| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 1158423 | Wendidiask | popa (BOI18_popa) | C++20 | 0 ms | 0 KiB |
/*
:---:--:--::::::--:-:----:
::-:::. ::-:::
.* -===.. :---
.-+*##=-------====-: :--:
:+##=.* -= :=+- .-=:
.+++-. :- -=. .=+- .--.
.++=- + -= :*- :-.
+*+- :- -=. -*: :=:
#*: * -=. +- .=
=:%+ -: -=. +- --
-%: * -=. *: +.
:%-.. =. -=. # -:
:=# .. + -=. == -=
---+ . = -=. .# .+
-: == . = :------::------:.-=. % .=
:- == .=. ---:. :--#+. .% :-
= .* .+.. :=- :++ -+ =:
+ #. + -=: :*%-*: *
:= :# :=:= . :-----: =+-+ .+
* -* *=- . .- :::::::. :* .#+. +
:- -+ :#. --::-----: :* ++= .-
+ :*. *. .:::::*%: ++ +.-=. +
+ -#**----: .= -*. + -= =
* +.#=+: + . .=+: =. -=. =
= -+ :++: + .. .=+= .= -=. -
= %- :====-:. =. .:--===- * -=. -
+ +-- .:--==#==-:.= + -=. =
* + * =. : .= -=. =
+ = * + = -=. +
:- = .= =: = -=. .-
* + =: * * -=. +
:- .+ =- :- .+ -=. .+
* * := * := -= *
= :- =: .- .=: -=. =:
-: * .=- + .=- -= :-
=: -: --: :- :--: -=. .=
=:+ ----. * :---. -= .+
--=*----------------------=*+=--=+--=+*+------------------------------------------------%+:.
:: := +:- -- :.
.=. =.
-= -=
.=: .=.
.=: .+:
:-- :-:
-- -=.
:--. --:
:--: .--:
.-:-: ::--.
:----. .:----
.:::::::::-::::::--::::::::.
*/
// Hello this is Tyx2019's clone
#include "popa.h"
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define debug(x) cerr << #x << " is " << x << "\n"
#define hehe ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define repb(i, a, b) for (int i = b; i >= a; i--)
#define pii pair<int, int>
#define linebreak cout << "---------------------------------------------------------\n"
#define f first
#define s second
#define pb push_back
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
// good luck
const ll MS = 1005, mod = 1e9+9, INF = 3e18, blk = 400;
int next[MS], L[MS], R[MS];
int build(int l, int r) {
if (l > r) return -1;
int m = l;
while (next[m] <= r) m = next[m];
L[m] = build(l, m-1); R[m] = build(m+1, r);
return m;
}
int solve(int n, int *left, int *right) {
stack<int> st;
// next[i] = first index j such that a[j] divides a[i], ..., a[j].
// i.e. first index j such that gcd(a[i, ... ,j]) = a[j]
for (int i = 0; i < n; i++) {
if (st.empty()) {
st.push(i); continue;
}
while (!st.empty() and query(st.top(), i, i, i)) {
next[st.back()] = i; st.pop();
}
st.push(i);
}
while (!st.empty()){
next[st.back()] = MS-1; st.pop();
}
int root = build(0, n-1);
for (int i = 0; i < n; i++) left[i] = L[i], right[i] = r[i];
return root;
}