1343B - Balanced Array.cpp

/*
You are given a positive integer n, it is guaranteed that n is even (i.e. divisible by 2).

You want to construct the array a of length n such that:

The first n2 elements of a are even (divisible by 2);
the second n2 elements of a are odd (not divisible by 2);
all elements of a are distinct and positive;
the sum of the first half equals to the sum of the second half (?i=1n2ai=?i=n2+1nai).
If there are multiple answers, you can print any. It is not guaranteed that the answer exists.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1=t=104) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2=n=2·105) — the length of the array. It is guaranteed that that n is even (i.e. divisible by 2).

It is guaranteed that the sum of n over all test cases does not exceed 2·105 (?n=2·105).

Output
For each test case, print the answer — "NO" (without quotes), if there is no suitable answer for the given test case or "YES" in the first line and any suitable array a1,a2,…,an (1=ai=109) satisfying conditions from the problem statement on the second line.

Example
inputCopy
5
2
4
6
8
10
outputCopy
NO
YES
2 4 1 5
NO
YES
2 4 6 8 1 3 5 11
NO
*/

#include<bits/stdc++.h>
using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
	int t;
	cin >> t;
	
	while (t--) {
		int n;
		cin >> n;
		
		if ((n / 2) % 2) {
			cout << "NO\n";
		} else {
			cout << "YES\n";
			int sum = 0;
			for (int i = 2; i <= n; i += 2) {
				cout << i << " ";
				sum += i;
			}
			for (int i = 1; i < n - 1; i += 2) {
				cout << i << " ";
				sum -= i;
			}
			cout << sum << "\n";
			
		}
	}
    
    return 0;
}