Description
The Dole Queue
| The Dole Queue |
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 3 0 0 0
Sample output
4 8, 9 5, 3 1, 2 6, 
10, 7
where represents a space.
/////uva好坑啊，就没输出换行就WA。
#include<iostream>
#include<iomanip>
#include<string.h>
using namespace std;
int main()
{
int n[10000],a,b,c,i,j,k;
int s,l,b1,c1;
while(cin>>a>>b>>c&&a)
{
n[0]=-9999;
l=0;b1=0,c1=0;i=1;j=a;
memset(n,0,sizeof(n));
n[0]=n[0];
s=a;
while(s)
{l--;c1=b1=0;
for(;i<=a;i++)
{
if(n[i]==0)
b1++;
if(b1==b)
{cout<<setw(3)<<i;n[i]=l;s--;break;}
if(i==a)
i=0;
}
if(s>=0)
{
for(;j>0;j--)
{
if(j==a+1)
j--;
if(n[j]==0||n[j]==l)
c1++;
if(c1==c&&n[j]!=l)
{cout<<setw(3)<<j;n[j]=l;s--;}
if(j==1)
j=a+1;
if(c1==c)
{if(s!=0)
cout<<',';
if(s==0)
cout<<endl;
break;}
}
}
}
}
return 0;
}
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