| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 416710 | jainbot27 | A Difficult(y) Choice (BOI21_books) | C++17 | 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.
// Difficult(y) Choice
#include <bits/stdc++.h>
#include "books.h"
using namespace std;
using ll=long long;
#define FOR(i, a, b) for(int i=a; i<b; i++)
#define ROF(i, a, b) for(int i=b-1; i>=a; i--)
#define F0R(i, n) FOR(i, 0, n)
#define R0F(i, n) ROF(i, 0, n)
#define f first
#define s second
#define pb push_back
#define siz(x) (int)x.size()
int n, k, a, s;
void solve(int N, int K, ll A, int S){
n=N, k=K, a=A, s=S;
assert(n==s);
// task is query 40 books and try to see if we can get a set of K books such that A<=\sum<=2A we have K is less than 10 so SMALL
// solve subtask by subtask:
/*
observation, problem has to require BINARY SEARCH
subtask 1: we can just like iterte over first 2 books, check which if we have a third one
subtask 2: if the last element we take is <=A then if the previous elements are >A, we can find a solution
*/
vector<ll> x(n);
F0R(i, n){
x[i]=skim(i+1);
}
{
ll p=0;
F0R(i, k-1) p+=x[i];
FOR(i, k-1, n){
if(p+x[i]>=A&&2*A<=p+x[i]){
vector<int> ans;
F0R(i, k-1) ans.pb(i+1);
ans.pb(i+1);
answer(ans);
}
}
}
ll sum=0;
F0R(i, k-1) sum+=x[i];
FOR(i, k-1, n){
sum+=x[i];
if(sum>=a&&sum<=2*a){
vector<int> ans;
FOR(j, i-k+1, i+1) ans.pb(j+1);
answer(ans);
}
sum-=x[i-k+1];
}
impossible();
}
int main(){
cin.tie(0)->sync_with_stdio(0);
}