Good Bye 2022: 2023 is NEAR
B
Koxia and Permutation
Codeforces
1000
原题
← AC →
翻译中…

Reve has two integers $n$ and $k$.

翻译中…

Let $p$ be a permutation$^\dagger$ of length $n$. Let $c$ be an array of length $n - k + 1$ such that $$c_i = \max(p_i, \dots, p_{i+k-1}) + \min(p_i, \dots, p_{i+k-1}).$$ Let the cost of the permutation $p$ be the maximum element of $c$.

翻译中…

Koxia wants you to construct a permutation with the minimum possible cost.

翻译中…

$^\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).

翻译中…

Input

翻译中…

Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2000$) — the number of test cases. The description of test cases follows.

翻译中…

The first line of each test case contains two integers $n$ and $k$ ($1 \leq k \leq n \leq 2 \cdot 10^5$).

翻译中…

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

翻译中…

Output

翻译中…

For each test case, output $n$ integers $p_1,p_2,\dots,p_n$, which is a permutation with minimal cost. If there is more than one permutation with minimal cost, you may output any of them.

翻译中…

Note

翻译中…

In the first test case,

翻译中…
  • $c_1 = \max(p_1,p_2,p_3) + \min(p_1,p_2,p_3) = 5 + 1 = 6$.
  • $c_2 = \max(p_2,p_3,p_4) + \min(p_2,p_3,p_4) = 3 + 1 = 4$.
  • $c_3 = \max(p_3,p_4,p_5) + \min(p_3,p_4,p_5) = 4 + 2 = 6$.
翻译中…

Therefore, the cost is $\max(6,4,6)=6$. It can be proven that this is the minimal cost.

翻译中…

样例

Input #1
3
5 3
5 1
6 6
Output #1
5 1 2 3 4
1 2 3 4 5
3 2 4 1 6 5
InputCopy OutputCopy
题目信息
题号
1770B
时间限制
1 s
内存限制
256 MB
标签
构造

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。