数据较水，直接用线段树暴算即可

code:

#include <iostream>
using namespace std;
const long long maxn=4*1e6+5;
long long val1[maxn],val2[maxn];
void update1(long long u,long long l,long long r,long long s,long long t,long long x) {
	if(s<=l&&r<=t) {
		val1[u]=x;
		return;
	}
	long long mid=(l+r)/2;
	if(s<=mid)update1(u*2,l,mid,s,t,x);
	if(t>mid)update1(u*2+1,mid+1,r,s,t,x);
	val1[u]=max(val1[u*2],val1[u*2+1]);
}
long long query1(long long u,long long l,long long r,long long s,long long t) {
	if(s<=l&&r<=t) {
		return val1[u];
	}
	long long mid=(l+r)/2;
	if(t<=mid)return query1(u*2,l,mid,s,t);
	if(s>mid)return query1(u*2+1,mid+1,r,s,t);
	return max(query1(u*2,l,mid,s,t),query1(u*2+1,mid+1,r,s,t));
}
void update2(long long u,long long l,long long r,long long s,long long t,long long x) {
	if(s<=l&&r<=t) {
		val2[u]=x;
		return;
	}
	long long mid=(l+r)/2;
	if(s<=mid)update2(u*2,l,mid,s,t,x);
	if(t>mid)update2(u*2+1,mid+1,r,s,t,x);
	val2[u]=min(val2[u*2],val2[u*2+1]);
}
long long query2(long long u,long long l,long long r,long long s,long long t) {
	if(s<=l&&r<=t) {
		return val2[u];
	}
	long long mid=(l+r)/2;
	if(t<=mid)return query2(u*2,l,mid,s,t);
	if(s>mid)return query2(u*2+1,mid+1,r,s,t);
	return min(query2(u*2,l,mid,s,t),query2(u*2+1,mid+1,r,s,t));
}
int main() {
	long long n,k;
	cin>>n>>k;
	for(long long i=1; i<=n; i++) {
		long long x;
		cin>>x;
		update1(1,1,n,i,i,x);
		update2(1,1,n,i,i,x);
	}
	for(int i=1;i<=n-k+1;i++){
		cout<<query2(1,1,n,i,i+k-1)<<" ";
	}
	cout<<endl;
	for(int i=1;i<=n-k+1;i++){
		cout<<query1(1,1,n,i,i+k-1)<<" ";
	}
}