A. Median of an Array
time limit per test: 1 second
memory limit per test: 256 megabytes
input: standard input
output: standard output
You are given an array a a a of n n n integers.
The median of an array q 1 , q 2 , … , q k q_1, q_2, \ldots, q_k q1,q2,…,qk is the number p ⌈ k 2 ⌉ p_{\lceil \frac{k}{2} \rceil} p⌈2k⌉, where p p p is the array q q q sorted in non-decreasing order. For example, the median of the array [ 9 , 5 , 1 , 2 , 6 ] [9, 5, 1, 2, 6] [9,5,1,2,6] is 5 5 5, as in the sorted array [ 1 , 2 , 5 , 6 , 9 ] [1, 2, 5, 6, 9] [1,2,5,6,9], the number at index ⌈ 5 2 ⌉ = 3 \lceil \frac{5}{2} \rceil = 3 ⌈25⌉=3 is 5 5 5, and the median of the array [ 9 , 2 , 8 , 3 ] [9, 2, 8, 3] [9,2,8,3] is 3 3 3, as in the sorted array [ 2 , 3 , 8 , 9 ] [2, 3, 8, 9] [2,3,8,9], the number at index ⌈ 4 2 ⌉ = 2 \lceil \frac{4}{2} \rceil = 2 ⌈24⌉=2 is 3 3 3.
You are allowed to choose an integer i i i ( 1 ≤ i ≤ n 1 \le i \le n 1≤i≤n) and increase a i a_i ai by 1 1 1 in one operation.
Your task is to find the minimum number of operations required to increase the median of the array.
Note that the array
a
a
a may not necessarily contain distinct
numbers.
Input
Each test consists of multiple test cases. The first line contains a single integer t t t ( 1 ≤ t ≤ 1 0 4 1 \le t \le 10^4 1≤t≤104) — the number of test cases. Then follows the description of the test cases.
The first line of each test case contains a single integer n n n ( 1 ≤ n ≤ 1 0 5 1 \le n \le 10^5 1≤n≤105) — the length of the array a a a.
The second line of each test case contains n n n integers a 1 , a 2 , … , a n a_1, a_2, \ldots, a_n a1,a2,…,an ( 1 ≤ a i ≤ 1 0 9 1 \le a_i \le 10^9 1≤ai≤109) — the array a a a.
It is guaranteed that the sum of the values of n n n over all test cases does not exceed 2 ⋅ 1 0 5 2 \cdot 10^5 2⋅105.
Output
For each test case, output a single integer — the minimum number of operations required to increase the median of the array.
Example
inputCopy
8
3
2 2 8
4
7 3 3 1
1
1000000000
5
5 5 5 4 5
6
2 1 2 3 1 4
2
1 2
2
1 1
4
5 5 5 5
outputCopy
1
2
1
3
2
1
2
3
Note
In the first test case, you can apply one operation to the first number and obtain the array [ 3 , 2 , 8 ] [3, 2, 8] [3,2,8], the median of this array is 3 3 3, as it is the number at index ⌈ 3 2 ⌉ = 2 \lceil \frac{3}{2} \rceil = 2 ⌈23⌉=2 in the non-decreasing sorted array [ 2 , 3 , 8 ] [2, 3, 8] [2,3,8]. The median of the original array [ 2 , 2 , 8 ] [2, 2, 8] [2,2,8] is 2 2 2, as it is the number at index ⌈ 3 2 ⌉ = 2 \lceil \frac{3}{2} \rceil = 2 ⌈23⌉=2 in the non-decreasing sorted array [ 2 , 2 , 8 ] [2, 2, 8] [2,2,8]. Thus, the median increased (KaTeX parse error: Expected 'EOF', got '&' at position 3: 3 &̲gt; 2) in just one operation.
In the fourth test case, you can apply one operation to each of the numbers at indices 1 , 2 , 3 1, 2, 3 1,2,3 and obtain the array [ 6 , 6 , 6 , 4 , 5 ] [6, 6, 6, 4, 5] [6,6,6,4,5], the median of this array is 6 6 6, as it is the number at index ⌈ 5 2 ⌉ = 3 \lceil \frac{5}{2} \rceil = 3 ⌈25⌉=3 in the non-decreasing sorted array [ 4 , 5 , 6 , 6 , 6 ] [4, 5, 6, 6, 6] [4,5,6,6,6]. The median of the original array [ 5 , 5 , 5 , 4 , 5 ] [5, 5, 5, 4, 5] [5,5,5,4,5] is 5 5 5, as it is the number at index ⌈ 5 2 ⌉ = 2 \lceil \frac{5}{2} \rceil = 2 ⌈25⌉=2 in the non-decreasing sorted array [ 4 , 5 , 5 , 5 , 5 ] [4, 5, 5, 5, 5] [4,5,5,5,5]. Thus, the median increased (KaTeX parse error: Expected 'EOF', got '&' at position 3: 6 &̲gt; 5) in three operations. It can be shown that this is the minimum possible number of operations.
In the fifth test case, you can apply one operation to each of the numbers at indices 1 , 3 1, 3 1,3 and obtain the array [ 3 , 1 , 3 , 3 , 1 , 4 ] [3, 1, 3, 3, 1, 4] [3,1,3,3,1,4], the median of this array is 3 3 3, as it is the number at index ⌈ 6 2 ⌉ = 3 \lceil \frac{6}{2} \rceil = 3 ⌈26⌉=3 in the non-decreasing sorted array [ 1 , 1 , 3 , 3 , 3 , 4 ] [1, 1, 3, 3, 3, 4] [1,1,3,3,3,4]. The median of the original array [ 2 , 1 , 2 , 3 , 1 , 4 ] [2, 1, 2, 3, 1, 4] [2,1,2,3,1,4] is 2 2 2, as it is the number at index ⌈ 6 2 ⌉ = 3 \lceil \frac{6}{2} \rceil = 3 ⌈26⌉=3 in the non-decreasing sorted array [ 1 , 1 , 2 , 2 , 3 , 4 ] [1, 1, 2, 2, 3, 4] [1,1,2,2,3,4]. Thus, the median increased (KaTeX parse error: Expected 'EOF', got '&' at position 3: 3 &̲gt; 2) in two operations. It can be shown that this is the minimum possible number of operations.
AC代码：
#include<map>
#include<set>
#include<stack>
#include<cmath>
#include<queue>
#include<string>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<numeric>
#include<iomanip>
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int, int>PII;
const int N=3e5+10;
const int MOD=998244353;
const int INF=0X3F3F3F3F;
const int dx[]={-1,1,0,0,-1,-1,+1,+1};
const int dy[]={0,0,-1,1,-1,+1,-1,+1};
const int M = 1e6 + 10;
int t;
int n;
int a[N];
int main()
{
cin >> t;
while(t --){
int n;
cin >> n;
for(int i = 1; i <= n; i++)
{
cin >> a[i];
}
sort(a + 1, a + 1 + n);
if(n & 1)
{
int cnt = 1;
for(int i = (n + 1) / 2; i <= n; i ++)
{
if(a[i] == a[i + 1] && i != n) cnt ++ ;
else break;
}
cout << cnt << endl;
}
else {
int cnt = 1;
for(int i = n / 2; i <= n; i ++)
{
if(a[i] == a[i + 1] && i != n) cnt ++;
else break;
}
cout << cnt << endl;
}
}
return 0;
}