제출 #211926

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
211926		mode149256	삶의 질 (IOI10_quality)	C++14	100 / 100	2769 ms	140636 KiB

quality

#include <bits/stdc++.h>
#include "quality.h"
using namespace std;

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<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}

int rectangle(int R, int C, int H, int W, int Q[3001][3001]) {

	auto gali = [&](int m) -> bool {
		vii sk(R, vi(C, 0));

		int reik = H * W / 2 + 1;

		for (int i = 0; i < R; ++i)
		{
			for (int j = 0; j < C; ++j)
			{
				sk[i][j] = int(Q[i][j] <= m);
				if (i) sk[i][j] += sk[i - 1][j];
				if (j) sk[i][j] += sk[i][j - 1];
				if (i and j) sk[i][j] -= sk[i - 1][j - 1];

				if (i >= H - 1 and j >= W - 1) {
					int dab = sk[i][j];
					if (i - H >= 0) dab -= sk[i - H][j];
					if (j - W >= 0) dab -= sk[i][j - W];
					if (i - H >= 0 and j - W >= 0)	dab += sk[i - H][j - W];

					// printf("m = %d, i = %d, j = %d, dab = %d, sk = %d\n",
					// m, i, j, dab, sk[i][j]);
					if (dab >= reik) return true;
				}
			}
		}

		return false;
	};

	int l = 1;
	int h = R * C;
	int m;
	while (l < h) {
		m = (l + h) / 2;
		// printf("l = %d, m = %d, h = %d, gali = %d\n", l, m, h, (int)gali(m));
		if (gali(m))
			h = m;
		else
			l = m + 1;
	}
	return l;
}

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 20 / 20
• Subtask 5: 20 / 20