题解：Codeforces Round 967 (Div. 2) B [思维/构造]

B. Generate Permutation

time limit per test: 1.5 seconds

memory limit per test: 256 megabytes

input: standard input

output: standard output

There is an integer sequence $a$ of length $n$, where each element is initially $-1$.

Misuki has two typewriters where the first one writes letters from left to right, with a pointer initially pointing to $1$, and another writes letters from right to left with a pointer initially pointing to $n$.

Misuki would choose one of the typewriters and use it to perform the following operations until $a$ becomes a permutation of $[1, 2, \ldots, n]$

write number: write the minimum positive integer that isn't present in the array $a$ to the element $a_i$, $i$ is the position where the pointer points at. Such operation can be performed only when $a_i = -1$.
carriage return: return the pointer to its initial position (i.e. $1$ for the first typewriter, $n$ for the second)
move pointer: move the pointer to the next position, let $i$ be the position the pointer points at before this operation, if Misuki is using the first typewriter, $i := i + 1$ would happen, and $i := i - 1$ otherwise. Such operation can be performed only if after the operation, $1 \le i \le n$ holds.

Your task is to construct any permutation $p$ of length $n$, such that the minimum number of carriage return operations needed to make $a = p$ is the same no matter which typewriter Misuki is using.

有一个长度为 $n$ 的整数序列 $a$ ，其中每个元素的初始值为 $-1$ 。

美雪有两台打字机，第一台打字机从左往右写字母，指针最初指向 $1$ ，另一台打字机从右往左写字母，指针最初指向 $n$ 。

美雪会选择其中一台打字机进行以下操作，直到 $a$ 变成 $[1, 2, \ldots, n]$ 的排列。

写数：将 $a$ 中未出现的最小正整数写入 $a_i$ ， $i$ 是指针指向的位置。这种操作只能在 $a_i = -1$ 时执行。
回车：将指针返回到初始位置(第一台打字机为 $1$ ，第二台打字机为 $n$ )。
移动指针：将指针移动到下一个位置，假设 $i$ 是指针在执行此操作前所指向的位置，如果美雪使用的是第一台打字机，则为 $i := i + 1$ ，否则为 $i := i - 1$ 。只有在操作之后， $1 \le i \le n$ 成立时，才能执行该操作。

你的任务是构造长度为 $n$ 的任意排列 $p$ ，使得无论美雪使用哪台打字机， $a = p$ 所需的最小回车操作次数都相同。

Input

Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the permutation.

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

输入

每个测试包含多个测试用例。第一行输入包含一个整数 $t$ ( $1 \le t \le 500$ ) - 测试用例的数量。测试用例说明如下。

每个测试用例的第一行都包含一个整数 $n$ ( $1 \le n \le 2 \cdot 10^5$ ) - 排列的长度。

保证所有测试用例的 $n$ 之和不超过 $2 \cdot 10^5$ 。

Output

For each test case, output a line of $n$ integers, representing the permutation $p$ of length $n$ such that the minimum number of carriage return operations needed to make $a = p$ is the same no matter which typewriter Misuki is using, or $-1$ if it is impossible to do so.

If there are multiple valid permutations, you can output any of them.

输出

对于每个测试用例，输出一行 $n$ 整数，表示长度为 $n$ 的排列 $p$ ，使得无论美雪使用哪台打字机，都能实现 $a = p$ 所需的最少回车操作次数相同，如果无法实现，则输出 $-1$ 。

如果存在多种有效的排列组合，则可以输出其中任何一种。

Example

Input

Output

1
-1
3 1 2

Note

In the first testcase, it's possible to make $a = p = [1]$ using $0$ carriage return operations.

In the second testcase, it is possible to make $a = p = [1, 2]$ with the minimal number of carriage returns as follows:

If Misuki is using the first typewriter:

Write number: write $1$ to $a_1$, $a$ becomes $[1, -1]$
Move pointer: move the pointer to the next position. (i.e. $2$)
Write number: write $2$ to $a_2$, $a$ becomes $[1, 2]$

If Misuki is using the second typewriter:

Move pointer: move the pointer to the next position. (i.e. $1$)
Write number: write $1$ to $a_1$, $a$ becomes $[1, -1]$
Carriage return: return the pointer to $2$.
Write number: write $2$ to $a_2$, $a$ becomes $[1, 2]$

It can be proven that the minimum number of carriage returns needed to transform $a$ into $p$ when using the first typewriter is $0$ and it is $1$ when using the second one, so this permutation is not valid.

Similarly, $p = [2, 1]$ is also not valid, so there is no solution for $n = 2$.

In the third testcase, it is possibile to make $a = p = [3, 1, 2]$ with $1$ carriage return with both the first and the second typewriter. It can be proven that, for both typewriters, it is impossible to write $p$ with $0$ carriage returns.

With the first typewriter it is possible to:

Move pointer: move the pointer to the next position. (i.e. $2$)
Write number: write $1$ to $a_2$, $a$ becomes $[-1, 1, -1]$
Move pointer: move the pointer to the next position. (i.e. $3$)
Write number: write $2$ to $a_3$, $a$ becomes $[-1, 1, 2]$
Carriage return: return the pointer to $1$.
Write number: write $3$ to $a_1$, $a$ becomes $[3,1,2]$

With the second typewriter it is possible to:

Move pointer: move the pointer to the next position. (i.e. $2$)
Write number: write $1$ to $a_2$, $a$ becomes $[-1, 1, -1]$
Carriage return: return the pointer to $3$.
Write number: write $2$ to $a_3$, $a$ becomes $[-1, 1, 2]$
Move pointer: move the pointer to the next position. (i.e. $2$)
Move pointer: move the pointer to the next position. (i.e. $1$)
Write number: write $3$ to $a_1$, $a$ becomes $[3, 1, 2]$

注

在第一个测试用例中，可以使用 $0$ 回车操作制作 $a = p = [1]$ 。

在第二个测试用例中，可以使用 $$$0$$$ 回车操作制作 $a = p = [1, 2]$ ，最小回车次数如下：

如果美树使用的是第一台打字机：

写数字：将 $1$ 写入 $a_1$ ， $a$ 变为 $[1, -1]$
移动指针：将指针移动到下一个位置。(即 $2$ )
写入数字：将 $2$ 写入 $a_2$ ， $a$ 变为 $[1, 2]$ 。

如果美雪使用的是第二台打字机：

移动指针：将指针移动到下一个位置。(即 $1$ )
写入数字：将 $1$ 写入 $a_1$ ， $a$ 变为 $[1, -1]$ 。
回车：返回指向 $2$ 的指针。
写入数字：将 $2$ 写入 $a_2$ ， $a$ 变为 $[1, 2]$ 。

可以证明，使用第一台打字机时，将 $a$ 转换为 $p$ 所需的最少回车次数为 $0$ ，而使用第二台打字机时则为 $1$ ，因此这种排列是无效的。

同样， $p = [2, 1]$ 也是无效的，因此 $n = 2$ 无解。

题意

你需要在一个水平的序列中从 $1$ 打印到 $n$
你有两台打字机

一台从最左边开始打印，每次只能打印/向右走一格/回到初始位置
一台从最右边开始打印，每次只能打印/向左走一格/回到初始位置

每次打印只能打印 $上一个数字+1$
你需要给出一个序列，使得打印这个序列时
不管选择哪一台打印机，回到初始位置的操作数相同

如果存在这样的序列，那就从左到右打印这个序列
如果不存在这样的序列，那就输出 $-1$

题解

这题实际上考察了思维和构造
怎么构造这样的一个序列呢？
其实样例给了我们提示
我们只要类似于把数字在两边交替放置，一直到中间即可
比如，当 $n = 11$ 时
序列为 $1,3,5,7,9,11,10,8,6,4,2$
这样的话，就可以不管从哪边开始打字，
返回初始位置次数都是一样的了

但是，如果 $n$ 是偶数的话就只能输出 $-1$
构造不了序列

代码

#include <bits/stdc++.h>
#define int long long
#define INF 0x3f3f3f3f
#define all(x) x.begin(),x.end()

using i64 = long long;
const int N = 2e5 + 10;
int t = 1;
int a[N];

void solve() {
	int n;
	std::cin >> n;
	if(n%2 == 1) {
		int st = 1;
		while(st <= n) {
			std::cout << st << " ";
			st += 2;
		}
		st = n-1;
		while(st) {
			std::cout << st << " ";
			st -= 2;
		}
		std::cout << "\n";
	}
	else std::cout << -1 << "\n";
}

signed main(){
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);

	std::cin >> t;
	while(t--){
		solve();
	}
	return 0;
}

有什么建议或者意见的欢迎在评论区留言！

posted on 2024-08-21 16:49 Jiejiejiang 阅读(59) 评论(0) 编辑收藏举报

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