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
题意：两个人从一个数字围成的圈中选数，一个顺时针选，另一个逆时针选，各自数一个固定的数后将那个数拿出去，如果两个人选到了同一个数，就只输出一个，否则先输出A的再输出B的。
题解：用一个大小为人数的数组表示这个圈，如果写数组元素的移动，代码就会变得很"丑陋"（个人这样认为），因此我们将选出去的数赋值为0，数数的时候跳过0去选。
代码：
#include<bits/stdc++.h>
using namespace std;
int n,k,m;
int a[21];
int read()//快速读入
{
int x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int go(int p,int d,int t)//选人，d是标识是正循环还是逆循环
{
while(t--)
do{p=(p+d+n-1)%n+1;}while(!a[p]);//找到下一个非0的数字
return p;
}
int main()
{
while(1)
{
n=read();k=read();m=read();
if(!n)break;
for(int i=1;i<=n;i++)a[i]=i;
int left=n,p1=n,p2=1;
while(left)
{
p1=go(p1,1,k);//正循环
p2=go(p2,-1,m);//逆循环
printf("%3d",p1);left--;//选出去一个人后，记得减去1
if(p2!=p1){printf("%3d",p2);left--;}
a[p1]=a[p2]=0;//将选出去的数字赋值成0，接下来的循环遇到了就跳过
if(left)printf(",");
}
puts("");
}
return 0;
}