deque<int> dq;
    for (int i = 1; i <= n; i++) {
        dq.push_back(i);
    }
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
        int rem = n - i;
        if (k >= rem) {
            ans.push_back(dq.back());
            dq.pop_back();
            k -= n - i;
        } else {
            ans.push_back(dq.front());
            dq.pop_front();
        }
    }

in case of possible events for example if n = 10 try printing the last k elements first then the last k elements before that and so on.

    int x = (n + k - 1) / k;
    if (x > k) {
        cout << "IMPOSSIBLE\n";
        return;
    }
    vector<int> ans(n);
    int j = 0;
    int p = n;
    while (x--) {
        for (int i = max(1, p - k + 1); i <= p; i++) {
            ans[j++] = i;
        }
        p -= k;
    }