Submission #556957

#	Time	Username	Problem	Language	Result	Execution time	Memory
556957		kym	A Difficult(y) Choice (BOI21_books)	C++14	45 / 100	79 ms	216 KiB

books

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

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 = skim(mid);
    	if(x*k >= a)hi=mid;
    	else lo=mid;
    }
    vector<pi> v;
    hi=max(hi,(ll)k);
    FOR(i,hi-k-1,hi+k+1) {
    	if(i<=0||i>n)continue;
    	v.pb(pi(skim(i),i));
    }
    //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);
    		//exit(0);
    	}
    }
    impossible();
    
}

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