原文 A stem-cell treatment put a London cancer patient's HIV into remission, marking the second such reported case and reinvigorating efforts to cure the AIDS-causing infection that afflicts some 37 million people globally. The patient has been in remi…

Title: London HIV patient's remission spurs hope for curing AIDS HIV 艾滋病毒  human immunodeficiency virus remission n.缓解,暂时康复 spur v.促进,激励 n.马刺(骑马的人鞋子上的一根小刺,用来刺激马快点跑) AIDS 艾滋病  HIV病毒感染的最后阶段 A stem-cell treatment put a London cancer(此处出现是因为艾滋病毒可能引发肿瘤) p…

Another OCD Patient Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 645    Accepted Submission(s): 238 Problem Description Xiaoji is an OCD (obsessive-compulsive disorder) patient. This mornin…

http://acm.hdu.edu.cn/showproblem.php?pid=4960 2014 Multi-University Training Contest 9 Another OCD Patient Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 181    Accepted Submission(s): 58 Pr…

BSS Audio® Introduces Full-Bandwidth Acoustic Echo Cancellation Algorithm for Soundweb London Conferencing Processors December 13, 2010       SALT LAKE CITY, Utah – BSS Audio®, a Harman International Company (NYSE-HAR), today introduced a new full-ba…

KDE声音服务器 arts arts介绍arts是KDE的核心声音系统,支持多音频流.全双工.网络声音请求.ALSA与OSS驱动后端.JACK声音服务器后端等扩展,它既是声音服务器,也 提供一套音频软件的开发类库.对于KDE3.x系统来说arts是不可缺少的一部分,KDE中几乎所有与声音有关的特性都和arts有平滑稳定的集成.由于arts是其作者的个人秀,在2004年底作者宣布终止对arts的维护后它已经很难再作出进一步的突破,这之后的更新大多都是其他志愿者的零敲碎 打,而且其艰涩的开发框架对后…

甲学者将HIV病毒的遗传物质彻底水解后得到A.B.C三种化合物,乙学者将组成T2噬菌体的遗传物质彻底水解后得到了A.B.D三种化合物.你认为C.D两种化合物分别指的是 A．尿嘧啶.胸腺嘧啶 B．胸腺嘧啶.尿嘧啶 C．核糖.脱氧核糖 D．尿嘧啶.腺嘌呤 答案: C [解析]HIV病毒的遗传物质是RNA,噬菌体病毒是DNA,RNA彻底水解的产物是磷酸.碱基.核糖,DNA水解的产物是碱基.磷酸和脱氧核糖,所以C.D分别为核糖和脱氧核糖,所以C选项正确.…

Another OCD Patient Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 716    Accepted Submission(s): 270 Problem Description Xiaoji is an OCD (obsessive-compulsive disorder) patient. This morni…

HDU 4960 Another OCD Patient pid=4960" target="_blank" style="">题目链接 记忆化搜索,因为每一个碎片值都是正数,所以每一个前缀和后缀都是递增的,就能够利用twopointer去找到每一个相等的位置,然后下一个区间相当于一个子问题,用记忆化搜索就可以,复杂度接近O(n^2) 代码: #include <cstdio> #include <cstring> #incl…

ARTS: Algrothm: leetcode算法题目 Review: 阅读并且点评一篇英文技术文章 Tip/Techni: 学习一个技术技巧 Share: 分享一篇有观点和思考的技术文章 Algorithm [leetcode]9. Palindrome Number https://leetcode.com/problems/palindrome-number/ 1)problem Determine whether an integer is a palindrome. An integ…