| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 705463 | YassineBenYounes | 동굴 (IOI13_cave) | 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.
/*
ID: Yassine BenYounes
TASK: guard
LANG: C++
*/
#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef double db;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pbds tree<int, null_type, less<int>,rb_tree_tag, tree_order_statistics_node_update>
using namespace __gnu_pbds;
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (PGCD)
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);} // least common multiple (PPCM)
int dx[8] = {1, 0, 0, -1, 1, 1, -1, -1};
int dy[8] = {0, 1, -1, 0, 1, -1, -1, 1};
#define endl "\n"
#define ss second
#define ff first
#define all(x) (x).begin() , (x).end()
#define pb push_back
#define vi vector<int>
#define vii vector<pair<int,int>>
#define vl vector<ll>
#define vll vector<pair<ll,ll>>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd pair<double,double>
#define vdd vector<pdd>
#define speed ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
////////////////////Only Clear Code//////////////////////////
vi primes;
const int ppppp = 3e6+9;
ll mod = 1e9+7;
bool prime[ppppp];
ll fact[ppppp];
void sieve(int n){
memset(prime, 1, sizeof prime);
prime[1] = 0;
for(int i = 2; i < n;i++){
if(!prime[i])continue;
primes.pb(i);
for(int j = 2*i;j < n;j+=i){
prime[j] = 0;
}
}
}
vi factors(int n){
vi res;
for(int i = 0; i < primes.size() && primes[i]*primes[i] <= n;i++){
if(n % primes[i] == 0)res.pb(primes[i]);
while(n % primes[i] == 0){
n /= primes[i];
}
}
if(n > 1)res.pb(n);
return res;
}
ll binpow(ll a, ll b){
if(b == 0)return 1;
ll x = binpow(a, b/2) % mod;
ll g = x*x;
g %= mod;
if(b%2 == 0)return g;
g*=a;
g%=mod;
return g;
}
ll modinv(ll a){
return binpow(a, mod-2);
}
ll sub(ll a, ll b){
ll c = (a%mod)-(b%mod);
while(c < 0)c+=mod;
return c;
}
ll add(ll a, ll b){
return (((a%mod)+(b%mod)) % mod);
}
ll mult(ll a, ll b){
a %= mod;
b %= mod;
ll c = a*b;
c %= mod;
return c;
}
ll divide(ll a, ll b){
a %= mod;
b = modinv(b) % mod;
ll c = a*b;
c%=mod;
return c;
}
void pre_fact(int n){
fact[0]=1;
for(int i = 1; i < n;i++){
fact[i] = mult(fact[i-1],i);
}
}
ll choose(ll k, ll n){
ll c = mult(mult(fact[n],modinv(fact[k])), modinv(fact[n-k]));
return c;
}
ll perm(ll n){
return fact[n];
}
ll arrange(ll k, ll n){
ll c = mult(fact[n],modinv(fact[n-k]));
return c;
}
void init(){
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif // ONLINE_JUDGE
}
const int mx = 2e5+9;
const int LOG = 25;
const ll inf = 1e15;
int pos[mx], door[mx], tr[mx];
int tryCombination(int pos[]);
void answer(int pos[], int door[]);
void exploreCave(int n){
memset(pos, -1, sizeof pos);
for(int i = 0; i < n;i++){
// iterate on doors
bool right = 0;
for(int j = 0;j < n;j++){
// iterate on switches
if(pos[j] == -1)tr[j] = right;
else tr[j] = pos[j];
}
int x = tryCombination(tr);
if(x == i)right = 1;
int l = 0, r = n-1;
while(l < r){
int md = (l+r)/2;
for(int j = 0; j < n;j++){
// iterate on switches
if(j >= l && j <= md){
if(pos[j] == -1)tr[j] = right;
else tr[j] = pos[j];
}
else{
if(pos[j] == -1)tr[j] = !right;
else tr[j] = pos[j];
}
}
int x = tryCombination(tr);
if(x == i){
l = md+1;
}
else{
r = md;
}
}
// switch for i-th door : l
door[l] = i;
pos[l] = right;
}
answer(pos, door);
}
/*
void run_case(){
}
int main(){
//freopen("promote.in", "r", stdin);
//freopen("promote.out", "w", stdout);
init();
speed;
//sieve(ppppp);
//pre_fact(ppppp);
int t;
cin >> t;
//t=1;
while(t--){
run_case();
}
}*/
/*
NEVER GIVE UP!
DOING SMTHNG IS BETTER THAN DOING NTHNG!!!
Your Guide when stuck:
- Continue keyword only after reading the whole input
- Don't use memset with testcases
- Check for corner cases(n=1, n=0)
- Check where you declare n(Be careful of declaring it globally and in main)
*/