133 - The Dole Queue 算法竞赛第四章

Time limit: 3.000 seconds

Problem Description

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.

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

---

#include<stdio.h>
#define maxn 25
int n,k,m,a[maxn];
int go(int p,int d,int t){
	while(t--){
		do{
			p=(p+d+n-1)%n+1;
		}while(a[p]==0);
	}
	return p;
}
int main(){
	while(scanf("%d%d%d",&n,&k,&m)&&n){
		for( int i=1; i<=n; i++ )a[i]=i;
		int p1=0,p2=n+1,left=n;
		while(left){
			p1=go(p1,1,k);p2=go(p2,-1,m);
			printf("%3d",p1);left--;
			if(p1!=p2){
				printf("%3d",p2);left--;
			}
			if(left) printf(",");
			a[p1]=a[p2]=0;
		}
		printf("\n");
	}
	return 0;
}