " << temp1;
				else
					cout << " " << temp1;
			}
			else
			{
				if (temp1 < 10)
					cout << "  " << temp1;
				else
					cout << " " << temp1;
				if (temp2 < 10)
					cout << "  " << temp2;
				else
					cout << " " << temp2;
			}
			if (num)
				cout << ",";
		}
		cout << endl;
	}
	return 0;
}

写的非常搓，看了刘汝佳的方法，感觉很简便，尝试学习一下，调试了才发现他写的暗含很多道理，比如顺时针开始从n开始，逆时针是1开始，主要是函数用得好，看起来简便多了。

#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int n, k, m, a[25];
int Go(int p,int d,int m)
{
	if (a[p] != 0)
		m--;
	while (m--)
	{
		do{
			p = (p + d + n) % n;
		} while (a[p] == 0);
	}
	return p;
}
int main()
{
	int i, j;
	while (scanf("%d%d%d", &n, &k, &m) != EOF)
	{
		if (n == 0 && k == 0 && m == 0)break;
		for (i = 0; i < n; i++)
			a[i] = i + 1;
		int left = n;
		int p1, p2;
		p1 = 0;
		p2 = n - 1;
		while (left)
		{
			p1 = Go(p1, 1, k);
			p2 = Go(p2, -1, m);
			printf("%3d", p1+1); left--;
			if (p1 != p2)
			{
				printf("%3d", p2+1); left--;
			}
			if (left)
				cout << ",";
			a[p1] = a[p2] = 0;
		}
		cout << endl;
	}
	return 0;
}