제출 #673675

#제출 시각아이디문제언어결과실행 시간메모리
673675stanislavpolynA Difficult(y) Choice (BOI21_books)C++17
0 / 100
2 ms1180 KiB
#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>;

void solve(int N, int K, long long A, int S) {
    int l = 1;
    int r = N;
    while (l < r) {
        int mid = l + ((r - l) >> 1);

        if (skim(mid) >= A) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    {
        ll sum = 0;
        sum += skim(l);
        if (sum >= A && l > K - 1) {
            ll sum = 0;
            fr (i, 1, K - 1) sum += skim(i);

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

    ve<int> what(N + 1);

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

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

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

    impossible();
}
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