286A - Lucky Permutation - CodeForces Solution

#include<bits/stdc++.h>
using namespace std;
int n,a[1000005];
int main()
{
    cin>>n;
    if(n%4!=0&&n%4!=1)
	{
        cout<<-1<<endl;
        return 0;
    }
    for(int i=1;i<n/2;i+=2)
	{
        a[i]=i+1;
        a[i+1]=n-i+1;
        a[n-i+1]=n-i;
        a[n-i]=i;
    }
    if(n%2==1) 
    {
    	a[(n/2)+1]=(n/2)+1;
	}
    for(int i=1;i<=n;i++)
	{
        cout<<a[i]<<' ';
    }
    return 0;
}