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F. Video Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course…
Video Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, V…
题目链接:http://codeforces.com/contest/731/problem/F F. Video Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Vlad is fond of popular computer game Bota-2. Recently, the developers…
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, Vlad immediately…
Codeforces 731 F. Video Cards 题目大意:给一组数,从中选一个数作lead,要求其他所有数减少为其倍数,再求和.问所求和的最大值. 思路:统计每个数字出现的个数,再做前缀和,用于之后快速求和.将原数组排序后去重,枚举每一个数做lead的情况下,其余数减少后再求和的结果.不断维护最大值即可. PS:这里用到了一个方便的函数unique()来去重,使用前需要先将数组排序,参数为数组的首地址和尾后地址,返回新的尾后地址.(该函数没有将重复的元素删除,只是放到了尾地址后面)可…
题目链接:http://codeforces.com/contest/731/problem/F 题意:有n个数,从里面选出来一个作为第一个,然后剩下的数要满足是这个数的倍数,如果不是,只能减小为他的倍数,否则就舍弃掉,然后把没有舍弃的数的值加起来,求和的最大值; 43 2 15 9 就拿这个来说,当拿3当做第一个数时结果是3+15+9=27因为2不是3的倍数:当拿2作为第一个数时,结果是2+2+14+8=26因为3,15,9都不是2的倍数,所以只能减小;同理...求最大的和; 我们可以记录每个…
题意:给定n个数字,你可以从中选出一个数A(不能对该数进行修改操作),并对其它数减小至该数的倍数,统计总和.问总和最大是多少? 题解:排序后枚举每个数作为选出的数A,再枚举其他数, sum += a[i]-a[i]%A;复杂度为O(n^2), 爆炸. 显然应该降至O(nlogn). 枚举每个数a[i]做为A,再枚举A的倍数j即可, sum += j*(lower_bound(a, a+n, j+a[i])-lower_bound(a, a+n, j)); 那么此时复杂度为O(nlognlogn)…
http://codeforces.com/contest/731/problem/F 注意到一个事实,如果你要找一段区间中(从小到大的),有多少个数是能整除左端点L的,就是[L, R]这样.那么,很明显,把这个区间分段.分成[L, 2 * L - 1],因为这段区间中,都不大于等于L的两倍,这样就使得这段区间的贡献就是sum_number * L了. 就是中间有多少个数,每个数贡献一个L值.然后到枚举[2 * L, 3 * L - 1]这段区间,贡献的值就是sum_number * (2 *…
考虑每个数在最大值内的倍数都求出来大概只有max(ai)ln(max(ai))个. 先排个序,然后对于每个数ai,考虑哪些数字可以变成ai*k. 显然就是区间[ai*k,ai*(k+1))内的数,这个二分一下就好了. #include <bits/stdc++.h> using namespace std; ]; int main() { int n; scanf("%d", &n); ; i < n; i++) scanf("%d", a…
题意:给定 n 个数,可以对所有的数进行缩小,问你找出和最大的数,使得这些数都能整除这些数中最小的那个数. 析:用前缀和来做,先统计前 i 个数中有有多少数,然后再进行暴力去找最大值,每次都遍历这一段就好. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #inclu…
BACKGROUND The present invention relates to video processing systems. Advances in imaging technology have led to high resolution cameras for personal use as well as professional use. Personal uses include digital cameras and camcorders that can captu…
The new hardware tessellation feature available on Direct3D 11 video cards has great potential, but using it effectively currently requires understanding higher-order surfaces as well as a myriad of performance implications. In addition to the Window…
学习 KVM 的系列文章: (1)介绍和安装 (2)CPU 和 内存虚拟化 (3)I/O QEMU 全虚拟化和准虚拟化(Para-virtulizaiton) (4)I/O PCI/PCIe设备直接分配和 SR-IOV (5)libvirt 介绍 (6)Nova 通过 libvirt 管理 QEMU/KVM 虚机 (7)快照 (snapshot) (8)迁移 (migration) 本文将分析 PCI/PCIe 设备直接分配(Pass-through)和 SR-IOV, 以及三种 I/O 虚拟化…
F. Video Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course…
原帖地址:http://www.anandtech.com/show/9969/jedec-publishes-hbm2-specification The high-bandwidth memory (HBM) technology solves two key problems related to modern DRAM: it substantially increases bandwidth available to computing devices (e.g., GPUs) and…
http://anandtech.com/show/2549 Now that NVIDIA’s has announced its newest GPU architecture (the GeForce GTX 200 series), interesting architectural details are popping up on the web. The best writeup I’ve found is by AnandTech. In the past, such detai…
台式机安装CentOS 6.4 x86_64位  集成显卡ati4290 CentOS的release notes上: The proprietary drivers for older AMD ( former ATI ) video cards, namely the 2xxx, 3xxx and 4xxx series ( both integrated in motherboards or standalone cards) are not compatible with the new…
Ural doctors worry about the health of their youth very much. Special investigations showed that a lot of clever students instead of playing football, skating or bicycling had participated in something like Programming Olympiads. Moreover, they call…
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Archeologists have found a secret pass in the dungeon of one of the pyramids of Cycleland. To enter the treasury they have to open an unusua…
Contents 1 "Low" memory (< 1 MiB) 1.1 Overview 1.2 BIOS Data Area (BDA) 1.3 Extended BIOS Data Area (EBDA) 1.4 ROM Area 2 "Upper" Memory (> 1 MiB) 3 See Also 3.1External Links 这篇文章主要内容是计算机启动时,BIOS跳转到你的bootloader代码后的计算机的物理内存. 1,&q…
目录 DirectX11 Study Note Create a DirectX graphics interface factory.创建一个DirectX图形界面工厂 CreateDXGIFactory function CreateDXGIFactory函数 Use the factory to create an adapter 使用工厂创建适配器 IDXGIFactory interface IDXGIFactory接口 Enumerate the primary adapter ou…
访问Hard Drive 使用-hda –hdb qemu-system-x86_64 -enable-kvm -name ubuntutest  -m 2048 -hda ubuntutest.img -hdb ubuntutest1.img -boot c -vnc :19 -net nic,model=virtio -net tap,ifname=tap0,script=no,downscript=no 访问CD-ROM/DVD-ROM 使用-cdrom 如果想直接挂载Host机器上的CD…
使用目的 避免阻塞主线程 提高程序响应能力 C#中使用 C# 中的 Async 和 Await 关键字是异步编程的核心. 疑惑 The async and await keywords don't cause additional threads to be created. Async methods don't require multithreading because an async method doesn't run on its own thread. The method ru…
一.DRM 简介 In computing, the Direct Rendering Manager (DRM), a subsystem of the Linux kernel, interfaces with the GPUs of modern video cards. DRM exposes an API that user-space programs can use to send commands and data to the GPU, and to perform opera…
How PCI Express Works | PCIe工作原理 PCI Express is a high-speed serial connection that operates more like a network than a bus. Learn how PCI Express can speed up a computer and replace the AGP and view PCI Express pictures. Peripheral Component Interco…
How PCI Works | PCI工作原理 Your computer's components work together through a bus. Learn about the PCI bus and PCI card, such as the one above. See more computer hardware pictures. The power and speed of computer components has increased at a steady rat…
出处:http://www.nirsoft.net/vc/change_screen_brightness.html SetDeviceGammaRamp API函数位于Gdi32.ll中,接收一个256*3 RGB值的数组.增加这个数组中的值会使屏幕更亮,而减少这些值会使屏幕变暗.可以通过增加或减少的红/绿/蓝成分的值来显示影响.例如:在所有RGB值中增加蓝色分量将增加整个显示的蓝色.下面的类封装调用GetDeviceGammaRamp和SetDeviceGammaRamp,并设置屏幕的亮度,…
(4):I/O 设备直接分配和 SR-IOV 本文将分析 PCI/PCIe 设备直接分配(Pass-through)和 SR-IOV, 以及三种 I/O 虚拟化方式的比较. 1. PCI/PCI-E 设备直接分配给虚机 (PCI Pass-through) 设备直接分配 (Device assignment)也称为 Device Pass-Through. 先简单看看PCI 和 PCI-E 的区别(AMD CPU): (简单点看,PCI 卡的性能没有 PCI-E 高,因为 PCI-E 是直接连在…
无意翻到之前收藏的一个文档,共享一下. Solaris/Linux 命令手册 1. 系统 # passwd:修改口令 # exit:退出系统 2. 文件 # cp:复制文件或目录,参数:-a递归目录,-i覆盖确认 # mv:改名移动 # rm:删除,参数:-r递归删除 3. 目录 # mkdir:创建目录 # rmdir:删除空目录 # cd:改变工作目录 # pwd:查看当前路径 # ls:列目录,参数:-a所有文件,-c按时间排序,-l详细信息 4. 文本 # sort:排序 # uniq:…
KVM详解,太详细太深入了,经典 2016-07-18 19:56:38 分类: 虚拟化 原文地址:KVM详解,太详细太深入了,经典 作者:zzjlzx KVM 介绍(1):简介及安装 http://www.cnblogs.com/sammyliu/p/4543110.html 学习 KVM 的系列文章: (1)介绍和安装 (2)CPU 和 内存虚拟化 (3)I/O QEMU 全虚拟化和准虚拟化(Para-virtulizaiton) (4)I/O PCI/PCIe设备直接分配和 SR-IOV…