| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 828030 | minhcool | Packing Biscuits (IOI20_biscuits) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "mushrooms.h"
//#define local
#ifndef local
#include ""
#endif
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
//#define int long long
#define fi first
#define se second
#define pb push_back
#define mp make_pair
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef pair<ii, ii> iiii;
const int N = 3e5 + 5;
const int oo = 1e18 + 7, mod = 1e9 + 7;
mt19937 rng(1);
int rnd(int l, int r){
int temp = rng() % (r - l + 1);
return abs(temp) + l;
}
int count_mushrooms(int n){
if(n <= 280){
int answer = 1;
for(int i = 1; i < n; i++){
vector<int> v;
v.pb(0), v.pb(i);
answer += (!use_machine(v));
}
return answer;
}
vector<int> v;
v.pb(0), v.pb(1);
vector<int> v1, v2;
v1.pb(0);
if(use_machine(v)) v2.pb(1);
else v1.pb(0);
v.clear();
v.pb(0), v.pb(2);
if(use_machine(v)) v2.pb(1);
else v1.pb(1);
int itr = 2;
while(v1.size() < 140 && v2.size() < 140){
vector<int> v;
if(v1.size() >= 2){
v.pb(v1[0]), v.pb(itr + 1), v.pb(v1[1]), v.pb(itr + 2);
int temp = use_machine(v);
if(temp & 1) v2.pb(itr + 2);
else v1.pb(itr + 2);
if(temp & 2) v2.pb(itr + 1);
else v1.pb(itr + 1);
}
else{
v.pb(v2[0]), v.pb(itr + 1), v.pb(v2[1]), v.pb(itr + 2);
int temp = use_machine(v);
if(temp & 1) v1.pb(itr + 2);
else v2.pb(itr + 2);
if(temp & 2) v1.pb(itr + 1);
else v2.pb(itr + 1);
}
itr += 2;
}
int answer = v1.size();
while(itr < n){
if(v1.size() >= v2.size()){
int nxt = min(n - 1, itr + (int)v1.size() - 2);
vector<int> v;
for(int i = itr; i <= nxt; i++){
v.pb(v1[i - itr]), v.pb(itr);
}
v.pb(v1[nxt - itr + 1]);
int temp = use_machine(v);
answer += temp/2;
itr = nxt + 1;
}
else{
int nxt = min(n - 1, itr + (int)v2.size() - 2);
vector<int> v;
for(int i = itr; i <= nxt; i++){
v.pb(v2[i - itr]), v.pb(itr);
}
v.pb(v2[nxt - itr + 1]);
int temp = use_machine(v);
answer += (v.size() - 1 - temp)/2;
itr = nxt + 1;
}
}
return answer;
}