https://codeforces.com/contest/1440/problem/B
Question: Divide the sequence, find the sum of the median in each block, and construct the largest one.
Idea: When n=2, n/2=1, greedy takes 2 from the front to the back, and does not waste the large numbers behind.
When n>2, for example, n=3. Take 1 at the front and 2 at the back, the most greedy.
When n=4, n/2=2; so there is 1 in the front and 3 in the back. The median is on the big side.
When n/5=, n/2=3, 2 in the front and 3 in the back. The median is on the big side.
Then sweep forward after O(n)
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e6+100;
typedef long long LL;
LL a[maxn];
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
LL t;cin>>t;
while(t--){
LL n,k;cin>>n>>k;
LL sum=0;LL ans=0;
for(LL i=1;i<=n*k;i++) cin>>a[i];
if(n==2){
for(LL i=1;i<=n*k;i+=n){
if(ans>=k){
break;
}
sum+=a[i];ans++;
}
cout<<sum<<endl;
}
else{
LL st=0;
if(n%2==0) st=n/2+1;
else st=ceil(1.0*n/2.0);
for(LL i=n*k-(st-1);i>=1;i-=st){
if(ans>=k){
break;
}
/// cout<<a[i]<<" ";
sum+=a[i];ans++;
}
cout<<sum<<endl;
}
}
return 0;
}