UVA 133 The Dole Queue 【约瑟夫环】

题面：

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

题目大意：

一共有n个人，坐成一个环。以1-n逆时针编号。有两个人A,B。A从1开始数k个人，B从n开始数m个人。被数到的人退场。然后继续数，直到所有人都退场。
输出退场次序，若A与B数的人编号不同，A数的先输出。

大致思路：

开一个数组保存编号。退场的编号为0，数到0的时候跳过。记得对A取膜，B小于零时变为n。

代码：

#include<bits/stdc++.h>
using namespace std;
bool isempty(int *s,int n)//判断是否全部退场
{
    for(int i=1;i<=n;++i)
        if(s[i]!=0)
            return true;
    return false;
}
int main()
{
    int n,k,m;
    int s[21];
    while(cin>>n>>k>>m&&(n||k||m))
    {
        int a=1,b=n,cnt=1;
        for(int i=1;i<=n;++i)
            s[i]=i;
        while(isempty(s,n))
        {
            if(cnt!=1)
                printf(",");
            int ca=1,cb=1;
            while(ca!=k||cb!=m)
            {
                if(ca<k){
                    a++;
                    if(a>n)
                        a%=n;
                    if(s[a]!=0)
                        ca++;
                }
                if(cb<m){
                    b--;
                    if(b<1)
                        b=n;
                    if(s[b]!=0)
                        cb++;
                }
              //  cout<<ca<<" "<<cb<<endl;
            }
            if(a!=b)
                printf("%3d%3d",s[a],s[b]);
            else
                printf("%3d",s[a]);
            s[a]=s[b]=0;
            if(isempty(s,n)){//数到人退场之后的a,b指针处理
                while(s[a]==0)
                {
                    ++a;
                    if(a>n)
                        a=1;
                }
                while(s[b]==0)
                {
                    --b;
                    if(b<1)
                        b=n;
                }
            }
            cnt++;
        }
        printf("\n");
    }
    return 0;
}