제출 #556980

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
556980		kym	A Difficult(y) Choice (BOI21_books)	C++14	0 / 100	40 ms	276 KiB

books

#include <bits/stdc++.h>
using namespace std;
#include "books.h"
#define all(x) (x).begin(), (x).end()
#define ll long long
#define s second 
#define f first 
#define pb push_back
#define FOR(i,s,e) for(ll i = s; i <= (ll)e; ++i)
typedef pair <long long ,int> pi;
#ifdef LOCAL
#define db(x) cerr << #x << "=" << x << "\n"
#define db2(x, y) cerr << #x << "=" << x << " , " << #y << "=" << y << "\n"
#define db3(a,b,c) cerr<<#a<<"="<<a<<","<<#b<<"="<<b<<","<<#c<<"="<<c<<"\n"
#define dbv(v) cerr << #v << ":"; for (auto ite : v) cerr << ite << ' '; cerr <<"\n"
#define dbvp(v) cerr << #v << ":"; for (auto ite : v) cerr << "{"  << ite.f << ',' << ite.s << "} "; cerr << "\n"
#define dba(a,ss,ee) cerr << #a << ":"; FOR(ite,ss,ee) cerr << a[ite] << ' '; cerr << "\n"
#define reach cerr << "LINE: " << __LINE__ << "\n";
#else
#define db(x)
#define db2(x,y)
#define db3(a,b,c)
#define dbv(v)
#define dbvp(v)
#define dba(a,ss,ee)
#define reach
#endif
//
// --- 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.
//


/*
SAMPLE GRADER for task BOOKS

USAGE:
place together with your solution and books.h in the same directory, then:
g++ <flags> sample_grader.cpp <solution_file>
e.g.:
g++ -std=c++17 sample_grader.cpp books.cpp

INPUT/OUTPUT:
The sample grader expects on standard input two lines. The first line should
contain the four integers N, K, A and S. The second line should contain a list
of N integers, the sequence of difficulties x_1 x_2 ... x_N which has to be
strictly increasing. Then, the grader writes to standard output a protocol of
all grader functions called by your program.
At the end, the grader prints your verdict.
*/
map<int,int> memo;
int query(int x) {
	if(memo.find(x) != memo.end())return memo[x];
	else return memo[x] = skim(x);
}
void solve(int n, int k, long long a, int s) {
    ll lo = 0, hi = n+1;
    while(lo<hi-1) {
    	ll mid = (lo+hi)/2;
    	ll x = query(mid);
    	if(x >= a)hi=mid;
    	else lo=mid;
    }
    vector<pi> v;
    ll r1 = 0;
    FOR(i,1,k) {
    	int x = query(i);
    	r1 += x;
    	v.pb(pi(x,i));
    }
    reach
    if(r1 > 2*a)impossible();
    if(r1 >= a){
    	vector<int> ans;
    	FOR(i,1,k)ans.pb(i);
    	answer(ans);
    }
    //r1<a;
    if(hi <= n &&  (hi>=k)&& query(hi) >= a && query(hi) <= 2*a) {
    	ll tot = 0;
    	FOR(i,1,k-1) tot+=query(i);
    	tot+=query(hi);
    	if(tot <= 2*a) {
    		vector<int> ans;
    		FOR(i,1,k-1)ans.pb(i);
    		ans.pb(hi);
    		answer(ans);
    	}
    }
    //we don't use hi
    FOR(i,hi-k,hi-1) {
    	if(i>0ll && i<=n)v.pb(pi(skim(i),i));
    }
    sort(all(v));
    v.resize(unique(all(v)) - v.begin());
    //assert(hi>=10 && hi<=n-10);
    FOR(i,0,(1<<(v.size()))-1) {
    	if(__builtin_popcount(i) != k)continue;
    	ll tot = 0;
    	vector<int> res;
    	FOR(j,0,v.size()-1) {
    		if(i&(1<<j)) {
    			tot += v[j].f;
    			res.pb(v[j].s);
    		}
    	}
    	if(tot >= a && tot <= 2*a){
    		answer(res);
    	}
    }
    impossible();
    
}

• 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