제출 #1224477

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1224477		SpyrosAliv	A Difficult(y) Choice (BOI21_books)	C++20	100 / 100	0 ms	408 KiB

books

#include "books.h"
#include <bits/stdc++.h>
using namespace std;
#define ll long long

int n, k, s;
ll a;

void solve(int N, int K, ll A, int S) {
    n = N;
    k = K;
    a = A;
    s = S;
    // binary search for first position of A
    // either use that and all k-1 smallest indices, or only use elements less than a
    // when using elements less than a, if their sum is less than a, just move the latst index to the back (will be less than a)
    int idx = -1;
    int lo = 1, hi = n;
    while (lo <= hi) {
        int mid = (lo + hi) / 2;
        ll val = skim(mid);
        if (val >= a) {
            hi = mid-1;
        }
        else {
            lo = mid+1;
            idx = mid;
        }
    }
    if (idx == -1) { // k >= 3
        impossible();
        return;
    }
    if (k - idx >= 2) { // a + a + 1
        impossible();
        return;
    }
    // try k-1 first vals and value >= a
    vector<ll> vals(k+1);
    ll tot = 0;
    for (int i = 1; i <= k; i++) {
        vals[i] = skim(i);
        tot += vals[i];
    }
    if (tot > 2 * a) {
        impossible();
        return;
    }
    if (idx + 1 <= n) {
        ll nextVal = skim(idx+1);
        tot -= vals[k];
        tot += nextVal;
        if (tot >= a && tot <= 2 * a) {
            vector<int> fin;
            for (int i = 1; i <= k-1; i++) fin.push_back(i);
            fin.push_back(idx+1);
            answer(fin);
            return;
        }
        tot -= nextVal;
        tot += vals[k];
    }
    if (k >= idx) {
        impossible();
        return;
    }
    int bound = idx;
    vector<int> index(k+1);
    for (int i = 1; i <= k; i++) index[i] = i;
    for (int i = k; i >= 1; i--) {
        if (tot >= a && tot <= 2*a) {
            vector<int> fin;
            for (int i = 1; i <= k; i++) fin.push_back(index[i]);
            answer(fin);
            return;
        }
        if (i == bound) {
            impossible();
            return;
        }
        ll nxtVal = skim(bound);
        tot += nxtVal;
        tot -= vals[i];
        index[i] = bound;
        bound--;
    }
    if (tot >= a && tot <= 2*a) {
        vector<int> fin;
        for (int i = 1; i <= k; i++) fin.push_back(index[i]);
        answer(fin);
        return;
    }
    impossible();
    /*
    vector<ll> arr(n+1);
    for (int i = 1; i <= k; i++) {
        arr[i] = skim(i);
    }
    ll currS = 0;
    vector<int> index(k+2);
    for (int i = 1; i <= k; i++) {
        currS += arr[i];
        index[i] = i;
    }
    index[k+1] = n+1;
    for (int curr = k; curr >= 1; curr--) {
        if (currS >= a && currS <= 2*a) break;
        int allow = index[curr+1] - 1;
        if (allow == curr) continue;
        int lo = curr+1, hi = allow;
        int idx = -1;
        while (lo <= hi) {
            int mid = (lo + hi) / 2;
            ll val = skim(mid);
            ll tot = currS - arr[curr] + val;
            if (tot >= a && tot <= 2 * a) {
                index[curr] = mid;
                vector<int> fin;
                for (int i = 1; i <= k; i++) {
                    fin.push_back(index[i]);
                }
                answer(fin);
                return;
            }
            if (tot > 2 * a) {
                hi = mid-1;
            }
            else {
                idx = mid;
                lo = mid+1;
            }
        }
        if (idx == -1) {
            impossible();
            return;
        }
        currS -= arr[curr];
        currS += arr[idx];
        index[curr] = idx;
    }
    if (currS > 2 * a || currS < a) {
        impossible();
        return;
    }
    else {
        vector<int> fin;
        for (int i = 1; i <= k; i++) {
            fin.push_back(index[i]);
        }
        answer(fin);
        return;
    }*/
}

• 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