제출 #1304918

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1304918		Nonoze	Hedgehog Daniyar and Algorithms (IZhO19_sortbooks)	C++20	0 / 100	116 ms	105300 KiB

sortbooks

/*
*	Author: Nonoze
*	Created: Tuesday 23/12/2025
*/
#include <bits/stdc++.h>
using namespace std;

#ifndef DEBUG
	#define dbg(...)
#endif

// #define cout cerr << "OUT: "
#define endl '\n'
#define endlfl '\n' << flush
#define quit(x) return (void)(cout << x << endl)

template<typename T> void read(T& x) { cin >> x; }
template<typename T1, typename T2> void read(pair<T1, T2>& p) { read(p.first), read(p.second); }
template<typename T> void read(vector<T>& v) { for (auto& x : v) read(x); }
template<typename T1, typename T2> void read(T1& x, T2& y) { read(x), read(y); }
template<typename T1, typename T2, typename T3> void read(T1& x, T2& y, T3& z) { read(x), read(y), read(z); }
template<typename T1, typename T2, typename T3, typename T4> void read(T1& x, T2& y, T3& z, T4& zz) { read(x), read(y), read(z), read(zz); }
template<typename T> void print(vector<T>& v) { for (auto& x : v) cout << x << ' '; cout << endl; }

#define sz(x) (int)(x.size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define make_unique(v) sort(all(v)), v.erase(unique(all(v)), (v).end())
#define pb push_back
#define mp(a, b) make_pair(a, b)
#define fi first
#define se second
#define cmin(a, b) a = min(a, b)
#define cmax(a, b) a = max(a, b)
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define QYES quit("YES")
#define QNO quit("NO")

// #define int long long
#define double long double
const int inf = numeric_limits<int>::max() / 4;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9+7, LOG=20;



void solve();

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int tt=1;
	// cin >> tt;
	while(tt--) solve();
	return 0;
}

struct node {
	int mx, res, sz;
	int* vals;
	node() : mx(0), res(0) {}
};


struct segtree {
	int n;
	vector<node> t;
	segtree(int sz) : n(sz) {
		t.resize(4*n);
	}
	void build(int v, int tl, int tr, vector<int>& a) {
		t[v].vals = new int[tr - tl + 1], t[v].sz = tr - tl + 1;
		if (tl == tr) {
			t[v].vals[0] = a[tl];
			t[v].mx = a[tl];
		} else {
			int tm = (tl + tr) / 2;
			build(v*2, tl, tm, a);
			build(v*2+1, tm+1, tr, a);
			int idxl=0, idxr=0, pos=0;
			while (pos < t[v].sz) {
				if (idxl<t[v*2].sz && idxr<t[v*2+1].sz && t[v*2].vals[idxl] == t[v*2+1].vals[idxr]) {
					t[v].vals[pos++] = t[v*2].vals[idxl++];
					idxr++, t[v].sz--;
					continue;
				}
				if (idxl < t[v*2].sz && (idxr == t[v*2+1].sz || t[v*2].vals[idxl] < t[v*2+1].vals[idxr])) {
					t[v].vals[pos++] = t[v*2].vals[idxl++];
				} else {
					t[v].vals[pos++] = t[v*2+1].vals[idxr++];
				}
			}
			for (int i=0; i<t[v].sz-1; i++) assert(t[v].vals[i]<t[v].vals[i+1]);
			merge_nodes(v);
		}
	}

	pair<int, int> query(int v, int tl, int tr, int l, int r, int mxleft) {
		if (tl > r || tr < l) return {-inf, -1};
		if (l <= tl && tr <= r) {
			if (mxleft!=-1) {
				int pos=lower_bound(t[v].vals, t[v].vals + t[v].sz, mxleft) - t[v].vals;
				return {max(t[v].res, mxleft + (pos!=0?t[v].vals[pos-1]:-inf)), t[v].mx};
			} else {
				return {t[v].res, t[v].mx};
			}
		}
		int tm = (tl + tr) / 2;
		pair<int, int> left_res = query(v*2, tl, tm, l, r, mxleft);
		pair<int, int> right_res = query(v*2+1, tm+1, tr, l, r, max(mxleft, left_res.second));
		return {max(left_res.first, right_res.first), max(left_res.second, right_res.second)};
	}

	void merge_nodes(int v) {
		t[v].mx = max(t[v*2].mx, t[v*2+1].mx);
		t[v].res = max(t[v*2].res, t[v*2+1].res);
		int pos=lower_bound(t[v*2+1].vals, t[v*2+1].vals + t[v*2+1].sz, t[v*2].mx) - t[v*2+1].vals;
		cmax(t[v].res, t[v*2].mx + (pos!=0?t[v*2+1].vals[pos-1]:-inf));
	}
};

int n, m, q;
vector<int> a;


void solve() {
	read(n, m);
	a.clear(), a.resize(n); read(a);
	segtree st(n); st.build(1, 0, n-1, a);
	for (int i=0; i<m; i++) {
		int l, r, k; read(l, r, k); l--, r--;
		auto res = st.query(1, 0, n-1, l, r, -1);
		cout << (res.first<=k?1:0) << endl;
	}
}

• Subtask 1: 0 / 8
• Subtask 2: 0 / 9
• Subtask 3: 0 / 13
• Subtask 4: 0 / 17
• Subtask 5: 0 / 30
• Subtask 6: 0 / 23