设 \(f_i\) 表示对于前 \(i\) 个数，\(i\) 选了， \(g_i\) 表示对于前 \(i\) 个数，\(i\) 没选，\(sum_i=\sum_{j=1}^{i}e_j\) 我们可以列出状态转移方程。
\[\begin{aligned}g_i&=\max\{g_{i-1},f_{i-1}\}\\f_i&=\max\limits_{i-k\le j<i}\{g_j+(sum_i-sum_{j})\}\end{aligned}
\]
然后化简一下
\[\begin{aligned}g_i&=\max\{g_{i-1},f_{i-1}\}\\f_i&=\max\limits_{i-k\le j<i}\{g_j-sum_{j}\}+sum_i\end{aligned}
\]
对于 \(g\) 直接做，对于 \(f\) 可以用滑动窗口，单调队列来做。
//#define FILE_INPUT
#include <iomanip>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define int long long
using namespace std;bool Mbe;#define rep(i, a, b) for (int i = a, END##i = b; i <= END##i; i++)
#define per(i, a, b) for (int i = a, END##i = b; i >= END##i; i--)
#define DEBUG(x) cerr << #x << " = " << x << '\n'using LL = long long;
using ULL = unsigned long long;inline LL read() {LL s = 0, fu = 1; char ch = getchar();while (ch < '0' || ch > '9') ch == '-' ? fu = -1 : 0, ch = getchar();while (ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();return s * fu;
}const int Mod = 1e9 + 7;
const int Inf = 0x3f3f3f3f;
const LL InfLL = 0x3f3f3f3f3f3f3f3f;const int N = 1e5 + 10;
int n, k, a[N], f[N], g[N], sum[N], q[N], hh, tt; // 留个下界 0void Init() {
}void Solve() {n = read(), k = read();rep(i, 1, n) a[i] = read(), sum[i] = a[i] + sum[i - 1];rep(i, 1, n) {g[i] = max(g[i - 1], f[i - 1]);while (hh <= tt && i - k > q[hh]) hh++;f[i] = g[q[hh]] - sum[q[hh]] + sum[i];while (hh <= tt && g[q[tt]] - sum[q[tt]] < g[i] - sum[i]) tt--;q[++tt] = i;}rep(i, 1, n) printf("%lld %lld\n", f[i], g[i]);printf("%lld\n", max(f[n], g[n]));
}bool Med;signed main() {#ifdef FILE_INPUTfreopen("input.in", "r", stdin);
#endifint T = 1;
// T = read();while (T--) {Init();Solve();}cerr << "Memory: " << fixed << setprecision(3) << (&Med - &Mbe) / 1048576.0 << " MB\n";cerr << "Time: " << fixed << setprecision(3) << 1e3 * clock() / CLOCKS_PER_SEC << " ms\n";return 0;
}
