제출 #673685

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
673685		stanislavpolyn	A Difficult(y) Choice (BOI21_books)	C++17	100 / 100	2 ms	1104 KiB

books

#include <bits/stdc++.h>

#include "books.h"

#define fr(i, a, b) for (int i = (a); i <= (b); i++)
#define rf(i, a, b) for (int i = (a); i >= (b); i--)
#define fe(x, y) for (auto& x : y)

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define pw(x) (1LL << (x))
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()

using namespace std;

mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define fbo find_by_order
#define ook order_of_key

template <typename T>
bool umn(T& a, T b) {
    return a > b ? a = b, 1 : 0;
}
template <typename T>
bool umx(T& a, T b) {
    return a < b ? a = b, 1 : 0;
}

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
using ve = vector<T>;

const int N = 1e5 + 5;

ll what[N];

ll get(int i) {
    if (what[i]) {
        return what[i];
    } else {
        return what[i] = skim(i);
    }
}

void solve(int N, int K, long long A, int S) {
    fill(what + 1, what + 1 + N, 0);

    int l = 1;
    int r = N;
    while (l < r) {
        int mid = l + ((r - l) >> 1);

        if (get(mid) >= A) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    {
        ll sum = 0;
        sum += get(l);

        if (sum < A) l++;

        if (sum >= A && l > K - 1) {
            fr (i, 1, K - 1) sum += get(i);

            if (sum <= 2 * A) {
                ve<int> ans;
                fr (i, 1, K - 1) ans.pb(i);
                ans.pb(l);
                answer(ans);
            }
        }
    }


    fr (i, 1, K) {
        what[i] = get(i);
    }

    fr (i, max(1, l - K), l - 1) {
        what[i] = get(i);
    }

    ve<pll> v;
    fr (i, 1, N) {
        if (what[i]) {
            v.pb({what[i], i});
        }
    }

    fr (i, 0, sz(v) - K) {
        ll sum = 0;
        fr (j, i, i + K - 1) {
            sum += v[j].fi;
        }
        if (sum >= A && sum <= 2 * A) {
            ve<int> ans;
            fr (j, i, i + K - 1) {
                ans.pb(v[j].se);
            }
            answer(ans);
        }
    }

    impossible();
}

• Subtask 1: 5 / 5
• Subtask 2: 15 / 15
• Subtask 3: 10 / 10
• Subtask 4: 15 / 15
• Subtask 5: 15 / 15
• Subtask 6: 20 / 20
• Subtask 7: 20 / 20