Codeforces Round #636 (Div. 3) D.Constant Palindrome Sum

Codeforces Round #636 (Div. 3) D.Constant Palindrome Sum

题目链接
You are given an array a consisting of n integers (it is guaranteed that n is even, i.e. divisible by 2). All ai does not exceed some integer k.

Your task is to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace ai with some integer in range [1;k]) to satisfy the following conditions:

after all replacements, all ai are positive integers not greater than k;
for all i from 1 to n2 the following equation is true: ai+an−i+1=x, where x should be the same for all n2 pairs of elements.
You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤1e4) — the number of test cases. Then t test cases follow.

The first line of the test case contains two integers n and k (2≤n≤2e5,1≤k≤2e5) — the length of a and the maximum possible value of some ai correspondingly. It is guratanteed that n is even (i.e. divisible by 2). The second line of the test case contains n integers a1,a2,…,an (1≤ai≤k), where ai is the i-th element of a.

It is guaranteed that the sum of n (as well as the sum of k) over all test cases does not exceed 2⋅105 (∑n≤2e5, ∑k≤2e5).

Output

For each test case, print the answer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.

Example

input

4
4 2
1 2 1 2
4 3
1 2 2 1
8 7
6 1 1 7 6 3 4 6
6 6
5 2 6 1 3 4

output

0
1
4
2

比较巧妙的一道题目~
我们任意想到暴力，即从 $[2,2*k]$ 里面一个个枚举，找最少替代次数，这样想肯定是会超时的，所以我们考虑用预处理的方式快速计算每个数需要的最少替代次数
首先用 $cnt[i]$ 记录每一对 $a[i]+a[n-i-1]$ 出现的次数
然后用 $pre[i]$ 记录每一对 $a[i],a[n-i-1]$ 变换一次得到的数的范围，很容易发现这个范围就是 $[min(a[i],a[n-i-1])+1,max(a[i],a[n-i-1])+k]$，通过记录端点的值控制区间修改，即 $pre[min(a[i],a[n-i-1])+1]+1$， $pre[max(a[i],a[n-i-1])+k+1]-1$，那么对每个数 $i$，所需的最小变换次数就是 $(n/2-pre[i])*2+pre[i]-sum[i]$，遍历更新一下答案即可，AC代码如下：

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

int main(){
    int t;
    cin>>t;
    while(t--){
        int n,k;
        cin>>n>>k;
        vector<int>a(n);
        for(auto &i:a) cin>>i;
        vector<int>cnt(2*k+1);
        for(int i=0;i<n/2;i++)
            cnt[a[i]+a[n-i-1]]++;
        vector<int>pre(2*k+2);
        for(int i=0;i<n/2;i++){
            int l1=a[i]+1,r1=a[i]+k;
            int l2=a[n-i-1]+1,r2=a[n-i-1]+k;
            pre[min(l1,l2)]++;
            pre[max(r1,r2)+1]--;
        }
        for(int i=1;i<=2*k+1;i++) pre[i]+=pre[i-1];
        int ans=1e9;
        for(int i=2;i<=2*k;i++){
            ans=min(ans,(pre[i]-cnt[i])+(n/2-pre[i])*2);
        }
        cout<<ans<<endl;
    }
    return 0;
}