UVA - 133 The Dole Queue

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

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.

注意分别移动就行了

#include <stdio.h>
#include <string.h>

int seq[21];

int main()
{
    int n,k,m;
    while(~scanf("%d%d%d",&n,&k,&m)&&n)
    {
		int left = n;
		memset(seq,0,sizeof(seq));
		int pa = n;
		int pb = 1;
        while(left>0)
        {
			int A = k;
			int B = m;

			while(A--)
			{
				do {pa = pa%n + 1;}while(seq[pa]);
			}
			while(B--)
			{
				do {pb = (pb+n-2)%n + 1;}while(seq[pb]);
			}

			printf("%3d",pa);left--;
			if(pa!=pb){printf("%3d",pb);left--;}
			seq[pa] = seq[pb] = 1;
			if(left) putchar(',');
        }
        putchar('\n');
    }
	return 0;
}