제출 #226952

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
226952		staniewzki	Unscrambling a Messy Bug (IOI16_messy)	C++17	100 / 100	8 ms	512 KiB

messy

#include<bits/stdc++.h>
using namespace std;
 
#define REP(i, n) for(int i = 0; i < n; i++)
#define FOR(i, a, b) for(int i = a; i <= b; i++)
#define ST first
#define ND second
 
ostream& operator<<(ostream &out, string str) {
	for(char c : str) out << c;
	return out;
}
 
template<class L, class R> ostream& operator<<(ostream &out, pair<L, R> p) {
	return out << "(" << p.ST << ", " << p.ND << ")";
}
 
template<class T> auto operator<<(ostream &out, T a) -> decltype(a.begin(), out) {
	out << "{";
	for(auto it = a.begin(); it != a.end(); it = next(it))
		out << (it != a.begin() ? ", " : "") << *it;
	return out << "}";
}
 
void dump() { cerr << "\n"; }
template<class T, class... Ts> void dump(T a, Ts... x) {
	cerr << a << ", ";
	dump(x...);
}
 
#ifdef DEBUG
#  define debug(...) cerr << "[" #__VA_ARGS__ "]: ", dump(__VA_ARGS__)
#else
#  define debug(...) false
#endif
 
template<class T> int size(T && a) { return (int) a.size(); }
 
using LL = long long;
using PII = pair<int, int>;

#include "messy.h"

vector<int> restore_permutation(int n, int w, int r) {
	function<void(int, int)> add = [&](int l, int r) {
		if(l == r) return;
		int m = (l + r) / 2;
		string b(n, '0');
		REP(i, n) if(i < l || r < i)
			b[i] = '1';
		FOR(i, l, (l + r) / 2) {
			b[i] = '1';
			add_element(b);
			b[i] = '0';
		}
		add(l, m), add(m + 1, r);
	};
	add(0, n - 1);

	compile_set();

	vector<int> ans(n);
	REP(i, n) ans[i] = i;
	auto solve = [&](auto self, int l, int r, vector<int> opt) {
		if(l == r) {
			ans[l] = opt[0];
			return;
		}

		string b(n, '1');
		for(int i : opt) b[i] = '0';
		vector<int> x, y;
		for(int i : opt) {
			b[i] = '1';
			if(check_element(b))
				x.emplace_back(i);
			else
				y.emplace_back(i);
			b[i] = '0';
		}

		int m = (l + r) / 2;
		self(self, l, m, x), self(self, m + 1, r, y);
	};
	solve(solve, 0, n - 1, ans);

	vector<int> p(n);
	REP(i, n) p[ans[i]] = i;
	return p;
}

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