害怕，帖被删了、、、

vp 时调了 $90 \min$ 一直 WA，现在又 WA#11，有没有做过的大佬救救菜鸡啊！

具体思想就是二进制分组然后背包一下。

#include <bits/stdc++.h>
#define i64 long long
#define sz(a) int((a).size())
#define all(a) (a).begin(), (a).end()
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
#define per(i, a, b) for (int i = (b) - 1; i >= (a); --i)
using namespace std;

const int N = 2e5 + 1, Nk = 20;
const i64 B = 1e5;
int nq, n, a[N], fa[Nk][N], lg[N];
bitset<N> f;

vector<int> split(int x) {
	vector<int> s;
	for (int p = 1; p <= x; p <<= 1) {
		s.push_back(p), x -= p;
	}
	x ? s.push_back(x) : void();
	return s;
}

int main() {
	// freopen("D.in", "r", stdin);
	// freopen("D.out", "w", stdout);
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cin >> nq >> n, n += 1;
	rep(i, 1, n) {
		a[i] = -1, lg[i] = __builtin_ctz(i);
	}
	f[0] = true, a[0] = 0;
	rep(i, 0, nq) {
		i64 x;
		int o, y;
		cin >> o >> x >> y;
		vector<int> s = split(y);
		if (o == 1) {
			x = (x + B - 1) / B;
			for (const int &u : s) {
				f |= f << x * u;
			}
		} else if (o == 2) {
			per(j, 0, n) {
				fa[0][j] = min<i64>((x * j + B - 1) / B, n);
				rep(k, 0, Nk - 1) {
					if (fa[k][j] >= n) {
						break;
					}
					fa[k + 1][j] = fa[k][fa[k][j]];
				}
			}
 			for (const int &u : s) {
				per(j, 0, n) {
					int k = j, v = u;
					while (k < n and v) {
						k = fa[lg[v & -v]][k];
						v ^= v & -v;
					}
					k < n and f[j] ? f[k] = true : 0;
				}
			}
		}
		rep(j, 0, n) {
			if (a[j] == -1 and f[j]) {
				a[j] = i + 1; 
			}
		}
	}
	rep(i, 1, n) {
		cout << a[i] << ' ';
	}
	cout << '\n';
	return 0;
}