제출 #581423

#제출 시각아이디문제언어결과실행 시간메모리
581423islingrHedgehog Daniyar and Algorithms (IZhO19_sortbooks)C++17
100 / 100
1251 ms137376 KiB
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

#include "bits/stdc++.h"

using namespace std;
using ll = long long;
using ld = long double;

#define rep(i, a, b) for (auto i{a}; i < (b); ++i)
#define per(i, a, b) for (auto i{b}; i-- > (a);)
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) static_cast<int>((x).size())

template <class T>
bool uin(T& a, const T& b) {
    return a > b ? a = b, true : false;
}
template <class T>
bool uax(T& a, const T& b) {
    return a < b ? a = b, true : false;
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

struct segment_tree {
    static constexpr int inf = 1e9;
    int n;
    vector<int> t;
    segment_tree() {}
    segment_tree(int n) : n{n}, t(2 * n, -inf) {}
    int query(int l, int r) {
        int res = -inf;
        for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
            if (l & 1) res = max(res, t[l++]);
            if (r & 1) res = max(res, t[--r]);
        }
        return res;
    }
    void update(int p, int x) {
        for (t[p += n] = x; p >>= 1;) t[p] = max(t[p << 1], t[p << 1 | 1]);
    }
};

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (auto& x : a) cin >> x;

    vector<int> p(n);  // p[i] = last (strictly) greater element than a[i]
    rep(i, 0, n) {
        p[i] = i - 1;
        while (p[i] >= 0 && a[i] >= a[p[i]]) p[i] = p[p[i]];
    }

    segment_tree t(n);
    vector<vector<int>> g(n);
    rep(i, 0, n) {
        if (p[i] < 0) continue;
        t.update(i, a[i] + a[p[i]]);
        g[p[i]].push_back(i);
    }

    vector<vector<tuple<int, int, int>>> queries(n);
    rep(i, 0, m) {
        int l, r, k;
        cin >> l >> r >> k;
        queries[l - 1].emplace_back(r, k, i);
    }

    vector<bool> ans(m);
    rep(l, 0, n) {
        for (auto [r, k, i] : queries[l]) ans[i] = t.query(l, r) <= k;
        for (auto c : g[l]) t.update(c, 0);
    }
    rep(i, 0, m) cout << ans[i] << '\n';
}
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