1 solutions

  • 0
    @ 2026-8-11 19:22:34
    #include<bits/stdc++.h>
    using namespace std;
    const int N=1e6+10;
    int q[N];
    int a[N];
    int main()
    {
        int n,k;
        cin>>n>>k;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        int hh=0,tt=-1;
        for(int i=1;i<=n;i++)
        {
            if(hh<=tt&&i-q[hh]+1>k) hh++; //超过窗口个数
            while(hh<=tt&&a[q[tt]]>=a[i]) tt--; //当前元素更小
            q[++tt]=i; //入队
            if(i>=k) cout<<a[q[hh]]<<" "; //最小值
        }
        cout<<endl;
        hh=0,tt=-1; //同理计算最大值
        for(int i=1;i<=n;i++)
        {
            if(hh<=tt&&i-q[hh]+1>k) hh++;
            while(hh<=tt&&a[q[tt]]<=a[i]) tt--;
            q[++tt]=i;
            if(i>=k) cout<<a[q[hh]]<<" ";
        }
        return 0;
    }

    Information

    ID
    337
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    5
    Tags
    (None)
    # Submissions
    16
    Accepted
    7
    Uploaded By