题目
A median of an array of integers of length n is the number standing on the ⌈n2⌉ (rounding up) position in the non-decreasing ordering of its elements. Positions are numbered starting with 1. For example, a median of the array [2,6,4,1,3,5] is equal to 3. There exist some other definitions of the median, but in this problem, we will use the described one.Given two integers n and k and non-decreasing array of nk integers. Divide all numbers into k arrays of size n, such that each number belongs to exactly one array.
You want the sum of medians of all k arrays to be the maximum possible. Find this maximum possible sum.
Input
The first line contains a single integer t (1≤t≤100) — the number of test cases. The next 2t lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, k (1≤n,k≤1000).
The second line of the description of each test case contains nk integers a1,a2,…,ank (0≤ai≤109) — given array. It is guaranteed that the array is non-decreasing: a1≤a2≤…≤ank.
It is guaranteed that the sum of nk for all test cases does not exceed 2⋅105.
Output
For each test case print a single integer — the maximum possible sum of medians of all k arrays.
思考过程
通过思考，可以很轻易地想到：先每一行地判断每一个数周围的数是否相等，若相等则对某一个数做出某些改变，而后再每一列地进行相同地判断，但稍微仔细一想，这样的方式并不能保证每个数至多+1，因此我们需要换一种思路。
基于每个数最多+1的条件，我们可以将奇偶判断代入到这个题里面，只要保证奇数的周围全是偶数，偶数的周围全是奇数便可。
代码如下
代码如下（示例）：
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
typedef long long LL;
int test=1;
int n,k;
int a[N];
LL res=0;
int main()
{
scanf("%d",&test);
while(test--)
{
res=0;
cin >> n >> k;
for(int i=1;i<=n*k;i++)
scanf("%d",&a[i]);
for(int i=k*n-n/2;i>(n-1)/2*k;i-=n/2+1)
res += a[i];
cout << res << '\n';
}
return 0;
}