E - Magic Points (0到4*n-5個點圍成正方形，最多交點)
Given an integer , we say a point on a 2D plane is a magic point, if and only if both and are integers, and exactly one of the following conditions is satisfied:
-
and ;
-
and ;
-
and ;
-
and .
It's easy to discover that there are magic points in total. These magic points are numbered from to in counter-clockwise order starting from .
DreamGrid can create magic lines from these magic points. Each magic line passes through exactly two magic points but cannot be parallel to the line or (that is to say, the coordinate axes).
The intersections of the magic lines are called dream points, and for some reason, DreamGrid would like to make as many dream points as possible. Can you tell him how to create these magic lines?
Input
There are multiple test cases. The first line of input contains an integer (about 100), indicating the number of test cases. For each test case, there is only one integer ().
Output
For each case output integers in one line separated by one space, indicating that in your answer, point and point is connected by a line for all .
If there are multiple answers, you can print any of them.
Sample Input
3
2
3
4
Sample Output
0 2 1 3
1 4 2 5 3 6
0 6 1 9 3 8 4 10
Hint
The sample test cases are shown as follow:
隊長做出來一臉懵逼~~~
#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
for(int i=0; i<n-1; i++)
{
printf("%d %d ",i,n+i);
}
if(n==2)
printf("%d %d\n",1,3);
else printf("%d %d\n",n-1,3*n-4);
}
return 0;
}