CF1682C.LIS or Reverse LIS?
普及/提高-
通过率:0%
AC君温馨提醒
该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。
题目描述
You are given an array a of n positive integers.
Let LIS(a) denote the length of longest strictly increasing subsequence of a . For example,
- LIS([2,1,1,3]) = 2 .
- LIS([3,5,10,20]) = 4 .
- LIS([3,1,2,4]) = 3 .
We define array a′ as the array obtained after reversing the array a i.e. a′=[an,an−1,…,a1] .
The beauty of array a is defined as min(LIS(a),LIS(a′)) .
Your task is to determine the maximum possible beauty of the array a if you can rearrange the array a arbitrarily.
输入格式
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤104) — the number of test cases. Description of the test cases follows.
The first line of each test case contains a single integer n (1≤n≤2⋅105) — the length of array a .
The second line of each test case contains n integers a1,a2,…,an (1≤ai≤109) — the elements of the array a .
It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 .
输出格式
For each test case, output a single integer — the maximum possible beauty of a after rearranging its elements arbitrarily.
输入输出样例
输入#1
3 3 6 6 6 6 2 5 4 5 2 4 4 1 3 2 2
输出#1
1 3 2
说明/提示
In the first test case, a = [6,6,6] and a′ = [6,6,6] . LIS(a)=LIS(a′) = 1 . Hence the beauty is min(1,1)=1 .
In the second test case, a can be rearranged to [2,5,4,5,4,2] . Then a′ = [2,4,5,4,5,2] . LIS(a)=LIS(a′)=3 . Hence the beauty is 3 and it can be shown that this is the maximum possible beauty.
In the third test case, a can be rearranged to [1,2,3,2] . Then a′ = [2,3,2,1] . LIS(a)=3 , LIS(a′)=2 . Hence the beauty is min(3,2)=2 and it can be shown that 2 is the maximum possible beauty.