| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 1218075 | thdh__ | Super Dango Maker (JOI22_dango3) | C++20 | 0 ms | 0 KiB |
#include <dango3.h>
#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define eb emplace_back
#define pu push
#define ins insert
#define fi first
#define se second
#define all(a) a.begin(),a.end()
#define bruh ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fu(x,a,b) for (auto x=a;x<=b;x++)
#define fd(x,a,b) for (auto x=a;x>=b;x--)
#define int ll
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
//mt19937 mt(chrono::steady_clock::now().time_since_epoch().count());
/*
Competitive Programming notes that I need to study & fix my dumbass self:
1. Coding:
- Always be sure to check the memory of arrays (maybe use vectors), for loops, I don't know
- Always try to maximize the memory if possible, even if you are going for subtasks
- Do not exploit #define int long long, it will kill you
2. Stress:
- Don't be cocky and think stressing with your dumbass brute-force solution will give you guaranteed AC.
- Always try generating big testcases and try if they run
3. Time management:
- Don't overcommit or undercommit, always spend a certain amount of time to think a problem, don't just look at it and say I'm fucked
- Do not spend too much time coding brute-force solutions, they should be easily-codable solutions that don't take up too much time
I hate offline because I am dumb
*/
typedef pair<int, int> ii;
const int N = 2e5+5;
const int mod = 1e9+7;
const int inf = 1e18;
using cd = complex<double>;
const long double PI = acos(-1);
int power(int a,int b) {ll x = 1;if (a >= mod) a%=mod; while (b) {if (b & 1) x = x*a % mod;a = a*a % mod;b>>=1;}return x;}
void Solve(int N, int M)
{
vector<bool> used(N*M+1);
vector<vector<int>> g(M+1);
for (int i = 1; i <= n*m; i++)
{
int l = 1, r = m, res = -1;
while (l <= r)
{
int mid = l+r>>1;
for (int j = 1; j <= n*m; j++) used[j] = 0;
used[i] = 1;
for (int j = 1; j <= mid; j++) for (auto k : g[j]) used[k] = 1;
vector<int> q;
for (int j = 1; j <= n*m; j++) if (!used[j]) q.pb(j);
if (Query(q) == m-mid) ans = mid, l = mid+1;
else r = mid-1;
}
g[res].pb(i);
}
for (int i = 1; i <= m; i++) Answer(g[i]);
}