UVa - 133 - The Dole Queue(救济金发放)

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).

Sample input

10 4 30 0 0

Sample output

 4  8,  9  5,  3  1,  2  6,  10,  7

where  represents a space.

#include<cstdio>#include<cstring>using namespace std;int n,k,m,left;int a[20];int main(){int go(int n,int k,int m);while(scanf("%d%d%d",&n,&k,&m)){if(n==0)break;left = n; //剩余的人数 memset(a,0,20);for(int i=1 ;i<=n ;i++){a[i] = i;}int p1 = n,p2 = 1;while(left){p1 = go(p1,1,k);p2 = go(p2,-1,m);printf("%3d",p1);left--;if(p2!=p1){printf("%3d",p2);left--;}a[p1] = a[p2] = 0;if(left)printf(",");}printf("\n");}return 0;}int go(int p ,int d ,int t){while(t--){//数不是0的个数t次 do{p = (p+d+n-1)%n+1; //***}while(a[p]==0);}return p;}