Submission #1302690

#	Time	Username	Problem	Language	Result	Execution time	Memory
1302690		WeIlIaN	Hedgehog Daniyar and Algorithms (IZhO19_sortbooks)	C++20	100 / 100	1421 ms	112048 KiB

sortbooks

#include <bits/stdc++.h>
using namespace std;

#define MOD1 1000000007
#define MOD2 998244353
#define fir first
#define sec second

#define pushf push_front
#define pushb push_back
#define popf pop_front
#define popb pop_back
#define mp make_pair
#define all(a) a.begin(), a.end()
#define lbound(v, x) lower_bound(all(v), x) - v.begin()
#define ubound(v, x) upper_bound(all(v), x) - v.begin()
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)

#define FOR1(a) for (int _ = 0; _ < (a); ++_)
#define FOR2(i, a) for (int i = 0; i < (a); ++i)
#define FOR3(i, a, b) for (int i = (a); i < (b); ++i)
#define RFOR1(a) for (int _ = (a)-1; _ >= 0; --_)
#define RFOR2(i, a) for (int i = (a)-1; i >= 0; --i)
#define RFOR3(i, a, b) for (int i = (b)-1; i >= (a); --i)
#define overload3(a, b, c, d, ...) d
// Always choose the fourth argument to call. Hence, which function to call is determined by the number of given arguments.
#define REP(...) overload3(__VA_ARGS__, FOR3, FOR2, FOR1)(__VA_ARGS__)
#define RREP(...) overload3(__VA_ARGS__, RFOR3, RFOR2, RFOR1)(__VA_ARGS__)

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<vb> vvb;
typedef vector<vc> vvc;

typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;
typedef queue<int> qi;
typedef queue<ll> qll;
typedef queue<pii> qpii;
typedef queue<pll> qpll;
typedef deque<int> dqi;
typedef deque<ll> dqll;
typedef deque<pii> dqpii;
typedef deque<pll> dqpll;
typedef priority_queue<int> pqi;
typedef priority_queue<ll> pqll;
typedef priority_queue<pii> pqpii;
typedef priority_queue<pll> pqpll;
typedef priority_queue<int, vi, greater<int> > r_pqi;
typedef priority_queue<ll, vll, greater<ll> > r_pqll;
typedef priority_queue<pii, vpii, greater<pii> > r_pqpii;
typedef priority_queue<pll, vpll, greater<pll> > r_pqpll;

const ll inf = 1e9;

struct node {
	int n;
	vll val, lazy;
	node(int size = 0) : n(size) {
		val.assign(4 * n, 0);
		lazy.assign(4 * n, 0);
	}
	void build(vll &arr, int ind = 1, int l = 0, int r = -1) {
		if(r == -1)
			r = n-1;
        if(l == r) {
            val[ind] = arr[l];
            return;
        }
        int mid = (l + r) >> 1;
        build(arr, 2 * ind, l, mid);
        build(arr, 2 * ind + 1, mid + 1, r);
        val[ind] = val[2 * ind] + val[2 * ind + 1];
    }
	void update(int q, ll v, int ind = 1, int l = 0, int r = -1) {
		if(r == -1)
			r = n-1;
		if(q > r || l > q)
			return;
		if(q <= l && r <= q) {
			val[ind] = v;
			return;
		}
		int mid = (l + r) / 2;
		update(q, v, 2 * ind, l, mid);
		update(q, v, 2 * ind + 1, mid + 1, r);
		val[ind] = max(val[2 * ind], val[2 * ind + 1]);
	}
	ll query(int ql, int qr, int ind = 1, int l = 0, int r = -1) {
		if(r == -1)
			r = n-1;
		if(ql > r || l > qr)
			return 0;
		if(ql <= l && r <= qr)
			return val[ind];
		int mid = (l + r) / 2;
		return max(query(ql, qr, 2 * ind, l, mid), query(ql, qr, 2 * ind + 1, mid + 1, r));
	}
};

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n, m;
	cin>>n>>m;
	vll arr(n+1);
	REP(i, 1, n+1) {
		cin>>arr[i];
	}
	vector<array<int, 4>> qry(m);
	REP(i, m) {
		cin>>qry[i][0]>>qry[i][1]>>qry[i][2];
		qry[i][3] = i;
	}
	sort(all(qry), [](array<int, 4>& a, array<int, 4>& b) {
		return a[1] < b[1];
	});
	vpll st;
	vpll val(n+1);
	REP(i, 1, n+1) {
		while(!st.empty() && st.back().fir <= arr[i]) {
			st.popb();
		}
		val[i] = (st.empty() ? mp(0ll, 0ll) : mp(st.back().fir + arr[i], st.back().sec));
		st.pushb(mp(arr[i], i));
	}
	int r = 0;
	node root(n+1);
	vll ans(m);
	REP(i, m) {
		while(r < qry[i][1]) {
			r++;
			root.update(val[r].sec, val[r].fir);
		}
		ans[qry[i][3]] = (root.query(qry[i][0], qry[i][1]) <= qry[i][2]);
	}
	REP(i, m) {
		cout<<ans[i]<<endl;
	}
}

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