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

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這題大意就是兩個人同時數數 被點到的人出列  算是約瑟夫環的升級版 

這題我思路是用兩個陣列去模擬選人過程 當每輪的人數選定後再同時在兩個數組裡標記出   一直重複直到所有人出列

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#include <stdio.h>

int main()
{
	int n, k, m;
	while(~scanf("%d %d %d", &n, &k, &m) && n)
	{
		int data1[25] = {0}, data2[25] = {0};
		for(int i = 0; i <= n; i++) { data1[i] = 1; data2[i] = 1; }
		int temp, x = 0, y = 0, pos1 = 1, pos2 = n, flag = 0;
		for( ; ; )
		{
			for( ; ; )
			{
				for(int i = pos1; ; i++)
				{
					if(data1[i]) { x++; } //計算有效次數
					if(x % k == 0 && data1[i])//每次出列的一定是k的倍數
					{
						printf("%3d", i);
						temp = i;
						data1[i] = 0;//0代表出列
						if(i == n) { pos1 = 1; }
						else { pos1 = i + 1;}
						flag = 1;
						break;
					}
					i %= n;
				}
				if(flag) { flag = 0; break; }
			}
			for( ; ; )
			{
				for(int i = pos2; ; i--)
				{
					if(data2[i]) { y++; }//計算有效次數
					if(y % m == 0 && data2[i])
					{
						if(temp != i)
						{ printf("%3d", i); data1[i] = 0; data2[temp] = 0; }//使出列的人在兩個數組裡都標記上
						data2[i] = 0;
						if(i == 1) { pos2 = n; }
						else { pos2 = i - 1; }
						flag = 1;
						break;
					}
					if(i == 1) { i = n + 1; }
				}
				if(flag) { flag = 0; break; }
			}
			int sum = 0;
			for(int i = 1; i <= n; i++)
			{ if(data1[i] == 0) sum++; }
			if(sum == n) { break; }
			printf(",");
		}
		printf("\n");
	}
	return 0;
 }