学了线段树回来打打暴力（不是）

时间复杂度 $O(nlogn)$， 常数比较大，需要吸一下氧才能AC

#pragma GCC optimize(3)
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1000010;
struct T{
    int l, r, mx, mi;
}tr[N * 4];
int n, k;
int a[N];

inline void pushup(int u) {
    tr[u].mi = min(tr[u << 1].mi, tr[u << 1 | 1].mi);
    tr[u].mx = max(tr[u << 1].mx, tr[u << 1 | 1].mx);
} 

void build(int u, int l, int r) {
    tr[u] = {l, r, a[l], a[l]};
    if(l == r) return;
    else {
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }

}

int querymi(int u, int l, int r) {
    if(tr[u].l >= l && tr[u].r <= r) return tr[u].mi;
    else {
        int mid = tr[u].l + tr[u].r >> 1;
        int v = 1e9;
        if(l <= mid) v = min(v, querymi(u << 1, l, r));
        if(r > mid) v = min(v, querymi(u << 1 | 1, l, r));
        return v;
    }
}

int querymx(int u, int l, int r) {
    if(tr[u].l >= l && tr[u].r <= r) return tr[u].mx;
    else {
        int mid = tr[u].l + tr[u].r >> 1;
        int v = -1e9;
        if(l <= mid) v = max(v, querymx(u << 1, l, r));
        if(r > mid) v = max(v, querymx(u << 1 | 1, l, r));
        return v;
    }
}

int main()
{
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    build(1, 1, n);
    for(int i = 1; i <= n - k + 1; i++) printf("%d%c", querymi(1, i, i + k - 1), " \n"[i == n - k + 1]); 
    for(int i = 1; i <= n - k + 1; i++) printf("%d ", querymx(1, i, i + k - 1)); 
    return 0;
}