Absolute Permutation

This is the most efficient solution of the Absolute Permutation question asked
at hackerrank.com .

Here is the question.

We define  to be a permutation of the first  natural numbers in the range . Let  denote the position of  in permutation  (please use -based indexing).

 is considered to be an absolute permutation if  holds true for every .

Given  and , print the lexicographically smallest absolute permutation, ; if no absolute permutation exists, print -1.

Input Format

The first line of input contains a single integer, , denoting the number of test cases. 
Each of the  subsequent lines contains  space-separated integers describing the respective  and  values for a test case.

Constraints

Output Format

On a new line for each test case, print the lexicographically smallest absolute permutation; if no absolute permutation exists, print -1.

Key observations.

1. The value of  k should be less than  or equal to n/2 other wise abs(pos¡  -i) =k  will not hold true for every i.
2. K should be a divisor of n (i.e.n%k=0) .Take a pen and a paper and experiment with different k (divisor and non-divisor of n). It's better for you if you try it yourself.
3. The quotient of n/k should be divisible by 2. (i.e. (n/k)%2=0) Again think!!

If you still didn't get this than a detailed explanation is there at the end of this post.I have not given the full explanation here since i want you all to try with these hint's first.

That's it . That was all the hidden tricks that was there in this problem.

Here is my code.

#include <iostream>
#include <vector>
using namespace std;
int main(){
                 int t,n,k,dev;
                 cin >> t;
for(int a0 = 0; a0 < t; a0++){
           cin >> n >> k;
           dev=n/k;
            vector<int> pos(n);
            if(k==0){      //if k=0 print the array itself without manipulating any thing
                            for(int i=1;i<=n;i++)cout<<i<<" ";
            }
           else if ((k<=n/2)&&(n%k==0)&&(dev%2 ==0)){   //else if k<=n/2 and n%k==0 and (n/k)%2==0 than do this
                         for(int i=0;i<n;i++){
                                   if((i/k)%2 == 0) pos[i] = i + k+1;  
                                  else   pos[i] = i - k+1;
                          }
                          for (int i = 0; i< n; i++)  cout << pos[i] << " ";
            }
            else   cout << "-1";
            cout <<endl;
}
return 0;
}

Here is the complete explanation in case someone is still struggling.

Note: k<=n/2 is very apparent so I have not discussed that here.

Please forgive me for explaining this in paper since I don't have enough time to type it in my computer.

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