最敏捷的机器人

题目描述

思路

代码

#include <cstdio>
#include <cstring>
//#include <algorithm>
//using namespace std;
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) > (b) ? (b) : (a))
int n, m;
int arr[100005], f1[100005][21], f2[100005][21], log[100005];
inline int read() {
	int s = 0, f = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return f * s;
}
int main() {
	n = read(), m = read();
	log[0] = -1;
	memset(f2, 0x3f, sizeof(f2));
	for (int i = 1; i <= n; ++i) {
		arr[i] = read();
		// printf("%d
", arr[i]);
		log[i] = log[i >> 1] + 1;
		f1[i][0] = f2[i][0] = arr[i];
	}
	for (int i = 1; i <= 20; ++i) {
		for (int j = 1; j + (1 << i) - 1 <= n; ++j) {
			f1[j][i] = max(f1[j][i - 1], f1[j + (1 << (i - 1))][i - 1]);
			f2[j][i] = min(f2[j][i - 1], f2[j + (1 << (i - 1))][i - 1]);
		}
	}
	int j = log[m];
	for (int i = 1; i + m - 1 <= n; ++i) {
		printf("%d %d
", max(f1[i][j], f1[i + m - (1 << j)][j]), min(f2[i][j], f2[i + m - (1 << j)][j]) );
	}
	return 0;
}