| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 556941 | kym | A Difficult(y) Choice (BOI21_books) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#include "books.h"
#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;
using namespace std;
#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;
}
cerr << hi << '\n';
vector<pi> v;
FOR(i,hi-k,hi+k) {
if(i<=0||i>n)continue;
v.pb(pi(skim(i),i));
}
FOR(i,0,(1<<(v.size()))-1) {
if(__builtin_popcount(i) != k)continue;
int tot = 0;
vector<int> res;
FOR(j,0,v.size()-1) {
if(i&(1<<j)) {
tot += v[j].f;
res.pb(v[j].s);
}
}
dbv(res);
db(tot);
if(tot >= a && tot <= 2*a){
answer(res);
//exit(0);
}
}
impossible();
}