제출 #470303

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
470303		Haruto810198	A Difficult(y) Choice (BOI21_books)	C++17	0 / 100	1 ms	200 KiB

books

#include <bits/stdc++.h>

#include "books.h"

#define ll long long
#define FOR(i, l, r, d) for(int i=(l); i<=(r); i+=(d))
#define vi vector<int>
#define pb push_back

using namespace std;
//
// --- Sample implementation for the task books ---
//
// To compile this program with the sample grader, place:
//     books.h books_sample.cpp sample_grader.cpp
// in a single folder and run:
//     g++ books_sample.cpp sample_grader.cpp
// in this folder.
//

const int MAX = 100010;

ll arr[MAX];
vector<int> res;

int query(int id){
    if(arr[id] == 0){
        arr[id] = skim(id);
    }
    return arr[id];
}

bool solve_for_small_items(int small_cnt, int K, int A){

    vi pref, suf;

    if(small_cnt < K){
        return 0;
    }

    FOR(i, 1, K, 1){
        pref.pb(i);
        suf.pb(small_cnt + 1 - i);
    }

    FOR(i, 0, K, 1){

        int sum = 0;
        res.clear();

        FOR(j, 0, i-1, 1){
            sum += query(pref[j]);
            res.pb(pref[j]);
        }
        FOR(j, 0, K-i-1, 1){
            sum += query(suf[j]);
            res.pb(suf[j]);
        }

        if(A <= sum and sum <= 2*A){
            return 1;
        }

    }

    return 0;

}

void solve(int N, int K, ll A, int Q){

    // find first large item
    int first_large_item;
    int L = 1, R = N, mid;

    if(query(N) < A){
        first_large_item = N + 1;
    }
    else{
        while(L < R){
            mid = (L + R) / 2;
            if(query(mid) >= A) R = mid;
            else L = mid + 1;
        }
        first_large_item = L;
    }

    // all small items
    if(solve_for_small_items(first_large_item - 1, K, A)){
        answer(res);
        return;
    }

    // if no solution for all small items:

    // 1 large item
    if(first_large_item == N + 1){ // no large item
        impossible();
        return;
    }

    if(first_large_item < K){ // not enough small items
        impossible();
        return;
    }

    int sum = 0;
    res.clear();

    FOR(i, 1, K-1, 1){ // solution for 1 large item and K-1 small items
        sum += query(i);
        res.pb(i);
    }
    sum += query(first_large_item);
    res.pb(first_large_item);

    if(A <= sum and sum <= 2*A){
        answer(res);
    }
    else{
        impossible();
    }
    return;

}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 15
• Subtask 3: 0 / 10
• Subtask 4: 0 / 15
• Subtask 5: 0 / 15
• Subtask 6: 0 / 20
• Subtask 7: 0 / 20