also my solution to q3 is here:

//this is code ~~~~~~~~~~~~~~~~~

include

include

include

using namespace std;

int main() { ios::sync_with_stdio(false); cin.tie(NULL);

int N, K;
cin >> N >> K;

int Q;
cin >> Q;

vector<vector<long long>> grid(N + 1, vector<long long>(N + 1, 0));
vector<vector<long long>> row_prefix(N + 1, vector<long long>(N + 1, 0));
long long max_attractiveness = 0;

for(int q = 0; q < Q; q++) {
    int r, c;
    long long v;
    cin >> r >> c >> v;

    grid[r][c] = v;


    for(int j = 1; j <= N; j++) {
        row_prefix[r][j] = row_prefix[r][j-1] + grid[r][j];
    }


    int i_min = max(1, r - K + 1);
    int i_max = min(N - K + 1, r);
    int j_min = max(1, c - K + 1);
    int j_max = min(N - K + 1, c);

    for(int i = i_min; i <= i_max; i++) {
        for(int j = j_min; j <= j_max; j++) {

            long long sum = 0;
            for(int di = 0; di < K; di++) {
                int row = i + di;
                sum += row_prefix[row][j + K - 1] - row_prefix[row][j - 1];
            }
            max_attractiveness = max(max_attractiveness, sum);
        }
    }

    cout << max_attractiveness << '\n';
}

return 0;

}

~~~~~~~~~~

also, sorry about the messy formatting, i have no idea how to fix them, the three giant include is actually hashtag include iostream, vector, and algorithm. they are seperate, like write the include three times.

yeah that makes sense because its always possible in <= 2 steps so you will fail k = 0 if your solution is <= 3 steps