| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 867963 | Lib | Rarest Insects (IOI22_insects) | 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.
#include <bits/stdc++.h>
#include "insects.h"
using namespace std;
deque <int> PrevIndex;
int InMachine[2003];
int n;
int TypeCount;
int PossibleMincount(int cnt){
PrevIndex.clear();
//Pushing insects inside the machine, keeping track of the ones being putted inside for this iteration of binary searching
for(int i=0;i<n;i++){
if(!InMachine[i]){
InMachine[i]=1;
move_inside(i);
PrevIndex.push_back(i);
if(press_button()>cnt){
move_outside(i);
InMachine[i]=0;
PrevIndex.pop_back();
}
}
}
int CurrentInMachine=0;
for(int i=0;i<n;i++){
CurrentInMachine+=InMachine[i];
}
if(CurrentInMachine<cnt*TypeCount){
//There clearly exist a type of insect with cardinality less than cnt, so the answer is less than cnt. Removing insects from last iteration.
while(!PrevIndex.empty()){
InMachine[PrevIndex.top()]=0;
move_outside(PrevIndex.top());
PrevIndex.pop_front();
}
return 0;
}else{
//As every type of insects has at least cnt insects, the answer is at least cnt. Just return true
return 1;
}
}
int min_cardinality(int N){
int l,r,mid;
n=N;
l=1;
//Phase 1: Determining the amount of different types of insects
for(int i=0;i<n;i++){
move_inside(i);
InMachine[i]=1;
PrevIndex.push_back(i);
if(press_button()>1){
move_outside(i);
InMachine[i]=0;
PrevIndex.pop_back();
}
}
TypeCount=PrevIndex.size();
r=N/TypeCount;
//Phase 2: Binary search for the answer
while(r-l>=1){
mid=(l+r)/2;
if(PossibleMincount(mid)){
l=mid;
}else{
r=mid-1;
}
}
if(PossibleMincount(l)){
return l;
}else{
return r;
}
}