| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 235003 | rajarshi_basu | 카멜레온의 사랑 (JOI20_chameleon) | 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.
/*
SOLUTION:
let pile be of form 110 110 (in binary). Then we need to disbalance it (1,1->0,2 and similar)
in all turns which lets us use 1 turn less overall for the heap. Else just remove from the highest pile. For player B, try to keep everything same same, and disbalance only if you have to.
*/
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <fstream>
#include <queue>
#include <deque>
#include <iomanip>
#include <cmath>
#include <set>
#include <stack>
#include <map>
#include <unordered_map>
#define FOR(i,n) for(int i=0;i<n;i++)
#define FORE(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
#define ld long double
//#define int short
#define vi vector<int>
#define pb push_back
#define ff first
#define ss second
#define ii pair<int,int>
#define iii pair<int,ii>
#define pll pair<ll,ll>
#define plll pair<ll,pll>
#define mp make_pair
#define vv vector
#define endl '\n'
#include "chameleon.h"
using namespace std;
bool query(vi all, int extra){
all.pb(extra);
return Query(all) == all.size();
}
int query3(int a,int b,int c){
vi all;all.pb(a);all.pb(b);all.pb(c);
return Query(all);
}
vi res;
void binsrch(vi allNum,int x){
if(allNum.size() == 1){
res.pb(allNum[0]);
return;
}
vi halves[2];
FOR(i,allNum.size())halves[i%2].pb(i);
FOR(j,2)if(query(halves[j],x))binsrch(halves[j],x);
}
vi indsets[4];
void Solve(int n){
indsets[0].pb(1);
FORE(i,2,n){
FOR(j,4){
if(indsets[j].empty() or query(indsets[j],i)){
indsets[j].pb(i);
}
}
}
// everything has been partitioned.
vi edges[n+1];
FOR(i,4){
for(auto e : indsets[i]){
res.clear();
FOR(j,4){
if(i != j){
binsrch(indsets[j],e);
for(auto x:res)edges[e].pb(x);
}
}
}
}
int other[n+1];
set<int> allSets[n+1];
FOR(i,n+1)for(auto x:edges[i])allSets[i].insert(x);
FORE(i,1,n){
if(edges[i].size() == 1){
other[i] = edges[i][0];
other[edges[i][0]] = i;
continue;
}
// has 3 edges;
int a = query3(i,edges[i][0],edges[i][1]);
int b = query3(i,edges[i][1],edges[i][2]);
int numTorem;
if(a == 1){
numTorem = edges[i][2];
}else if(b == 1){
numTorem = edges[i][0];
}else{
numTorem = edges[i][1];
}
allSets[i].erase(numTorem);
allSets[numTorem].erase(i);
}
FORE(i,1,n){
int x = *(allSets[i].begin());
if(i < x)Answer(i,x);
}
}
int main(){
}