D. Constant Palindrome Sum

题目链接：https://codeforces.com/contest/1343/problem/D

想法：

假设 j = n - i + 1

我们知道有三种情况

1、 a[i] 和 a[j] 都不改变，

2、a[i] 或 a[j] 改变一个

3、a[i] 和 a[j] 都改变

我们考虑这些改变的范围对于此时a[i] + a[j] 的贡献，最后取枚举 a[i] + a[j] 的值找出里面贡献和最小的那个就可以了

#pragma GCC optimize(3, "Ofast", "inline")//O3优化
#pragma GCC optimize(2)//O2优化

#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <ctime>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>

#define ll long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f3f3f3f3f
#define max(a, b) (a>b?a:b)
#define min(a, b) (a<b?a:b)

const double eps = 1e-10;
const int maxn = 4e5 + 10;
const int MOD = 998244353;

int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }

using namespace std;

ll a[maxn],sum[maxn];

int main() {
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
        int n,k;
        cin >> n >> k;
        for (int i = 1;i <= n;i++)
            cin >> a[i];
        for (int i = 1;i <= 2 * k;i++)
            sum[i] = 0;
        for (int i = 1;i <= n/2;i++) {
            int j = n - i + 1;
            sum[2] += 2;
            sum[min(a[i],a[j]) + 1] -= 2;
            sum[min(a[i],a[j]) + 1] += 1;
            sum[a[i] + a[j]] -= 1;
            sum[a[i] + a[j] + 1] += 1;
            sum[max(a[i],a[j]) + k + 1] -= 1;
            sum[max(a[i],a[j]) + k + 1] += 2;
            sum[2 * k + 1] -= 2;
        }
        ll ans = INF;
        for (int i = 2;i <= 2 * k;i++) {
            sum[i] = sum[i-1] + sum[i];
            ans = min(ans,sum[i]);
        }
        cout << ans << endl;
    }
    return 0;
}