RT，最坏能被卡到 $\mathcal{O}(n^2)$ 的代码过了。

实际上这份代码是正解用堆维护改成暴力求 $\min$ 的产物，代码如下：

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
    cin.tie(nullptr) -> sync_with_stdio(false);
    int n; i64 m;
    cin >> n >> m;
    vector <i64> a(n + 1), f(n + 1, 1e18);
    for (int i = 1; i <= n; i ++) {
        cin >> a[i];
        if (a[i] > m) return cout << -1 << endl, 0;
    }
    deque <i64> q;
    f[0] = 0;
    i64 sum = 0;
    for (int i = 1, j = 1; i <= n; i ++) {
        sum += a[i];
        while (!q.empty() && a[q.back()] < a[i]) q.pop_back();
        q.push_back(i);
        for (; sum > m; j ++) {
            sum -= a[j];
            if (q.front() == j) q.pop_front();
        }
        int r = j - 1;
        for (auto &idx : q) {
            f[i] = min(f[i], f[r] + a[idx]);
            r = idx;
        }
    }
    return cout << f[n] << endl, 0;
}

下面这份代码可以生成 hack：

#include <bits/stdc++.h>
using std::cin;
using std::cout;
static const unsigned RS = 
    std::chrono::steady_clock::now().time_since_epoch().count();
int main() {
    std::mt19937_64 gen_64(RS);
    int n; cin >> n;
    int64_t m = 1e11, lim = m / n;
    cout << n << ' ' << m << std::endl;
    std::uniform_int_distribution <int64_t> dis(0, lim);
    std::vector <int64_t> a(n);
    for (int i = 0; i < n; i ++) a[i] = dis(gen_64);
    std::sort (a.begin(), a.end(), std::greater <int64_t> ());
    for (int i = 0; i + 1 < n; i ++) cout << a[i] << ' ';
    cout << gen_64() % (int)m << std::endl;
    return 0;
}