//freopen("/Users/zhangzhichao/Documents/HDOJ/HDOJ/1.txt", "r", stdin);
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <string.h>
#include <memory>
#include <queue>
using namespace std;
int dp[303][303][9][9];//dp[k][l][i][j]----->从第i行第j列，到第k + (1<<i)行 第l + (1<<j)列 最大的element
int map[303][303];
int n, m;
void init() {
for (int i = 0; i <= (int)log(n*1.0)/log(2.0); ++i) {
for (int j = 0; j <= (int)log(m*1.0)/log(2.0); ++j) {
if (i + j)
for (int k = 1; k <= n - (1<<i) + 1; ++k) {
for (int l = 1; l <= m - (1<<j) + 1; ++l) {
if (i == 0) {
dp[k][l][i][j] = max(dp[k][l][i][j-1], dp[k][l + (1<<(j-1))][i][j-1]);// 因为 dp 范围是 [j, j + 1<<j] 所以二分一下就是这样above
} else {
dp[k][l][i][j] = max(dp[k][l][i-1][j], dp[k + (1<<(i-1))][l][i-1][j]);
}
}
}
}
}
}
int RMQ(int x1, int y1, int x2, int y2) {
int k1 = (int)log(n*1.0)/log(2.0);
int k2 = (int)log(m*1.0)/log(2.0);
x2 = x2 - (1<<k1) + 1;
y2 = y2 - (1<<k2) + 1;
return max(max(dp[x1][y1][k1][k2], dp[x1][y2][k1][k2]), max(dp[x2][y1][k1][k2], dp[x2][y2][k1][k2]));
}
