C. Challenging Cliffs

time limit per test

2 seconds

memory limit per test

256 megabytes

You are a game designer and want to make an obstacle course. The player will walk from left to right. You have n heights of mountains already selected and want to arrange them so that the absolute difference of the heights of the first and last mountains is as small as possible.

In addition, you want to make the game difficult, and since walking uphill or flat is harder than walking downhill, the difficulty of the level will be the number of mountains i (1≤i<n) such that hi≤hi+1 where hi is the height of the i-th mountain. You don't want to waste any of the mountains you modelled, so you have to use all of them.

From all the arrangements that minimize |h1−hn|, find one that is the most difficult. If there are multiple orders that satisfy these requirements, you may find any.

Input

The first line will contain a single integer t (1≤t≤100) --- the number of test cases. Then t test cases follow.

The first line of each test case contains a single integer n (2≤n≤2⋅105) --- the number of mountains.

The second line of each test case contains n integers h1,...,hn (1≤hi≤109), where hi is the height of the i-th mountain.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.

Output

For each test case, output n integers --- the given heights in an order that maximizes the difficulty score among all orders that minimize |h1−hn|.

If there are multiple orders that satisfy these requirements, you may output any.

Example

Input

Copy

复制代码
2
4
4 2 1 2
2
3 1

Output

Copy

复制代码
2 4 1 2 
1 3

Note

In the first test case:

The player begins at height 2, next going up to height 4 increasing the difficulty by 1. After that he will go down to height 1 and the difficulty doesn't change because he is going downhill. Finally the player will go up to height 2 and the difficulty will increase by 1. The absolute difference between the starting height and the end height is equal to 0 and it's minimal. The difficulty is maximal.

In the second test case:

The player begins at height 1, next going up to height 3 increasing the difficulty by 1. The absolute difference between the starting height and the end height is equal to 2 and it's minimal as they are the only heights. The difficulty is maximal.

解题说明:此题是一道模拟题,采用贪心算法, 首先从小到大排序后找到2个下标x,y 使得abs(h[x]-h[y])最小 ,可以发现 1 <= i < x 时 h[i] <= h[i+1] y <= i < n 时h[i] <= h[i+1] 所以把y到n中的数放前面 把1到x中的数放后面 当h[1]!=h[n]的时候 h[i] <= h[i+1]的i的个数最大为n-2 ,当h[1]==h[n]的时候 为n-1。

cpp 复制代码
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
	int t, n, i; 
	cin >> t;
	while (t--)
	{
		cin >> n; 
		int a[n], m = INT_MAX, j;
		for (i = 0; i < n; i++)
		{
			cin >> a[i];
		}
		sort(a, a + n);
		for (i = 1; i < n; i++) 
		{
			if (a[i] - a[i - 1] < m)
			{
				m = a[i] - a[i - 1];
				j = i;
			}
		} 
		cout << a[j - 1] << " ";
		for (i = j + 1; i < n; i++)
		{
			cout << a[i] << " ";
		}
		for (i = 0; i < j - 1; i++)
		{
			cout << a[i] << " ";
		}
		cout << a[j] << endl;
	}
	return 0;
}
相关推荐
侃侃_天下3 天前
最终的信号类
开发语言·c++·算法
echoarts3 天前
Rayon Rust中的数据并行库入门教程
开发语言·其他·算法·rust
Aomnitrix3 天前
知识管理新范式——cpolar+Wiki.js打造企业级分布式知识库
开发语言·javascript·分布式
每天回答3个问题3 天前
UE5C++编译遇到MSB3073
开发语言·c++·ue5
伍哥的传说3 天前
Vite Plugin PWA – 零配置构建现代渐进式Web应用
开发语言·前端·javascript·web app·pwa·service worker·workbox
小莞尔3 天前
【51单片机】【protues仿真】基于51单片机的篮球计时计分器系统
c语言·stm32·单片机·嵌入式硬件·51单片机
小莞尔3 天前
【51单片机】【protues仿真】 基于51单片机八路抢答器系统
c语言·开发语言·单片机·嵌入式硬件·51单片机
liujing102329293 天前
Day03_刷题niuke20250915
c语言
我是菜鸟0713号3 天前
Qt 中 OPC UA 通讯实战
开发语言·qt
JCBP_3 天前
QT(4)
开发语言·汇编·c++·qt·算法