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 aa of nn positive integers.

Let LIS(a)\text{LIS}(a) denote the length of longest strictly increasing subsequence of aa . For example,

  • LIS([2,1,1,3])\text{LIS}([2, \underline{1}, 1, \underline{3}]) = 22 .
  • LIS([3,5,10,20])\text{LIS}([\underline{3}, \underline{5}, \underline{10}, \underline{20}]) = 44 .
  • LIS([3,1,2,4])\text{LIS}([3, \underline{1}, \underline{2}, \underline{4}]) = 33 .

We define array aa' as the array obtained after reversing the array aa i.e. a=[an,an1,,a1]a' = [a_n, a_{n-1}, \ldots , a_1] .

The beauty of array aa is defined as min(LIS(a),LIS(a))min(\text{LIS}(a),\text{LIS}(a')) .

Your task is to determine the maximum possible beauty of the array aa if you can rearrange the array aa arbitrarily.

输入格式

The input consists of multiple test cases. The first line contains a single integer tt (1t104)(1 \leq t \leq 10^4) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer nn (1n2105)(1 \leq n \leq 2\cdot 10^5) — the length of array aa .

The second line of each test case contains nn integers a1,a2,,ana_1,a_2, \ldots ,a_n (1ai109)(1 \leq a_i \leq 10^9) — the elements of the array aa .

It is guaranteed that the sum of nn over all test cases does not exceed 21052\cdot 10^5 .

输出格式

For each test case, output a single integer — the maximum possible beauty of aa 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, aa = [6,6,6][6, 6, 6] and aa' = [6,6,6][6, 6, 6] . LIS(a)=LIS(a)\text{LIS}(a) = \text{LIS}(a') = 11 . Hence the beauty is min(1,1)=1min(1, 1) = 1 .

In the second test case, aa can be rearranged to [2,5,4,5,4,2][2, 5, 4, 5, 4, 2] . Then aa' = [2,4,5,4,5,2][2, 4, 5, 4, 5, 2] . LIS(a)=LIS(a)=3\text{LIS}(a) = \text{LIS}(a') = 3 . Hence the beauty is 33 and it can be shown that this is the maximum possible beauty.

In the third test case, aa can be rearranged to [1,2,3,2][1, 2, 3, 2] . Then aa' = [2,3,2,1][2, 3, 2, 1] . LIS(a)=3\text{LIS}(a) = 3 , LIS(a)=2\text{LIS}(a') = 2 . Hence the beauty is min(3,2)=2min(3, 2) = 2 and it can be shown that 22 is the maximum possible beauty.

首页