Submission #256607

#	Time	Username	Problem	Language	Result	Execution time	Memory
256607		niyu	Unscrambling a Messy Bug (IOI16_messy)	C++14	100 / 100	2 ms	512 KiB

messy

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>
#include "messy.h"

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define beg(x) x.begin()
#define en(x) x.end()
#define all(x) beg(x), en(x)
#define resz resize

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
const ld PI = 4*atan((ld)1);

vi tr[512];

void add(int n, int l, int r) {
	if (l == r) return;
	string s = "";
	F0R(i, n) {
		if (i < l || i > r) s += "1";
		else s += "0";
	}
	FOR(i, l, (l + r)/2 + 1) {
		s[i] = '1';
		add_element(s);
		s[i] = '0';
	}
	add(n, l, (l + r)/2);
	add(n, (l + r) / 2 + 1, r);
}

void query(int n, int pos) {
	string s = "";
	F0R(i, n) s += "0";
	int j = pos;
	while (j) {
		for (int i : tr[j ^ 1]) s[i] = '1';
		j /= 2;
	}
	F0R(i, n) {
		if (s[i] == '1') continue;
		s[i] = '1';
		if (check_element(s)) tr[pos * 2].pb(i);
		else tr[pos * 2 + 1].pb(i);
		s[i] = '0';
	}
}

vi restore_permutation(int n, int w, int r) {
	//string s = "";
	add(n, 0, n - 1);
	compile_set();
	FOR(i, 1, n) query(n, i);
	vi res;
	F0R(i, n) {
		res.pb(0);
	}
	FOR(i, n, 2 * n) {
		res[tr[i][0]] = i - n;
	}
    return res;
}

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