一.velodyne-VLP16使用教程 推荐网址: http://blog.csdn.net/littlethunder/article/details/51920681 https://www.cnblogs.com/williamc17/p/9705492.html 源于用户手册的情报: 非常重要的一点:velodyne-VLP16的ros-driver所使用的坐标系有别于用户手册所说明的坐标系. 为了让所有开发者围绕相同的坐标系开发软件,ros中制定了标准的坐标系定义方案. 详情如以下链

/** * Copyright (C) 2010 Square, Inc. * * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http://www.apache.org/lice

Linux显示系统的诊断信息 youhaidong@youhaidong-ThinkPad-Edge-E545:~$ dmesg [ 1.492480] EDD information not available. [ 1.492521] PM: Hibernation image not present or could not be loaded. [ 1.506437] input: AT Translated Set 2 keyboard as /devices/platform/i80

Disruptor入门   获得Disruptor 可以通过Maven或者下载jar来安装Disruptor.只要把对应的jar放在Java classpath就可以了. 基本的事件生产和消费 我们从一个简单的例子开始学习Disruptor:生产者传递一个long类型的值给消费者,而消费者消费这个数据的方式仅仅是把它打印出来.首先声明一个Event来包含需要传递的数据: 1 public class LongEvent { 2 private long value; 3 public long

传送门 Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1, 2, . . . , n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle sho

  Prime Ring Problem  A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always

Season 1, Episode 16 -Burrows:Don't be. It's not your fault. 不要,不是你的错 -Fernando: Know what I like? 知道我喜欢什么? -Maricruz: Hmm? 恩 -Fernando: How after we make love... 我们做完以后 you get this little puddle of water in your belly button. puddle: 水坑  belly: 腹部,

Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always be 1. Inp

UVA - 524 Prime Ring Problem Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  into each circle separately, and the sum of number

题意  把1到n这n个数以1为首位围成一圈  输出全部满足随意相邻两数之和均为素数的全部排列 直接枚举排列看是否符合肯定会超时的  n最大为16  利用回溯法 边生成边推断  就要快非常多了 #include<cstdio> using namespace std; const int N = 50; int p[N], vis[N], a[N], n; int isPrime(int k) { for(int i = 2; i * i <= k; ++i) if(k % i == 0)

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 20105    Accepted Submission(s): 9001 Problem Description A ring is compose of n circles as shown in diagram. Put natural numb

1. 网卡处理数据包流程 网卡处理网络数据流程图: 图片来自参考链接1 上图中虚线步骤的解释: DMA 将 NIC 接收的数据包逐个写入 sk_buff ,一个数据包可能占用多个 sk_buff , sk_buff 读写顺序遵循FIFO(先入先出)原则. DMA 读完数据之后,NIC 会通过 NIC Interrupt Handler 触发 IRQ (中断请求). NIC driver 注册 poll 函数. poll 函数对数据进行检查,例如将几个 sk_buff 合并,因为可能同一个数据可能

Problem UVA524-Prime Ring Problem Accept:6782  Submit:43814 Time Limit: 3000 mSec  Problem Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1,2,...,n into each circle separately, and the sum of number

题目OJ地址: https://www.luogu.org/problemnew/show/UVA524 hdu oj 1016:  https://vjudge.net/problem/HDU-1016 zoj 1457  :https://vjudge.net/problem/ZOJ-1457 题意翻译 输入正整数n,把整数1,2,...,n组成一个环,使得相邻两个整数之和均为素数.输出时,从整数1开始逆时针排列.同一个环恰好输出一次.. 多组数据,读入到EOF结束. 第i组数据输出前加上一

http://www.cnblogs.com/yuxc/archive/2012/06/22/2558312.html 简介 OpenStack是一个美国国家航空航天局和Rackspace合作研发的开源云计算项目,并成为Apache下的一个重要开源项目,目前已经发展到了180家公司参与其中. OpenStack Object Storage(Swift)是OpenStack开源云计算项目的子项目之一.Swift的目的是使用普通硬件来构建冗余的.可扩展的分布式对象存储集群,存储容量可达PB级.Op

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25156    Accepted Submission(s): 11234 Problem Description A ring is compose of n circles as shown in diagram. Put natural num

A. Ring road time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The g

1 Command类中的rebalnace方法 在上篇文章中解说了,创建Ring已经为Ring加入设备.在加入设备后须要对Ring进行平衡,平衡 swift-ring-builder object.builder rebalance 首先会调用swift/cli/ringbuilder.py中方法,在main方法中首先会判读/etc/swift目录下是否有object.builder文件假设有就反序列化来初始化RingBuilder类.然后依据命令中的 第三个參数rebalance调用Comma

/* Name: NYOJ--488--素数环 Author: shen_渊 Date: 15/04/17 15:30 Description: DFS,素数打个表,37以内就够用了 */ #include<cstring> #include<iostream> #include<cstdio> #include<algorithm> using namespace std; void dfs(int); int n; //int prime[25] = {

Description 输入正整数n,把整数1,2,-,n组成一个环,使得相邻两个整数之和均为素数.输出时,从整数1开始逆时针排列.同一个环恰好输出一次.n<=16. A ring is composed of n (even number) circles as shown in diagram. Put natural numbers1,2--n into each circle separately, and the sum of numbers in two adjacent circl

参考自DPDK官方文档原文:http://doc.dpdk.org/guides-20.02/prog_guide/ring_lib.html 针对自己的理解做了一些辅助解释. 1 前置知识 1.1 CAS 学习无锁队列前先看一个基本概念,CAS原子指令操作. CAS(Compare and Swap,比较并替换)原子指令,用来保障数据的一致性. 指令有三个参数,当前内存值V.旧的预期值A.更新的值B,当且仅当预期值A和内存值V相同时,将内存值修改为B并返回true,否则什么都不做,并返回fal