Uva-133 The Dole Queue

In a serious attempt to downsize (reduce) the dole queue, The New National Green Labor 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 labor 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 shoshould 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 shocould 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.

This question is to give you a round-robin queue with a length of N. A person starts counting counter-clockwise from 1, and the number of th person is listed in the column. A person starts counting clockwise from the nth person, when the M column is selected, the two selected persons must be listed at the same time (so as not to affect the number of other persons), and the same person should be selected only once.

Solution: Simulate with arrays. When arr [I] is equal to 0, it indicates the I-th Column.

 

#include <iostream>#include<deque>#include<algorithm>#include<cstdio>using namespace std;int main(){    int n,k,m,size;    int arr[20];    while(cin>>n>>k>>m)    {        if(n==0&&k==0&&m==0)break;        int p=0,q=n-1;        size=n;        for(int i=0; i<n; i++)            arr[i]=i+1;        while(size>0)        {            for(int i=1; i<k; i++)            {                p=(p+1)%n;                if(!arr[p])i--;            }            for(int i=1; i<m; i++)            {                q=(q-1+n)%n;                if(!arr[q])i--;            }            if(arr[p])            {                printf("%3d",arr[p]);                arr[p]=0;                size--;            }            if(arr[q])            {                printf("%3d",arr[q]);                arr[q]=0;                size--;            }            for(int i=0; i<n; i++)            {                p=(p+1)%n;                if(arr[p])break;            }            for(int i=0; i<n; i++)            {                q=(q-1+n)%n;                if(arr[q])break;            }            if(size)printf(",");            else printf("\n");        }    }    return 0;}