| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 310280 | nonthaphat | 버섯 세기 (IOI20_mushrooms) | 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;
typedef long long ll;
typedef long double ld;
typedef double db;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef pair<pii, int> piipi;
typedef pair<int, pii> pipii;
typedef pair<pii, pii> piipii;
typedef pair<ll, pii> plpii;
typedef pair<db, db> pdd;
typedef pair<ld, ld> pldd;
typedef vector<int> vi;
typedef vector<pii> vii;
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define FOR2(i, a, b) for(int i=(a);i<=(b);++i)
#define ROF(i, a, b) for(int i=(b)-1;i>=(a);--i)
#define ROF2(i, a, b) for(int i=(b);i>=(a);--i)
#define GO(i, x) for(auto &i : x)
#define mp make_pair
#define fi first
#define se second
#define sz(x) (int)x.size()
#define all(x) (x).begin(), (x).end()
#define eb emplace_back
#define pf push_front
#define pb push_back
#define lb lower_bound
#define up upper_bound
template<typename T> inline bool min2(T &a, const T &b) { return b < a ? a = b, 1 : 0; }
template<typename T> inline bool max2(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
const int mod = 1e9 + 7;
// const int mod = 998244353;
const int P1 = 999983, P2 = 999979;
const ld PI = acos((ld)-1);
const int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
const ll INF = 1e18;
const int N = 1e6;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int sq = 281;
int use_machine(vector<int> x);
int count_mushrooms(int n){
int ansA = 1, ansB = 0;
vector<int> A, B;
if(n <= sq){
A.eb(0);
for(int i=1;i<n;i++){
vector<int> x;
x.eb(0), x.eb(i);
int result = use_machine(x);
if(result == 0){
A.eb(i);
ansA++;
}
else{
B.eb(i);
ansB++;
}
}
}
else{
A.eb(0);
for(int i=1;i<3;i++){
vector<int> x;
x.eb(0), x.eb(i);
int result = use_machine(x);
if(result == 0){
A.eb(i);
ansA++;
}
else{
B.eb(i);
ansB++;
}
}
if(sz(A) >= 2){
for(int i=3;i<sq;i+=2){
vector<int> x;
x.eb(i), x.eb(A[0]), x.eb(i+1), x.eb(A[1]);
int result = use_machine(x);
if(result%2 == 0){
A.eb(i);
ansA++;
}
else{
B.eb(i);
ansB++;
}
if(result <= 1){
A.eb(i+1);
ansA++;
}
else{
B.eb(i+1);
ansB++;
}
}
}
else{
for(int i=3;i<sq;i+=2){
vector<int> x;
x.eb(i), x.eb(B[0]), x.eb(i+1), x.eb(B[1]);
int result = use_machine(x);
if(result%2 == 0){
B.eb(i);
ansB++;
}
else{
A.eb(i);
ansA++;
}
if(result <= 1){
B.eb(i+1);
ansB++;
}
else{
A.eb(i+1);
ansA++;
}
}
}
int cur = sq;
while(cur != n){
if(sz(A) > sz(B)){
int cnt = min(n-cur, sz(A));
vector<int> x;
for(int i=0;i<cnt;i++){
x.eb(cur+i), x.eb(A[i]);
}
int result = use_machine(x);
if(result%2 == 0){
A.eb(cur+i);
ansA++;
}
else{
B.eb(cur+i);
ansB++;
}
ansA += cnt-result/2;
ansB += result/2;
cur += min(n-cur, sz(A)-1);
}
else{
int cnt = min(n-cur, sz(B));
vector<int> x;
for(int i=0;i<cnt;i++){
x.eb(cur+i), x.eb(B[i]);
}
int result = use_machine(x);
if(result%2 == 0){
B.eb(cur+i);
ansB++;
}
else{
A.eb(cur+i);
ansA++;
}
ansA += result/2;
ansB += cnt-result/2;
cur += min(n-cur, sz(B)-1);
}
}
}
return ansA;
}