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.

/////uva好坑啊，就没输出换行就WA。

#include<iostream>

#include<iomanip>

#include<string.h>

using namespace std;

int main()

{

	int n[10000],a,b,c,i,j,k;

	int s,l,b1,c1;

	while(cin>>a>>b>>c&&a)

	{

		n[0]=-9999;

		l=0;b1=0,c1=0;i=1;j=a;

		memset(n,0,sizeof(n));

		n[0]=n[0];

		s=a;

		while(s)

		{l--;c1=b1=0;

	for(;i<=a;i++)

	{

		if(n[i]==0)

			b1++;

		if(b1==b)

		{cout<<setw(3)<<i;n[i]=l;s--;break;}

		if(i==a)

			i=0;

	}

	if(s>=0)

	{

			for(;j>0;j--)

	{

				if(j==a+1)

					j--;

		if(n[j]==0||n[j]==l)

			c1++;

		if(c1==c&&n[j]!=l)

		{cout<<setw(3)<<j;n[j]=l;s--;}

		if(j==1)

			j=a+1;

			if(c1==c)

		{if(s!=0)

		cout<<',';

			if(s==0)

		cout<<endl;

			break;}

	}

	}

	}

	}

return 0;

}