Description

 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 30 0 0

Sample output

4 8, 9 5, 3 1, 2 6, 10, 7

where represents a space.

 

#include <iostream>
#include <iomanip>
using namespace std;
int a[1000];
int main()
{int n,l,r,i,j,k;
while(cin>>n,n!=0)
{int N=n;
for(i=0;i<=n+1;i++)
a[i]=0;
cin>>l>>r;
int q=0,p=n+1;
while(N!=0)
{
 for(j=1;j<=l;)
 {
  q++;
  if(q==n+1)q=1;
  if(a[q]!=1)j++;}
 
 for(k=1;k<=r;)
 { p--;
 
 if(p==0)p=n;
 if(a[p]!=1)k++;
 }
 a[q]=1;
 a[p]=1;
 if(q!=p){cout<<setw(3) <<q<<setw(3) <<p;N=N-2;}
 else if(q==p){cout<<setw(3) <<q;N--;}
 if(N!=0)cout<<",";
 else cout<<endl;
}

 

 

}
return 0;

}

注意输出格式