題意：
思路：先欽定這個序列的和為4n^{2} ，那麼差值就是2n
考慮一個初始序列1,2,3,⋯,n−1,2n+1.現在我們要做的就是將這個序列變成合法的我可以進行整體都+x的操作使得這個序列的和儘量逼近4n^{2}。直接算出每個數應該加上的x，還會有一點剩餘，加到a[n-1]上即可，a[n-1]+n==a[n],而剩餘數<n，顯然他們不會相等，構造完成。
#include <bits/stdc++.h>
#include<bits/extc++.h>
using namespace __gnu_pbds;
using namespace std;
using i64 = long long;
using u64 = unsigned long long;
using PII = pair<i64, i64>;
const int inf = 0x3f3f3f3f;
const i64 INF = 0x3f3f3f3f3f3f3f3f;
#define Z cout << "\n"
#define lb lower_bound
#define ub upper_bound
#define D(x) cerr << #x << ": " << (x) << "\n"
#define DV(v) cerr<<#v<<": ";for(int i=0;i<(v).size();i++)cerr<<((v)[i])<<",";cerr<<"\n"
#if 1
#define int i64
#endif
signed main() {
ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int>a(n + 5);
for (int i = 1; i <= n - 1; i++)a[i] = i;
a[n] = 2 * n + 1;
int S = 4 * n * n;
int sum = 0;
for (int i = 1; i <= n; i++)sum += a[i];
int d = (S - sum) / n;
int s = (S - sum) - d * n;
for (int i = 1; i <= n; i++)a[i] += d;
a[n - 1] += s;
for (int i = 1; i <= n; i++)cout << a[i] << ' ';
Z;
}
return 0;
}