1 solutions

  • 1
    @ 2024-11-24 23:55:58
    #include <iostream>
    using namespace std;
    typedef long long LL;
    const int N = 100010;
    LL s[N];
    LL n, k;
    LL res;
    int main()
    {
        cin >> n >> k;
        for(int i = 1; i <= n; ++ i)
        {
            cin >> s[i];
            s[i] += s[i - 1];
        }
        for(int i = k; i <= n; ++ i)
        {
            res = max(res, s[i] - s[i - k]);//所有长度为k的区间
        }
        cout << res << '\n';
        return 0;
    }
    
    • 1

    Information

    ID
    5515
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    8
    Tags
    # Submissions
    34
    Accepted
    5
    Uploaded By