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\n", 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\n", max(f1[i][j], f1[i + m - (1 << j)][j]), min(f2[i][j], f2[i + m - (1 << j)][j]) );
    }
    return 0;
}

转载于:https://www.cnblogs.com/liuzz-20180701/p/11491103.html