传送门
题意： 试构造一个数列，使得相邻两数值差的绝对值>= 2 且 <= 4；若无法构造直接输出-1。
思路：
- 显然可看成，只有n >= 4才能构造成功
- 例如n = 4时，排列可为 3 1 4 2 或 2 4 1 3 ，即左右两边以奇一偶(或一偶一奇)排列。
- 当n为偶数是以 2 4 1 3 为基础，而n为奇数时当以 3 1 4 2 为基础
代码实现：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <iostream>
#include <sstream>
#include <string>
#include <list>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{
1, 0}, {
-1, 0}, {
0, 1}, {
0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;int t, n;signed main()
{
IOS;cin >> t;while(t --){
cin >> n;if(n < 4) cout << -1 << endl;else{
int m = n;if(m % 2){
while(m > 0){
cout << m << " ";m -= 2;}cout << "4 2 ";for(int i = 6; i < n; i += 2)cout << i << " ";}else{
while(m > 5){
cout << m << " ";m -= 2;}cout << "2 4 ";for(int i = 1; i < n; i += 2)cout << i << " ";}cout << endl;}}return 0;
}