IBM 3090 with Vector Facility

Prior to the publication of the IBM technical paper behind what commonly known today as the Doherty Threshold[1] "... it was thought that a relatively slow response, up to two seconds, was acceptable because the person was thinking about the next tas

http://ianmurdock.com/post/how-i-came-to-find-linux/ lan murdock August 17, 2015 I saw my first Sun workstation in the winter of 1992, when I was an undergraduate at Purdue University. At the time, I was a student in the Krannert School of Management

存储管理子系统(DFSMS) 存储管理也是系统的一项基本服务. 系统存储管理子系统(Storage Management System,SMS)定义了系统中的各种输入/输出设备,并控制这些设备的操作:同时,它还提供文件系统的基本功能,使用户可以方便.快速地访问数据.系统的各种程序及数据都保存在硬盘上,如果希望执行某些程序,必须用到存储管理的服务,因此,SMS一般先于其他子系统启动,以提供基本的存储管理服务. 存储管理软件组件(Data Facility Storage Management Su

Frank is a portfolio manager of a closed-end fund for Advanced Commercial Markets (ACM ). Fund collects money (cash) from individual investors for a certain period of time and invests cash into various securities in accordance with fund's investment

链接:http://poj.org/problem?id=3090 题意:在坐标系中,从横纵坐标 0 ≤ x, y ≤ N中的点中选择点,并且这些点与(0,0)的连点不经过其他的点. 思路:显而易见,x与y只有互质的情况下才会发生(0,0)与(x,y)交点不经过其他的点的情况,对于x,y等于N时,可以选择的点均为小于等于N并且与N互质的数,共Euler(N)个,并且不重叠.所以可以得到递推公式aa[i]=aa[i]+2*Euler(N). 代码: #include <iostream> #in

这两天学习无锁的并发模式,发现相比于传统的 同步加锁相比,有两点好处1.无锁 模式 相比于 传统的 同步加锁  提高了性能 2.无锁模式 是天然的死锁免疫 下来介绍无锁的Vector--- LockFreeVector 它的结构是: private final AtomicReferenceArray<AtomicReferenceArray<E>> buckets; 从这里我们可以看到,它的内部是采用的是 无锁的引用数组, 数组嵌套数组 相当于一个二维数组,它的大小可以动态的进行

http://poj.org/problem?id=3090 法雷级数 法雷级数的递推公式很简单:f[1] = 2; f[i] = f[i-1]+phi[i]. 该题是法雷级数的变形吧,答案是2*f[i]-1. #include <stdio.h> #include <iostream> #include <map> #include <set> #include <stack> #include <vector> #include

3478题目意思是Farey序列是由一系列不能约分的分数a/b(0<a<b<=n且gcd(a,b)=1)按照递增的顺序组成的,然后给出了Fn的前几项,让你计算Fn中有多少个分数. 很明显Fn中的数的个数就是分别与(2,3,4.....n-1,n)互质的数的个数的和.也就是求欧拉函数的前n项和. 1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <st

Description The HDWBP Inc. has n clients and needs to service these clients by opening k facilities. Each opened facility can serve any number of clients and each client must be served by an open facility. There are m potential locations for these k