uva 133

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

#include<iostream>  
#include<string.h>  
#include<stdio.h>  
#include<ctype.h>  
#include<algorithm>  
#include<stack>  
#include<queue>  
#include<set>  
#include<math.h>  
#include<vector>  
#include<map>  
#include<deque>  
#include<list>  
using namespace std;  
#define maxn 25
int n,k,m,a[maxn];
int go(int p,int d,int t)//构建走动函数，跳过数值为0的位置 
{
    while(t--)
    {
        do
        {
            p=(p+d+n-1)%n+1;
        }
        while(a[p]==0);
    }
    return p;
} 
int main()
{
    while(scanf("%d%d%d",&n,&k,&m)==3&&n)
    {
        for(int i=1;i<=n;i++)
        a[i]=i;
        int left =n;
        int p1=n,p2=1;//初始化移动位置，逆时针用p1，顺时针用p2
        while(left)
        {
            p1=go(p1,1,k);
            p2=go(p2,-1,m);
            printf("%d",p1);
            left--;
            if(p1!=p2)
            {
                 printf(" %d",p2);
                 left--;
            }
            if(left)
            printf(",");
            a[p1]=a[p2]=0;;
        } 
        printf("
");
    }
    return 0;
}