The Dole Queue UVA - 133

The Dole Queue UVA - 133

题目描述
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 counterclockwise
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 counterclockwise
official first. Separate successive pairs (or singletons) by commas (but there should not be a
trailing comma).
Note: The symbol ⊔ in the Sample Output below represents a space.

Sample Input
10 4 3
0 0 0

Sample Output
␣␣4␣␣8,␣␣9␣␣5,␣␣3␣␣1,␣␣2␣␣6,␣10,␣␣7 （‘␣’表示空格）

大致题意
有N个人，编号1到N，逆时针围成一个圈，一个人从1开始逆时针数k出列一人，输出对应的编号，一个人从N开始顺时针数m次出列一人，输出对应的编号，直到所有人都出列。两人同时开始数，如果数到同一个人则只需输出一次编号。注意最后结果的输出格式。

思路
简单的模拟。

下面是代码

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <fstream>
#include <math.h>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<sstream>
#include<ctime>
using namespace std;

int main()
{
    int n,k,m,a[20],l,r,num,i;
    while(1)
    {
        cin>>n>>k>>m;
        if(!(n+k+m))
        return 0;
        memset(a,0,sizeof(a));
        a[0]=1;
        num=n;
        l=n;
        r=1;
        while(1)
        {
            for(i=0;;r++)
            {
                //cout<<r<<endl;
                if(r==n+1) r=1;
                if(a[r]==0)
                {
                    i++;
                    if(i==k)
                    break;
                }
            }
            for(i=0;;l--)
            {
                if(l==0) l=n;
                if(a[l]==0)
                {
                    i++;
                    if(i==m)
                    break;
                }
            }
            if(l!=r)
            {
                printf("%3d%3d",r,l);
                a[r]=1;
                a[l]=1;
                num-=2;
            }
            else if(l==r)
            {
                printf("%3d",r);
                a[r]=1;
                num--;
            }

            if(num)
            cout<<',';
            else
            break;
        }
        cout<<endl;
    }
    return 0;
 }