UVa 133 - The Dole Queue

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.

 

 

#include <string.h>
#include <stdlib.h>
#include <stdio.h>
int k,m,N;
int left[30],right[30];

void deletea (int x)
{
right[left[x]]=right[x];
left[right[x]]=left[x];
}

int main()
{
while((scanf("%d%d%d",&N,&k,&m))&&N&&k&&m)
{
for(int i=1;i<=N;i++)
{
left[i]=(i==1?N:i-1);
right[i]=(i==N?1:i+1);
}
int off1=1,off2=N,p=N;
while(p)
{
int a=k-1,b=m-1;
while(a)
{
off1=right[off1];
a--;
}
while(b)
{
off2=left[off2];
b--;
}
if(off1==off2)
{
if(p==1)
printf("%3d",off1);
else
printf("%3d,",off1);
p--;
}
else
{
if(p==2)
printf("%3d%3d",off1,off2);
else
printf("%3d%3d,",off1,off2);
p-=2;
}
deletea (off1);
deletea (off2);
int off=off1;
off1=right[off1];if(off1==off2)off1=right[off1];
off2=left[off2];if(off==off2)off2=left[off2];
}
printf("\n");
}
}