| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 1070396 | andrei_iorgulescu | 버섯 세기 (IOI20_mushrooms) | C++14 | 7 ms | 600 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 "mushrooms.h"
#warning That's not FB, that's my FB
using namespace std;
int count_mushrooms(int n)
{
vector<int> A, B;
A.push_back(0);
if (n <= 227)
{
int ans = 1;
for (int i = 1; i < n; i++)
ans += 1 - use_machine({0,i});
return ans;
}
int hh = use_machine({0,1}), hhh = use_machine({0,2});
if (hh == 0)
A.push_back(1);
else
B.push_back(1);
if (hh == 1)
{
if (hhh == 0)
A.push_back(2);
else
B.push_back(2);
}
int a = A.size(), b = B.size();
///caut cel mai bun max(a,b) la care sa ma opresc din spamat
int best, omin = 1e9;
for (int i = 5; i <= 1000 and 2 * i <= n; i++)
{
int cur = i - 1;
int cate = n - 2 * i;
int j = -1;
while (cate > 0)
{
j++;
cur++;
cate -= (i + j / 2);
}
if (cur < omin)
{
omin = cur;
best = i;
}
}
best--;
best--;
best--;
vector<int> unsolved;
if (hh == 1)
{
for (int i = 3; i < n; i++)
unsolved.push_back(i);
}
else
{
for (int i = 2; i < n; i++)
unsolved.push_back(i);
}
reverse(unsolved.begin(),unsolved.end());
while (max(a,b) < best)
{
if (a >= 2)
{
vector<int> mac = {unsolved[unsolved.size() - 1], A[0], unsolved[unsolved.size() - 2], A[1]};
int xxx = use_machine(mac);
if (xxx == 0)
{
A.push_back(unsolved.back());
unsolved.pop_back();
A.push_back(unsolved.back());
unsolved.pop_back();
}
else if (xxx == 1)
{
B.push_back(unsolved.back());
unsolved.pop_back();
A.push_back(unsolved.back());
unsolved.pop_back();
}
else if (xxx == 2)
{
A.push_back(unsolved.back());
unsolved.pop_back();
B.push_back(unsolved.back());
unsolved.pop_back();
}
else
{
B.push_back(unsolved.back());
unsolved.pop_back();
B.push_back(unsolved.back());
unsolved.pop_back();
}
}
else
{
vector<int> mac = {unsolved[unsolved.size() - 1], B[0], unsolved[unsolved.size() - 2], B[1]};
int xxx = use_machine(mac);
if (xxx == 0)
{
B.push_back(unsolved.back());
unsolved.pop_back();
B.push_back(unsolved.back());
unsolved.pop_back();
}
else if (xxx == 1)
{
A.push_back(unsolved.back());
unsolved.pop_back();
B.push_back(unsolved.back());
unsolved.pop_back();
}
else if (xxx == 2)
{
B.push_back(unsolved.back());
unsolved.pop_back();
A.push_back(unsolved.back());
unsolved.pop_back();
}
else
{
A.push_back(unsolved.back());
unsolved.pop_back();
A.push_back(unsolved.back());
unsolved.pop_back();
}
}
a = A.size();
b = B.size();
}
int ans = A.size();
while (!unsolved.empty())
{
int k = max(a,b);
vector<int> cr;
for (int i = 0; i < k; i++)
{
if (!unsolved.empty())
{
cr.push_back(unsolved.back());
unsolved.pop_back();
}
}
if (a >= b)
{
vector<int> tq;
for (int i = 0; i < cr.size(); i++)
{
tq.push_back(cr[i]);
tq.push_back(A[i]);
}
int xxx = use_machine(tq);
if (xxx % 2 == 1)
B.push_back(cr[0]), xxx--;
else
A.push_back(cr[0]), ans++;
ans += (2 * ((int)cr.size() - 1) - xxx) / 2;
}
else
{
vector<int> tq;
for (int i = 0; i < cr.size(); i++)
{
tq.push_back(cr[i]);
tq.push_back(B[i]);
}
int xxx = use_machine(tq);
if (xxx % 2 == 1)
A.push_back(cr[0]), ans++, xxx--;
else
B.push_back(cr[0]);
ans += xxx / 2;
}
a = A.size();
b = B.size();
}
return ans;
}
/**
Aflu sa zicem doua A-uri, dau spamez AxAy ca unic determina x si y, candva ma opresc
Sa zicem ca am a A-uri si b B-uri
While (mai am nedescoperite)
{
k = max(a,b);
Bag x1Ax2A...xkA, astfel aflu cate din x sunt A-uri dar aflu si exact cum e x1, il bag in A-uri sau B-uri si dau a++ sau b++
}
Fara ultima optimizare era fix 2sqrt, asa idk, doamne ajuta
**/
컴파일 시 표준 에러 (stderr) 메시지
| # | Verdict | Execution time | Memory | Grader output |
|---|---|---|---|---|
| Fetching results... | ||||