Submission #621197

#	Time	Username	Problem	Language	Result	Execution time	Memory
621197		IWTIM	Unscrambling a Messy Bug (IOI16_messy)	C++14	100 / 100	2 ms	480 KiB

messy

#include "messy.h"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = long double; // or double, if TL is tight
using str = string; // yay python! 
 
// pairs
using pii = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;
#define mp make_pair
#define f first
#define s second
 
#define tcT template<class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
using vi = V<int>;
using vb = V<bool>;
using vl = V<ll>;
using vd = V<db>;
using vs = V<str>;
using vpi = V<pii>;
using vpl = V<pl>;
using vpd = V<pd>;
 
// vectors
// oops size(x), rbegin(x), rend(x) need C++17
#define sz(x) int((x).size())
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define pb push_back
#define eb emplace_back
#define ft front()
#define bk back()
 
#define lb lower_bound
#define ub upper_bound
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define rep(a) F0R(_,a)
#define each(a,x) for (auto& a: x)
const int MOD = 998244353;
const int MX = 2e5+5;
const ll BIG = 1e18; // not too close to LLONG_MAX
const db PI = acos((db)-1);
const int dx[4]{1,0,-1,0}, dy[4]{0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;
struct DSU {
	vi e; void init(int N) { e = vi(N,-1); }
	int get(int x) { return e[x] < 0 ? x : e[x] = get(e[x]); } 
	bool sameSet(int a, int b) { return get(a) == get(b); }
	int size(int x) { return -e[get(x)]; }
	bool unite(int x, int y) { // union by size
		x = get(x), y = get(y); if (x == y) return 0;
		if (e[x] > e[y]) swap(x,y);
		e[x] += e[y]; e[y] = x; return 1;
	}
};
/*
inline namespace Helpers {
	//////////// is_iterable
	// https://stackoverflow.com/questions/13830158/check-if-a-variable-type-is-iterable
	// this gets used only when we can call begin() and end() on that type
	tcT, class = void> struct is_iterable : false_type {};
	tcT> struct is_iterable<T, void_t<decltype(begin(declval<T>())),
	                                  decltype(end(declval<T>()))
	                                 >
	                       > : true_type {};
	tcT> constexpr bool is_iterable_v = is_iterable<T>::value;
 
	//////////// is_readable
	tcT, class = void> struct is_readable : false_type {};
	tcT> struct is_readable<T,
	        typename std::enable_if_t<
	            is_same_v<decltype(cin >> declval<T&>()), istream&>
	        >
	    > : true_type {};
	tcT> constexpr bool is_readable_v = is_readable<T>::value;
 
	//////////// is_printable
	// // https://nafe.es/posts/2020-02-29-is-printable/
	tcT, class = void> struct is_printable : false_type {};
	tcT> struct is_printable<T,
	        typename std::enable_if_t<
	            is_same_v<decltype(cout << declval<T>()), ostream&>
	        >
	    > : true_type {};
	tcT> constexpr bool is_printable_v = is_printable<T>::value;
}*/
#define pb push_back
vector <int> pp;
int nn,rr,ww;
void solve(int le, int ri) {
	if (le == ri) return ;
	string s = "";
	for (int i = 0; i < nn; i++) {
		s += "0";
	}
	for (int i = 0; i < le; i++) {
		s[i] = '1';
	}
	for (int i = ri + 1; i < nn; i++) {
		s[i] = '1';
	}
	int mid = (le + ri) / 2;
	for (int i = le; i <= mid; i++) {
		s[i] = '1';
		add_element(s);
		s[i] = '0';
	}
	solve(le,mid);
	solve(mid + 1, ri);
}
void solve1(int le, int ri, vector <int> v) {
	if (le == ri) {
		pp[v[0]] = le;
		return ;
	}
	int mid = (le + ri) / 2;
	string s;
	for (int i = 0; i < nn; i++) {
		s += "1";
	}
	for (int i = 0; i < (int)v.size(); i++) {
		s[v[i]] = '0';
	}
	vector <int> lef,rig;
	for (int i = 0; i < (int)v.size(); i++) {
		s[v[i]] = '1';
		if (check_element(s)) lef.pb(v[i]);
		else rig.pb(v[i]);
		s[v[i]] = '0';
	}
	solve1(le,mid,lef);
	solve1(mid + 1, ri, rig);
}
std::vector<int> restore_permutation(int n, int w, int r) {
	nn = n;
	ww = w;
	rr = r;
	pp = vector <int>(nn, 0);
    solve(0,nn - 1);
    compile_set();
    vector <int> v;
    for (int i = 0; i < nn; i++) {
    	v.pb(i);
	}
	solve1(0, nn - 1,v);
    return pp;
}

• Subtask 1: 20 / 20
• Subtask 2: 18 / 18
• Subtask 3: 11 / 11
• Subtask 4: 21 / 21
• Subtask 5: 30 / 30