제출 #628158

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
628158		jwvg0425	죄수들의 도전 (IOI22_prison)	C++17	0 / 100	10 ms	596 KiB

prison

#include <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>
#include <bitset>
#include <map>
#include <set>
#include <tuple>
#include <string.h>
#include <math.h>
#include <random>
#include <functional>
#include <assert.h>
#include <math.h>
#define all(x) (x).begin(), (x).end()
#define xx first
#define yy second

using namespace std;

template<typename T, typename Pr = less<T>>
using pq = priority_queue<T, vector<T>, Pr>;
using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;

vector<vector<int>> devise_strategy(int N) {

	const int BIT_SIZE = 13;
	vector<vector<int>> res(BIT_SIZE * 2, vector<int>(N + 1, 0));

	res[0][0] = 0;
	for (int ai = 1; ai <= N; ai++)
	{
		int bit = 1 << BIT_SIZE;
		if ((ai & bit) == 0)
			res[0][ai] = 1;
		else
			res[0][ai] = 2;
	}

	for (int order = 1; order <= BIT_SIZE - 2; order++)
	{
		res[order * 2 - 1][0] = res[order * 2][0] = order % 2;
		int lbit = 1 << (BIT_SIZE - order + 1);
		int bit = 1 << (BIT_SIZE - order);
		int mine = order % 2 == 1 ? -2 : -1;
		int other = order % 2 == 1 ? -1 : -2;

		// order + 1번 비트가 같은지 보고 -> 다르면 order번 비트에 따라 다음 상태로 진행
		for (int x = 1; x <= N; x++)
		{
			if ((x & lbit) == 0)
			{
				res[order * 2][x] = mine;
				res[order * 2 - 1][x] = x & bit ? order * 2 + 2 : order * 2 + 1;
			}
			else
			{
				res[order * 2 - 1][x] = other;
				res[order * 2][x] = x & bit ? order * 2 + 2 : order * 2 + 1;
			}
		}
	}

	{
		int order = BIT_SIZE - 1;

		res[order * 2 - 1][0] = order % 2;
		int lbit = 1 << (BIT_SIZE - order + 1);
		int bit = 1 << (BIT_SIZE - order);
		int mine = order % 2 == 1 ? -2 : -1;
		int other = order % 2 == 1 ? -1 : -2;

		for (int x = 1; x <= N; x++)
		{
			if ((x & lbit) == 0)
			{
				res[order * 2][x] = mine;
				res[order * 2 - 1][x] = (x & bit) ? order * 2 + 1 : mine;
			}
			else
			{
				res[order * 2 - 1][x] = other;
				res[order * 2][x] = (x & bit) ? order * 2 + 1 : mine;
			}
		}
	}

	{
		int order = BIT_SIZE;
		res[order * 2 - 1][0] = order % 2;
		int lbit = 1 << (BIT_SIZE - order + 1);
		int bit = 1 << (BIT_SIZE - order);
		int mine = order % 2 == 1 ? -2 : -1;
		int other = order % 2 == 1 ? -1 : -2;

		for (int x = 1; x <= N; x++)
		{
			if ((x & lbit) == 0)
			{
				res[order * 2 - 1][x] = mine;
			}
			else
			{
				res[order * 2 - 1][x] = (x & bit) ? other : mine;
			}
		}
	}

	return res;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 5
• Subtask 3: 0 / 90