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.
新成立的国家绿色劳工犀牛党（National Green Labour Rhinoceros Party）在认真尝试缩减（减少）领取失业救济金的队伍时，决定了以下战略。每天，所有申请领取失业救济金的人都会围成一个大圈，面朝内。任意选择一个人作为数字1，其余的人按逆时针方向编号，直到N（他将站在1的左边）。从1开始逆时针移动，一个劳工官员对第K个申请人进行计数，另一个劳工官员从N开始顺时针移动，对第M个申请人进行计数。两个被选中的人将被送走接受再培训；如果两个官员都选择同一个人，她（他）将被送走成为一名政治家。然后，每个官员开始在下一个可用的人处重新计数，这个过程将继续进行，直到没有人离开。请注意，两名受害者（抱歉，受训者）会同时离开圆圈，因此一名官员可以统计另一名官员已经选择的人员。（百度翻译）
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).
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
【代码】
#include <iostream>
#include <stdio.h>
#include <string.h>
int peo[30],n;
int go(int post,int d,int step)
{
int cnt=0;
while(cnt!=step)
{
post=(post+d+n)%n;
if(peo[post]!=0)
cnt++;
}
return post;
}
int main()
{
int k,m;
while(scanf("%d%d%d",&n,&k,&m)&&n&&k&&m)
{
memset(peo,0,sizeof(peo));
int p1=n-1,p2=0,left=n;
for(int i=0;i<n;i++)
peo[i]=i+1;
while(left)
{
p1=go(p1,1,k);
p2=go(p2,-1,m);
left--;
printf("%3d",peo[p1]);
if(p1!=p2)
{
printf("%3d",peo[p2]);
left--;
}
if(left)
{
printf(",");
}
peo[p1]=peo[p2]=0;
}
printf("\n");
}
}
来源：https://blog.csdn.net/weixin_43869730/article/details/100120273