Let's create a window of size $$$k + 1$$$ that ends (inclusively) at the last position ($$$i = n$$$). Now, notice that an element can be at the last position if and only if it's inside that window. Therefore, we're forced to choose an element from that window and place it at that position.

Next, let's move back one position, so now we're at position $$$i = n - 1$$$. At this point, the elements we can place at the penultimate position ($$$i$$$) are all of the elements from the previous window except for the element we were forced to put at the $$$n$$$-th position plus the element at position $$$i - k = n - 1 - k$$$ (if this position doesn't exist, we can simply assume it's $$$0$$$). And once again, we are forced to choose one of the elements from this alternate window.

Inductively, this pattern continues. Thus, the solution consists of iterating from the last position to the first with a starting window based on a multiset of size $$$k$$$ of the last elements. Then, at each iteration you extend the window by the element at position $$$i - k$$$ and choose either the largest element of the multiset or the smallest depending on whether the position when indexed by one is divisible by $$$x + 1$$$ or not. This chosen element is removed from the multiset but is also added to the answer (well, divided by two of course).

#include <bits/stdc++.h>

using namespace std;
using ll = int64_t;

void solv() {
	int n, x, k; cin >> n >> x >> k;
	vector<ll> a(n); for (auto &i : a) cin >> i;
	ll ans = 0;
	multiset<ll> b(a.begin() + n - k, a.end());
	for (int i = n; i--;) {
		b.insert(i - k < 0 ? 0 : a[i - k]);
		if ((i + 1)%(x + 1) == 0) {
			ans += *--b.end()/2;
			b.erase(--b.end());
		} else b.erase(b.begin());
	}
	cout << ans << '\n';
}

int main() {
	return ios_base::sync_with_stdio(0), cin.tie(0), solv(), 0;
}

There are a number of problems akin to this one where you find some fixed perspective that always forces you to make a move but that you are free to choose which move it is. There was a really nice one from the 2025 OCPC Winter training camp, which I'll add to this comment if they give me permission.

Anyways, hope this comment helped :)