Black hole picture captured for first time in space ‘breakthrough’ Astronomers have captured the first image of a black hole, heralding a revolution in our understanding of the universe’s most enigmatic objects. The picture shows a halo of dust and g…

Crixalis's Equipment Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3957    Accepted Submission(s): 1625 Problem Description Crixalis - Sand King used to be a giant scorpion(蝎子) in the deserts…

Crixalis's Equipment Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1854    Accepted Submission(s): 770 Problem Description Crixalis - Sand King used to be a giant scorpion(蝎子) in the deserts o…

http://acm.hdu.edu.cn/showproblem.php? pid=3177 Crixalis's Equipment Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2562    Accepted Submission(s): 1056 Problem Description Crixalis - Sand Kin…

Crixalis's Equipment Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2795    Accepted Submission(s): 1141 Problem Description Crixalis - Sand King used to be a giant scorpion(蝎子) in the deserts…

</pre><pre name="code" class="cpp">#include <iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Equipment { int a; int b; }equip[1000]; bool cmp(Equipment x,…

在这里放几道Steps里的题目把. find your present (2) Time Limit: 1000/2000 MS (Java/Others) Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 4570 Accepted Submission(s): 1343 Problem Description In the new year party, everybody will get a "special pr…

主题链接:http://acm.hdu.edu.cn/showproblem.php? pid=3177 Problem Description Crixalis - Sand King used to be a giant scorpion(蝎子) in the deserts of Kalimdor. Though he's a guardian of Lich King now, he keeps the living habit of a scorpion like living und…

对于x86_64来说,逻辑地址由16位选择子和64位偏移量组成(而32位时,逻辑地址由16位段选择符和32位偏移量组成),段寄存器仅仅存放选择子.CPU的分段单元(SU)执行以下操作:[1] 先检查选择子的TI字段,以决定描述子对应的描述子保存在哪一个描述符表中.TI字段指明描述子是在GDT中(在这种情况下,分段单元从gdtr寄存器中得到GDT的线性基地址)还是在激活的LDT中(在这种情况下,分段单元从ldtr寄存器中得到LDT的线性基地址).[2] 从选择子的13位index字段计算描述子的地…

2017-02-23 一.伙伴系统 LInux下用伙伴系统管理物理内存页,伙伴系统得益于其良好的算法,一定程度上可以避免外部碎片为何这么说?先回顾下Linux下虚拟地址空间的分布. 在X86架构下,系统有4GB的虚拟地址空间,其中0-3GB作为用户空间,而3-4GB是系统地址空间.linux系统系统地址空间理论上应该不可换出,即每个虚拟页面均会对应一个物理页帧.如果这样的话,系统地址空间就能使用1GB,如果系统有多余的内存,这里仍然使用不上,这就限制了其性能的发展.为了解决这一问题,就有了高端内…