barriers / 阅读 / 详情

面试后,猎头是不是都会将feedback 告知candidates呢?

2023-05-19 15:51:47
共1条回复
黑桃云

分析下你是这个岗位被猎头找到的前几个人还是最后几个人

猎头的工作特性是这样

先期:根据雇主的要求找些基本符合条件的垫底,通过最好,通不过可以降低雇主对人才预期的高度

后期:最有把握的猎头认为符合要求的人才在最后期推荐 这样通过率很高

如果最开始 就把最合适的推荐给雇主 那么雇主一定还要再看看别人 也给自己增加了寻找人才的难度

如果你是这个岗位的前期面试者 你可以不用打电话了

要是这个岗位托猎头找了好长时间了 你倒是可以打个电话给猎头问问

接口就是下周要出国培训4天 担心联络不方便 还显得很礼貌很自然

祝你成功

呵呵这种心情能理解 当年我也参加过猎头的面试 猎头其实是很希望他们找的人能成功的 所以你要是真的着急 可以给猎头打电话 给他们打电话不会显得你“着急”他们是帮着你太高你的身价的 所以 不会和雇主反映你有点着急的 话说回来 其实能被猎头推荐成功的几率还是比较小的 因为参加面试的人都是很强的 最好的还是调整心态 抱着不一定能成功的心态 做能成功的努力 我当时就是报了很大的希望 因为当时开的年薪太诱人了 后来就真的没成 事后想想却觉得是好事 现在我跳的地方薪水没那么多但是很有发展 领导也重视 所以 调整心态吧 分析再多 也是一个已经固定的结果 在你两次面试的时候才能改变 现在改变不了了 就安心等待吧 祝顺利

相关推荐

名列前茅的英文?

名列前茅 【拼音】:míng liè qián máo 【解释】:比喻名次列在前面。 【出处】:《左传·宣公十二年》:“蔿敖为宰,择楚国之令典,军行,右辕,左追蓐,前茅虑无,中权,后劲。” 【示例】:去年弟在京师,听说二公子中了秀才,且~,颇为学台赏识,实在可贺可贺。 ★姚雪垠《李自成》第二卷第十七章 【近义词】:首屈一指 【反义词】:名落孙山 【语法】:作谓语、定语;指名次列在前面 【英文】:come out top 【成语故事】:春秋时期,晋国派大将荀林父做统率援助郑国,还没到黄河边,郑国就投降了楚国,他们执行楚国的好法典,楚军分右、左、前、中、后五军,由前军拿着茅草当旌旗开路。中军辅佐先不听荀林父的指挥而攻楚,结果大败而归
2023-01-02 16:12:4413

中文是什么lnstructionstocandidates?

我那里可以条件要牛津字典,牛津词典里面都是有的,它里面都是有翻译的,或者到百度里面都是有翻译的,你可以去看嘛。
2023-01-02 16:13:5815

candidates 和 applicants 的区别?

一个是参加者,可以是高考考生,另一个是申请者,申请工作的那种
2023-01-02 16:14:541

考场第二号考生”,用英语应该怎么说

考场第二号考生_有道翻译翻译结果:Test no. 2 candidates.candidates_有道词典candidates英 ["kændɪdəts]美 ["kændədets]n. 候选人;申请者;应征人员(candidate的复数)更多释义>>[网络短语]The papers have been handed out to the candidates. 考卷已发给考生。
2023-01-02 16:15:021

博士研究生的英文怎么说?

doctor
2023-01-02 16:15:0812

Among the candidates, John is the very youngest.

去掉veryAmong the candidates, John is the youngest.在这些候选人当中,约翰是最年轻的。
2023-01-02 16:16:242

a slate of candidates是什么意思

就是很多候选人,这里应该有特殊语境,估计你被slate迷糊了,其实意思和a bunch of people的中bunch类似。
2023-01-02 16:16:372

teacher candidates 什么意思

教师考生
2023-01-02 16:16:462

pipeline candidates 是什么意思?以下是这个词组出现的地方

管道的选择品,可供选择的管道
2023-01-02 16:16:554

doctorial candidates和doctorial students的区别?

Doctorial. candidates. 是指已经完成博士阶段课程,但是未完成毕业论文的博士生。Doctorial students.就是指一般的在读博士生。
2023-01-02 16:17:141

麻烦翻译下.` 不要电脑翻译`谢谢

一般指示 7。候选人必须把他们的入学证明和身份证/护照/外国鉴定 文件如入境/再入境许可在整个文件的时间(附图)桌上。 10。考生不得在任何情况下任何,互相沟通,在 文件。 11。考生必须熟悉在每个封面印有自己的说明 纸和回答问题的小册子。 12。考生必须立即告知监考人员,如果他们是: 字母i.发表论文的问题,他们不应该坐下来,或 二。问题发出文件,不包含的网页或一些问题上说明 封面页,或 三。没有发出(如书写纸)在一份文件中指出的问题的补充材料。 17。考生无须复制问题,他们正尝试。他们只需要写明质询号码(包括字母字符,如果有的话)在书写纸的左边距。没有答案应该是写在书写纸的左边和右边距。 18。除非该指示说,候选人之间应留2行的答案 连续的问题,或某一问题连续部分。 19。考生应写在信纸双方除非另有规定 指示。 20。考生不应该在任何时候,他们的回答倍脚本。考生不应该分成列的页面。 23。在一份文件结束时,考生应立即停止写作时说,这样做的invigilator.If书面文件,以便使用,它们应安排在顶部最少的数字序列的问题的答案。 24。答案脚本,加上封面页提供的,应在顶部并列左上角与字符串提供。或者,答题可以提供。 医疗事故/不诚实 37。弊端是指候选人的行动,无论是有预谋或以其他方式,违反了去 补充安置运动(资格阶段)的规则和条例,它为 考生在测试中的优势还是使他们能够破坏其他候选人的表现。 不当行为的一个例子就是考生勾结或与他人勾结,企图; 复制从另一名候选人,在他人作品提交仿佛是候选人的个人和剽窃。 退出及退款 43。没有就业的补充练习费退款将候选人谁 退出整个运动的补充文件或就业采取的任何部分 文件。
2023-01-02 16:17:211

英语求助:翻译

最好用人翻译,不要用软件之类的,那样会很死板。青鸟,我是不行了,不过还是祝愿你早点实现。
2023-01-02 16:17:275

请准确翻译这句:Candidates with less experien……

如果候选人经验不丰富,将被考虑安排担任柜台主任助理的职位。
2023-01-02 16:17:553

请英语好的人帮忙翻译一段话

1. Candidates employees: As the candidates interview techniques on the market of books increased, the majority of candidates have mastered the skills a large number of candidates, therefore, improve the interviewer"s interviewing skills to become a very important topic. Our company will conduct studies, psychology and other point of the interviewer training, interviewing skills, from a huge number of Asian applicants who really need to recruit talent.2.360 appraisal system development: 360 effective evaluation system in foreign countries, but in some Asian countries, due to cultural differences in ways of thinking with people, 360 appraisal system must be modified with national realities. We have summarized a set of user experience to help businesses plan for internal and external mechanisms of corporate appraisal system.3. Emotional debugging and stress management methods of work: in some Asian countries, especially China and Japan, the staff was not as good as a lot of the West, like individuals, will work as a pleasure, but because the population and intense competition, and work as made a kind of pressure and burden. Corporate officers of our business to help customers in the face of the company and external pressure and the pressure of life circumstances, to maintain a healthy and positive mental attitude in order to cope with stress, improve work efficiency and the success of the enterprise push4. Management Communication Skills: Communication is an important part of leadership, in any country are the same. Training, incentives and penalties customer enterprise applications, to written and oral form, body language, body language and non-approach, strengthen customer communication and feedback from employeesThese are the need to increase training programs do not literally mean to write as long as the full, formal point, will not have a syntax error could be a thank you!
2023-01-02 16:18:082

the candidates held a heated_as to whether

discussion
2023-01-02 16:18:261

帮我给这句英语挑挑毛病

Candidates who falled to pass the interview are either talked too little or never stoped talking. they are either overpolite or overly rude.
2023-01-02 16:18:397

英语句式分析:Such characteristics make them perfect candidates for Dr. Brosnan"s

应该是目的状语吧
2023-01-02 16:19:093

高分求翻译,短篇汉译英,解决了另有加分

Computer with its efficient and convenient to infiltrate into the merits of our lives in various fields, the state education so that the development of office automation are universal. To all types of schools to test that Test Paper as an example, to the traditional manual method that not only time-consuming effort, and low efficiency. The use of computers to develop a management questions, and can automatically to that, Test Paper of the need to become a software system. Network of information technology courses Q & A management system is based on a set of B / S system, using Microsoft SQL Server2000 as the background database, JSP prospects for development tools, and in Microsoft Windows XP Professional systems to MyEclipse6.01GA environment for the development of tools to realize. The system uses MVC design pattern Struts2 Application Design and Implementation of the framework. The online system is a branch of the examination system, mainly to complete the examination system of management background. System includes four modules: Q & A management, the completion of the main questions of management; candidates management, responsible for the management of information on the candidates; examination administration, completion of examinations, difficulty level and scores of settings, system management, and the system User management. The biggest feature is its openness, convenience and flexibility. The system"s features is that it is not against a particular subject, that certain Questions to Test Paper designed, have a certain commonality. With the system and the applications to add a dynamic subjects Questions. The system basically achieved the background of online examination system management functions. Can facilitate the management of information and questions the candidates, setting papers and the management of user information. But there are still a lot of system flaws and shortcomings, there is still a need to improve. There are many features on the need to improve. Hope that the opportunity to upgrade their sound.
2023-01-02 16:19:234

their half 为什么不可以跟candidates are girls.

their half 为什么不可以跟candidates are girls.原因如下一。 half用作名词,意为"一半,半",是可数名词,其复数形式是halves.二。表示某物的一半"用"halfa(an)+单数名词"结构。三。表示半公斤的"用"halfakilo of+名词。
2023-01-02 16:19:561

c语言的投票程序如何设计?

给你写一个:#include<stdio.h>#define M 100 //投票人数上限#define N 3 //候选人数struct{ char code; //代号 int score; //得分} candidates[N], st; //候选人结构体数组int main(){ char candidate[M][N]; //投票 int n, c=0; //n: 实际投票人数,c: 有效选票 int i,j,k,t; char tmp[N+1]; //存放投票的中间变量 int s[N]={5,3,2}; for(i=0; i<N; ++i) { printf("输入第 %d 个候选人的代号:", i+1); scanf("%c", &candidates[i].code); fflush(stdin); //清除回车符 candidates[i].score=0; } printf("输入投票人数:"); scanf("%d",&n); for(i=0; i<n; ++i) //输入选票 { printf("输入第 %d 张选票:", i+1); scanf("%s",tmp); candidate[i][0]=tmp[0]; candidate[i][1]=tmp[1]; candidate[i][2]=tmp[2]; if(tmp[0]==tmp[1] || tmp[1]==tmp[2] || tmp[0]==tmp[2]) printf("该票无效:%s ", tmp); else { //tmp[0]号候选人加5分; //tmp[1]号候选人加3分; //tmp[2]号候选人加2分; c++; for(j=0; j<N; ++j) for(k=0; k<N; ++k) if(candidates[k].code==tmp[j]) candidates[k].score+=s[j]; } } //得分排序 for(i=0; i<N-1; ++i) { k=i; for(j=i+1; j<N; ++j) if(candidates[j].score>candidates[k].score) k=j; st=candidates[i]; candidates[i]=candidates[k]; candidates[k]=st; } if(candidates[0].score==candidates[1].score) printf("前两名得分是否相同,都是 %d 分,重新投票。", candidates[0].score); else { printf("候选人 得分 "); for(i=0; i<N; ++i) printf("%c %d ", candidates[i].code, candidates[i].score); printf("祝贺 %c 当选冠军! ", candidates[0].code); } return 0;}
2023-01-02 16:20:061

candidates are: Student::Student(const Student&)

#include<iostream>#include<fstream>#include<stdlib.h>using namespace std;class Student{private: long No; char *Name; int Score;public: StStudent(long No=0,char* Name=NULL,int Score=0); long GetNO(); char *GetName(); int GetScore(); void ShowStudent();}Student::Student(long stu_no,char *stu_name,int stu_score){ No=stu_no; Name=stu_name; Score=stu_score;}void Student::ShowStudent(){ cout<<No<<" "<<Name<<" "<<Score<<endl;}long Student::GetNO(){return No;}char *Student::GetName(){return Name;}int Student::GetScore(){return Score;}int main(){ Student stu[3]= { Student(200607001,"Li ming",70), Student(200607002,"Hu jun",80), Student(200607003,"Wang tian",90) }; int i,k,pos; fstream infile,outfile; outfile.open("Student.dat",ios::out|ios::binary|ios::trunc); if(!outfile) { cerr<<"File open error!"<<endl; exit(1); } for(i=0;i<3;i++) { outfile.write((char *)&stu[i],sizeof((stu[i]))); } outfile.close(); cout<<"Students.dat:"<<endl; infile.open("Student.dat",ios::in|ios::binary); for(i=0;i<3;i++) { infile.read((char *)&stu[i],sizeof(Student)); cout<<stu[i].GetNO()<<" "<<stu[i].GetName()<<" "<<stu[i].GetScore()<<endl; } infile.close(); infile.open("Student.dat",ios::in|ios::binary); cout<<"Please input the number of record:"; cin>>k; pos=(k-1)*sizeof(Student); infile.seekg(pos); infile.read((char *)&stu[i],sizeof(Student)); cout<<stu[i].GetNO()<<" "<<stu[i].GetName()<<" "<<stu[i].GetScore()<<endl; infile.close(); cout<<"Delete the second record!"<<endl; outfile.open("Students.dat",ios::out|ios::binary|ios::trunc); for(i=0;i<3;i++) { if(i!=1) outfile.write((char *)&stu[i],sizeof(stu[i])); } outfile.close(); infile.open("Students.dat",ios::in|ios::binary); for(i=0;i<2;i++) { infile.read((char *)&stu[i],sizeof(Student)); cout<<stu[i].GetNO()<<" "<<stu[i].GetName()<<" "<<stu[i].GetScore()<<endl; } infile.close(); system("pause"); return 0;}
2023-01-02 16:20:123

英语大神 路过的 请点进来啊啊啊啊?

qualified 这里是形容词,意思是合格的,坐后面candidate的定语。再放图时请注意方向,起码考虑到答题者的阅读方便。
2023-01-02 16:20:233

有谁能帮我做一下高一的英语翻译???

Be my 1 place of readjusting oneself to a certain extent generally with my dictionary in me being at arm"s length taller than one translator. (Whthin arm" s reach) 2, we have broken into laugh start as soon as seeing his fantastic garb. (Sight of) 3, he likes the to ahead of time accomplishing a job. (Schedule) That 4 makes me breathed a sigh of relief is that he has submitted project design on time. (On schedule) 5, he has no the mind doing business. (Sense) This 6 agreements give the peculiar position exporting a commodity toward Eastern Europe to this company. (Unique) 7 evidence is described, this male person tall person , deep colour skin, age controls in 20-year-old. (Describe) He tries to need to explain everything 8 , I do not listen to but. (Attempt) 9 children and parents"s exchange are playing important role in the field of character culture. (Communication) All 10 traffic with east is interrupted because of earthquake without exception already. (Communication) 11 touches this language environment if you need to learn this language , you accommodate oneself to. (Exposed) This 12 candidates change self view going ahead in these problem unceasingly. (Shift) He is short of 13 interest to English studying , this is piece of headache thing to his parents. (Lack) 14 ought to be based on behavior but be not to judge one people coming self"s words. (Judge) He always the first arrives at 15 school , final one is away from school. (First) When my 16 first time comes to Shanghai, I apologize to my stranger bumping against even on a bus in street unceasingly. (The first time) 17 I think that the first matter composing is to print cover with first name on my book. (Stamp) 18 increasing by because of big market quantity, a lot of small shop is expelled from a marketplace. (Drive out) 19, child"s body and mind is all developed by playing. (Physically) He performs 20 pop music and jazz , performs at the same time also classical music. (Classical) 21, we ought to make use of the once the thing that we had studied is reviewed during the past summer holidays well. (Make use of) 22 should see that he is in a frenzy of rage when too bad school report card of son. (Explode) 23, they talk about film only in general field. (General) 24 is completely not necessary for your most delighted star feels being anxious for. (There be) 25 is very queer he actually does not come to be at school this morning. (It" s strange that) 26 government makes every effort to cut down 50%"s total unemployment. (Aim) 27 is undisputed , he is to show consideration for person most in all children. (Considerate) 28 pawns students when it is said that they will have an once to go out and use personal influence. Cheer and shout frenzied. (At the top of) Tell me 29 as if it be him , I am jubilantly happy when I take over that too bad work unnecessarily. (Wild with joy) There are palm of complicated two three branch Ye Mao inside 30 park at an intersection; garden in the city center. (Flourish) 31 has to improve him in image in the public mental view if he needs to win an election campaign. (The general public) 32, during the period of the spring outing, this little boy has been given as many as fifty yuan. (Outing)
2023-01-02 16:20:344

"候选人已经到了,我现在要把他们带到你的办公室吗?一个一个还是两个一起"英语翻译,谢谢

The candidates are already here. Should I bring them to your office now? Do you want to see them one by one or together?
2023-01-02 16:20:483

英语高手进~~~~~~~~

1、UK stands for_____United Kingdom_______________.2、修正句子(三选二)A.The foreman, together with two labourers, IS responsible for the complete job.B.The test WITH which they choose the successful candidates CONSISTS of six pages.C.He has therefore to work hard to earn a higher income TO PROVIDE for his family.3、Please choose two from the given words, and explain your understanding towards history. [past] [representation] [historians] [historical evidence] [truth]Historical evidence is a representation of past truth.
2023-01-02 16:20:592

高分求英文翻译

1. A brief self-introduction, says: (2) afternoon! Thank you for inviting me to participate in the interview. First of all, I introduce themselves to. My name is XXX, 24-year-old, who lives XXX, XXX graduated from the University of time in 2006, the accounting profession. I have XXX in the company cashier, financial assistant manager, network administrators, drivers and other duties. I prudent character, work conscientiously strict, principled, strong sense of team, love the financial work. A high level of computer, able to skillfully use the windows as well as word, excel office software, will use financial software, and a 2-year cars Jialing, a business trip. 2. Candidates departments: the Ministry of Finance professional candidates: professional accounting The basic personal situation photos Name Sex date of birth Health status of health height (cm) 178 ethnic Han Members of the current political landscape duties without expectations of 2-4 million annual salary Hunfou account whether the location of origin ID cards are leaving work, Type of talent, social talents, participate in the working hours of two years work experience, 2006 Current place of residence, Zip code 102200 E-mail, mobile phone, fixed telephone The highest education, tertiary, the maximum degree, no, professional, accounting, file location, Shunyi Graduate institutions, title, no English Language, English, grade level, one, typing speed, "the level of computer Description and grading of "proficient Professional learning and work experience (from low to high fill) "Learning experience" start and end time graduate professional degree institutions No 2000-2004 financial accounting No Accounting 2004-2006 "Work experience" beginning and end of time the name main duties 2006.6-2006.10 Beijing-Shanghai Express Services Limited in charge of financial, accounting, network management Tai Fung Industrial 2006.11-2007.3 Accounting Services Limited Field accounting, finance assistant, network management 2007.3-2008.4 Shunxinnongye Pengcheng Food Branch (Florida Pengcheng) cash, accounting, assistant manager, the driver Information construction project planning, network management Family members and major social relations Name relations work units work in the political landscape Workshop director of the masses Masses of workers Training and Certification Beginning and end of time training certificate name names training institutions 2003.3-2003.10 truck driving Beijing Beijing North Jiajiao driver"s license (B) 2004.11 accounting and the quality of economic management school accounting business card 2005.9-2005.11 Abacus Abacus five technical Shunyi TVU Certificate Self-evaluation (500 characters)
2023-01-02 16:21:0815

高手翻译

1约会的执行董事,依本法第二十六条(1)条的规定,欧盟委员会提议机关管理董事会名单合适的候选人担任执行主席。2。有关信息提出候选人名单上,包括信息委员会关于选择准则和程序使用,应被传送到董事会的成员至少10个日历日开始之前,会议提出执行董事的提名被考虑3。董事会将决定哪一个候选人提出的要求参加委员会是。秘书处应邀请那些候选人出席下次董事会会议。4。面试候选人后,董事会就决定是否应该进行提名。董事会应当由无记名投票为了提名最适合的人选。经董事会一致决定、保密性要求可以放弃投票。提名程序应当如下:1我哪里有一位候选人撤回或在其他候选人,留下一个唯一的候选人选举,候选人应提供他或她获得多数票的选票。2哪里有几个候选人,选举应按照下列步骤。一)收到至少一个候选人的选票的议席的所有成员被选定。在所有的考生收到的议席的选票的所有成员在每一轮候选人(■),最低投票数不包括在内。在只有两个候选人左、轮跑,直至有一名候选人的票数收到议席有利的管理的董事会成员b)在只有两个候选人,剩下3回合后剩下的两个候选人都得到一个议席,候选人接受多数选举c)考生可以随时撤回他们的候选人的过程中。5。在被正式由董事会任命,提名候选人应立即做出声明之前,欧洲议会。6。投票的动议移的执行主任机关搁置由至少一个第三的管理委员会。运动应当说明原因促使成员要求去除执行董事。大多数的新公司董事会决定将采取行动。
2023-01-02 16:21:583

请英语高手帮忙翻译

叉!~
2023-01-02 16:22:107

National Entrance Test Of English For MA/MS Candidates中MA/MS是什么的缩写?

硕士是统称,其下分为MA (Master of Arts 文学硕士) 和 MS (Master of Science科学硕士) 两类
2023-01-02 16:22:392

C++错误求大神改

你的函数呢?
2023-01-02 16:22:473

一句英语单选

B quit 退出参加总统候选人竞赛denounce 带有告发,揭发意思change allegiances 带有改变效忠对象意思abstain from 是弃权
2023-01-02 16:22:585

It has been announced that candidates _______...

shall
2023-01-02 16:23:174

c语言 投票程序

问题解决了吗
2023-01-02 16:23:312

_______ all of them are strong candidates, only one will be chosen for the post. A.Since B..

D 试题分析:句意:尽管他们都是强有力的候选人,但是仅仅有一个将被选上担任这个职务。since既然;if如果;as 因为;while尽管,根据句意故选D。点评:本题难度适中。连词是近几年高考的热点,主要考查从句中的连词所作的成分,它们的区别及意义。需要考生会正确分析复杂的句子结构的能力,能够正确理解语境。即学即练:I do every single bit of housework__________my husband Bob just does the dishes now and then. A. since      B. while     C. when        D. as解析:B。考查并列转折连词while的用法。这里表示“我”和丈夫做家务的对比。根据句意,选B。
2023-01-02 16:23:391

代码反平庸吗?

代码反混淆(deobfuscation)和代码混淆(obfuscation)对应,是其逆过程。维基百科将代码混淆定义为故意生成人类难以理解的源代码或机器码的过程("In software development, obfuscation is the deliberate act of creating source or machine code that is difficult for humans to understand.")。代码反混淆可以理解为将原本人类难以理解的代码转化为简单的、可理解的、直观的代码的过程。这篇文章主要介绍一下 "Big Code" 在代码反混淆领域的应用。更具体一点就是介绍一下提出 "JSNice" 和 "Deguard" 的两篇文章,这两篇文章虽然已经发表快五年了,但至今没有文章Follow这两份工作,因为文章已经将使用 "Big Code" 做代码命名反混淆做到了极致。后来的人无法在这个问题上推陈出新,脱颖而出。"Big Code": 代码托管网站如GitHub上的大量免费可用的高质量代码被称为 "Big Code" ,这些数据结合统计推理或深度学习为新兴的开发工具的出现提供了契机。概率图模型:概率图模型是用图来表示变量概率依赖关系的理论,结合概率论与图论的知识,利用图来表示与模型有关的变量的联合概率分布。问题为了项目的安全,开发者在打包发布项目时会对代码进行混淆加密,包括但不限于用无意义的短变量去重命名类、变量、方法,以免代码被轻易破解泄露。另外由于JS脚本主要用于Web开发,对其进行混淆还能压缩脚本的大小,使得浏览器下载、加载更加快速,提升用户的浏览体验。这一类通过对类、变量、方法重命名的混淆方案确实能加大其他开发者对代码的理解难度。其他开发者不干了,为了能方便理解他人混淆后的代码,学习(抄袭)他人的经验,针对这一类混淆方法的反混淆方法也应运而生。下面先展示一下安卓APP的代码混淆技术:图1. Java程序的代码混淆过程经过混淆的代码在功能上是没有变化的,但是去掉了部分名称中的语义信息。因为种种限制,这类混淆也不可能对所有的名称都进行替换。上图中的SQLiteHelper、SQLiteDatabase和Cursor就是一个证明,这些名称都是安卓API,如果将这些类名混淆会影响代码的功能。理论上一个有经验的安卓开发者可以在这些有限的提示下为所有的名称找到富含语义的表示,所以反混淆只需要一个有经验的开发者(有经验的开发者:???我做错了什么)。退一步,找不到有经验的开发者怎么办,没关系,GitHub有高质量的各种项目,现训练一个有经验的开发者也行。不过为了人道主义,消灭996剥削,程序员表示可以用程序代替人,正好可以用GitHub数据训练一个程序做这个反混淆嘛!理论存在,实践开始。JSNice[1]用程序代替人其实并不简单,针对上图中的反混淆问题,程序需要具有“联想”和“推理”能力,比如从a extends SQLiteHelper这一句中,a应该很可能也是Helper类,结合类中有SQLiteDatabase实例推理出比较符合a的语义的类名是DBHelper。针对以上两点,程序需要先有关系的概念,能从一个词“联想”到另一个词,然后还要有推理的能力,能通过约束从几个候选词中找到最符合的那个词。怎么做这个事呢?虽然有很多的未混淆的JS脚本供程序学习,但在反混淆JS脚本时,程序无法理解复杂的JS脚本,所以需要将JS脚本表示成一种可以利用已知属性推理未知属性的结构,JSNice采用了概率图模型。概率图模型相比其他学习算法的优势在于可以利用图结构将已知信息带入知识网络,在使用概率图模型之前,往往要求图结构是已知的。现实中我们没有这些先验知识,但是有大量的样本(未混淆代码)。通过样本学习出未混淆JS脚本的概率图就是JSNice的核心。具体到JSNice,这个工具想要做的是为JS脚本推测名称(name)和类型(type)。先通过一张图看看JSNice的推理过程。图2. 一段推理出名称和类型的JavaScript程序及推理名称的推理过程由于JSNice推理名称和推理类型的过程类似,本文就只阐述推理名称的过程。1. 确定已知和未知属性JS脚本中包含了各类代码元素(elements)比如变量,常量,类,方法名,域等。对推理名称问题,元素的属性即带有语义的名称,一个需要推理名称的元素,称其属性未知。不需要推理名称的元素其属性已知。首先需要确定JS脚本中的元素属性是否已知。很容易看出图2(a)代码中属性已知的元素包括:常量0和[]、feild名称length和方法名称push,其他的局部变量如e,t,n,r和i的属性都是未知的。将问题泛化,如何判断任意的JS脚本中的任意元素的属性是否已知是需要解决的第一个问题。JSNice采用程序分析和人工指定的方式确定元素属性是否已知,简单来说,JSNice认为JS脚本中的常量(constants)、对象属性名(objects properties)、方法名(methods)和全局变量名(global variables)都是属性已知的元素,而所有的局部变量的属性都是未知的。值得注意的是JSNice将对象属性名称和API名称直接看做是常量。这个划分是否合理暂不去讨论,但是其确实适用与大部分的JS脚本。有兴趣的读者可以自行研究。2. 建立依赖网络第一步获取了JS脚本中的所有元素(属性已知或未知),接下来需要建立元素之间的关系,好方便后续的推理;图2(c)中简单的给出了一些关系,比如"var r = e.length"中可以得到(r, length, L=_.R)的关系。JSNice实际考虑的元素关系十分复杂,主要有三类,这里简单的进行描述:句法关系,这一类关系主要通过AST得到,例子如下:图3. (a) 语句i + j &amp;amp;amp;amp;amp;lt; k的AST. (b) 根据AST建立用于名称预测的依赖网络. (c) 用于类型预测的网络.这类关系的形式化定义如下别名关系,这类关系通过standard alias analysis得到,将方法调用时传入的arguments和方法声明时的parameters进行关联,形成ARG_TO_PM关系。函数名称关系,由于JS的语言特性,很多的局部变量本身是方法定义,MAY_CALL和MAY_ACCESS表示一个方法的可能调用方法关系和可能访问对象域关系,这类关系可以通过program analysis得到。类型推理采用的关系有所不同,但这里也不再详述,有兴趣的读者直接移步原文。3. 训练和推理现在整理一下对某个JS脚本x进行上述两步分析能得到什么?得到了一个依赖网络  ,其中  为节点集,分别表示未知(unknow)属性元素和已知属性元素。  为边集,表示两个程序元素之间的关系以及关系类型。现在不去考虑训练的过程,直接看一下推理过程,JSNice采用贪婪算法,对未知属性元素遍历其所有的属性可能,寻找到使score最大的属性作为结果。图4. 建立网络寻找未知属性元素y的最佳属性的模式具体的算法如下:JSNice在具体实现时对算法有所优化,但是其基本思想和主要框架都没变。其实对比前文提到人进行反混淆时需要的“联想”和“推理”两种能力,candidates函数担负了“联想”的重任,利用scoreEdges函数对不同的候选属性计算score并选择最大score对应属性的过程就是”推理“。将图2的推理部分摘出来看:r的candidates有len和length,t的candidates有step、j和q,i的candidates只有i。推理得到的(r、i、t)的属性是(len、i、step)而不是(length、i、step);是因为前者的综合score是0.4+0.8+0.5=1.7,而后者的综合score只有0.5+0.6+0.5=1.6。那么怎么得到scoreEdges和candidates函数呢?JSNice定义了一个条件随机场:x为给定某个JS脚本,y为未知属性元素(复数)的任意分配属性,score为指示属性y和脚本x的匹配程度的函数,其返回值为实数,值越大越匹配。Z是对应JS脚本x的一个惩罚系数,用来保证其Pr和为1将score定义为k个特征函数的加权平均,得到最终条件随机场的表示形式为:写到这里,出现了第一个问题,score为k个特征函数的加权平均,如何确定k呢?JSNice是在训练阶段的预处理步骤得到k的,实际上不仅仅这一步不仅获得了k,还直接定义了k个 pairwise feature functions  。往前回一步,本文前面一直说GitHub有未混淆的代码可供概率图模型学习,这里定义训练集  由t个未混淆的JS脚本组成。对  中的任意  元组,JSNice定义其特征的集合为整个训练集的所有特征集为JSNice直接定义pairwise feature functions为每个特征三元组的指示函数:所以训练集有多少特征三元组,k的值就有多大。但说了这么多,还是没有提到scoreEdges和candidates。别急,直接定义如此把前面的公式都串起来了,整个公式组其实只有  是未知,条件随机场的训练过程其实就是计算  的过程。请点击输入图片描述至于candidates,假设现在概率图模型中的  已经训练完成,根据前面的定义,  和  其实是一一对应的,  本身是特征三元组的指示函数,也和三元组一一对应,所以可以使用权重  直接限制节点  的可能取值。定义  函数对输入的特征三元组集合基于此权重返回top s的三元组。定义  。定义如下辅助函数:最后:candidates函数其实就是先在  中找到和  有  关系的  ,然后利用  和  在训练集中找到和  最相似的词作为候选。比较方便的是辅助函数其实可以在预测之前提前计算并缓存下来。由于笔者本身没有不研究概率图模型,所以训练模型得到  的内容就省略了,有兴趣的读者可以阅读原文[1],本文只简单的描述:JSNice采用判别式训练,由于最大似然需要计算  ,JSNice采用max-margin training,使用Structured Support Vector Machine (SSVM)并用scalable subgradient descent algorithm优化。请点击输入图片描述Deguard[2]相对JSNice做的对JS脚本进行反混淆,Deguard对安卓APK做反混淆的难度要大了很多,放在眼前的一个问题就是项目规模,JSNice的应用scope其实是Web上的JS脚本,考虑网站的加载等限制,单个JS脚本必不会太大,而安卓APK不同,由于安卓本身事件驱动的编程方式,一个简单的安卓APK的复杂度可能就能比得上十分复杂的JS脚本。并且安卓APK的大小一般也没有限制。还有一些安卓或者说Java需要的约束是建模时需要考虑的,比如一个Java类中的feilds名称必须不一样,一个package中的classes名称必须不一样。不满足这些约束,对APK进行反混淆的结果就失去了其实用性。考虑到安卓application的复杂性,选取合适的粒度建模是首要的问题,关系过于复杂不利于概率图模型的学习,关系过于简略可能导致无法预测准确的属性,必须有一个权衡。1. 确定程序元素(图的节点  )要为安卓APK定义一个依赖图,首先确定图上的节点,Deguard考虑了APK中的如下元素:Types,不管是基本类型(int, long, float),引用类型(Object, ArrayList),还是数组类型(int[], Object[])都能做为节点Feilds,APK中的任意类中定义的任意Feild都能做为节点Packages,APK中的任意package能做为节点,像package a.b就可以做为两个节点,a和a.b。Methods,APK中的任意方法都能做为节点,但是如果有overrides关系,那么子类和父类的method公用一个节点表示。Constant values and nullAccess Modifiers,比如static,synchronizedOperations这里没有考虑Java语言中的泛型机制是因为在编译过程中会消除泛型,APK本身就是编译过后的文件。另外和JSNice不一样的是,Deguard没有考虑局部变量名和参数名,但考虑了局部变量的类型和参数的类型,一方面减少规模,另一方面就是变量名和参数名本身不在APK中。2. 确定元素属性是否已知这里将元素属性定义为节点是否被混淆,属性已知说明节点名称未被混淆,不需要预测名称,属性未知说明节点名称已被混淆,需要预测名称。已知属性元素包括节点如果代表pakages,classes,methods和feilds且在安卓API中,那么这些节点是已知的,因为基于名称的混淆工具并不会混淆这些名称构造函数名称都是已知的节点代表对安卓API方法重写的方法也是已知的,因为前文已经提过,重写关系其实是用一个节点表示。剩下的元素都是未知的,需要预测名称3. 构建依赖网络Deguard用一张图详细描述了其用于构建依赖网络的关系图5. 安卓application采用的关联程序元素的关系,第二列定义了边的类型,如果满足第三列提出的条件,一个关系就成立相对JSNice中大部分的关系来自AST,Deguard选择的关系明显融合进了人的经验,更加的抽象。实际上本文的精华也是这一张图,某种程度上这图中展现出来了人类对具体问题具体分析的思考,而不是仅仅简单的复用已有工作提出的各种关系。剩下的内容比如模型的训练和推理其实和JSNice差不多,这里不会重复一遍,后续会讲不一样的地方,也就是Deguard如何处理Java程序带来的关于各种名称的约束。Java中的名称约束还是比较复杂的,这里拿一个例子讲一下:图6. 展示方法命名显示的例子A.a(A)没有任何约束,因为A作为参数的方法只有它一个A.b(Object)不能被重命名为equals,因为这样会override   java.lang#www.scqiuyi.com#Object.equals(Object)B.g()和B.h()方法重命名名称必须不一样,因为这两方法的特征一样B.g()和A.c()方法重命名名称必须不一样,如果一样就是隐式的继承B.h()和A.c()方法重命名名称必须不一样,也是由于method overriding的原因C.x()和B.h()却没有冲突,因为B.h()是private的相等约束(继承重写机制)可通过共用节点表示,不等约束也需要明确表示,所以Deguard提出了一个检测方法名称不等约束的算法其他元素,比如类名,Feilds名称的不等约束比较简单,直接处理就行。所有不等约束以集合  表示,  ,  中任意两个节点的名称必须不一样。注意这个约束只用与预测阶段,因为训练数据(未混淆)本身满足这些约束。很容易可以把这些约束结合到JSNice的算法1中。Deguard的概率图优化算法和JSNice也不一样,采用的是pseudo likelihood estimation。具体阐述推荐阅读文章[3]。值得注意的是,为什么JSNice就没有Deguard中提到的相等约束和不等约束,笔者个人认为还是由问题和语言特性共同决定,JSNice的名称预测其实只预测了局部变量,而JS的语言特性导致其本身不需要检测局部变量的名称冲突,只有执行结果报错才会说明程序出错。也就是说其实JS本身语言特性就没有这类约束,自然不需要建模。
2023-01-02 16:23:451

c++高高手来看看

#include <stdio.h> #include <string.h> #include <io.h> #include <sys/stat.h> #include <fcntl.h> #include "Huffman.h" int wh; int rh; bool Write(unsigned char *s,int len){ _write(wh,s,len); return true; } bool OpenFile(char* source,char* target){ int w_flag=_O_WRONLY | _O_CREAT | _O_EXCL | _O_BINARY; int r_flag=_O_RDONLY | _O_BINARY; rh=_open(source,r_flag,_S_IREAD | _S_IWRITE); wh=_open(target,w_flag,_S_IREAD | _S_IWRITE); if(rh==-1 || wh==-1){ if(rh!=-1){ _close(rh); printf(" 打开文件:"%s"失败!",target); } if(wh!=-1){ _close(wh); printf(" 打开文件:"%s"失败!",source); } return false; }else{ return true; } } void PrintUsage(){ printf(" 以动态哈夫曼算法压缩或解压缩文件。 "); printf(" hfm -? 显示帮助信息 "); printf(" hfm -e -i source -o target 压缩文件 "); printf(" hfm -d -i source -o target 解压缩文件 "); } void main(int argc,char *args[]){ int mode,i,j,K=0; char src[4096]; char target[4096]; unsigned char buffer[BUFFER_SIZE]; Huffman *h; mode=0; for(i=1;i<argc;i++){ if(args[i][0]=="-" || args[i][0]=="/"){ switch(args[i][1]){ case "?": mode=0;//帮助 break; case "e": case "E": mode=1;//压缩 break; case "d": case "D": mode=2;//解压缩 break; case "o": case "O": if(i+1>=argc){ mode=0; }else{//输出文件 j=0; while(args[i+1][j]!="" && j<4096){ target[j++]=args[i+1][j]; } if(j==4096){ mode=0; }else{ target[j]=""; K |= 1; } } break; case "i": case "I": if(i+1>=argc){ mode=0; }else{//输入文件 j=0; while(args[i+1][j]!="" && j<4096){ src[j++]=args[i+1][j]; } if(j==4096){ mode=0; }else{ src[j]=""; K |=2; } } break; } } } if(K!=3)mode=0; switch(mode){ case 0: PrintUsage(); return; case 1://压缩 if(!OpenFile(src,target))return; h=new Huffman(&Write,true); i=BUFFER_SIZE; while(i==BUFFER_SIZE){ i=_read(rh,buffer,BUFFER_SIZE); h->Encode(buffer,i); } delete h; _close(rh); _close(wh); printf("压缩完毕!"); break; case 2://解压缩 if(!OpenFile(src,target))return; h=new Huffman(&Write,false); i=BUFFER_SIZE; while(i==BUFFER_SIZE){ i=_read(rh,buffer,BUFFER_SIZE); h->Decode(buffer,i); } delete h; _close(rh); _close(wh); printf("解压缩完毕!"); break; } } =======end of hfm.cpp======================= =======Huffman.cpp============================= // Huffman.cpp: implementation of the Huffman class. // ////////////////////////////////////////////////////////////////////// #include "Huffman.h" ////////////////////////////////////////////////////////////////////// // Construction/Destruction ////////////////////////////////////////////////////////////////////// Huffman::Huffman(Output *output,bool mode) { Hbtree *tmp; int i; this->mode=mode; //设置输出函数,当缓冲区满时,将调用该函数输出 this->output=output; //初始化列表 for(i=0;i<LIST_LENGTH;i++)this->list[i]=NULL; //初始化哈夫曼树 this->root=this->NewNode(NOT_CHAR,LEFT,NULL); this->current=this->root; tmp=this->NewNode(CODE_ESCAPE,RIGHT,root); tmp->count=1; tmp=this->NewNode(CODE_FINISH,LEFT,root); tmp->count=0; root->count=root->child[LEFT]->count+root->child[RIGHT]->count; //设置缓冲区指针 this->char_top=BOTTOM_BIT; this->bit_top=TOP_BIT; this->buffer[0]=0; //重构哈夫曼树的最大计数值 this->max_count=MAX_COUNT; this->shrink_factor=SHRINK_FACTOR; this->finished=false; } Huffman::~Huffman() { if(this->mode==true){//如果是编码 //输出结束码 this->OutputEncode(CODE_FINISH); this->char_top++; } //强制清空缓冲区 this->Flush(); //释放空间 this->ReleaseNode(this->root); } Hbtree * Huffman::NewNode(int value, int index, Hbtree *parent) { Hbtree *tmp=new Hbtree; tmp->parent=parent; tmp->child[0]=NULL; tmp->child[1]=NULL; tmp->count=(1 << SHRINK_FACTOR); tmp->index=(index==0) ? 0 : 1; tmp->value=value; if(value!=NOT_CHAR)this->list[tmp->value]=tmp; if(parent!=NULL)parent->child[tmp->index]=tmp; return tmp; } void Huffman::ReleaseNode(Hbtree *node) { if(node!=NULL){ this->ReleaseNode(node->child[LEFT]); this->ReleaseNode(node->child[RIGHT]); delete node; } } //输出一位编码 int Huffman::OutputBit(int bit) { unsigned char candidates[]={1,2,4,8,16,32,64,128}; if(bit!=0) this->buffer[this->char_top] |= candidates[this->bit_top]; this->bit_top--; if(this->bit_top < BOTTOM_BIT){ this->bit_top=TOP_BIT; this->char_top++; if(this->char_top >= BUFFER_SIZE){//输出缓冲区 this->output(this->buffer,BUFFER_SIZE); this->char_top=0; } this->buffer[this->char_top]=0; } return 0; } //输出缓冲区 int Huffman::Flush() { this->output(this->buffer,this->char_top); this->char_top=0; return 0; } int Huffman::Encode(unsigned char c) { int value=c, candidates[]={128,64,32,16,8,4,2,1}, i; if(this->list[value]==NULL){//字符不存在于哈夫曼树中 //输出转义码 this->OutputEncode(CODE_ESCAPE); //输出字符 for(i=0;i<8;i++)this->OutputBit(value & candidates[i]); this->InsertNewNode(value); }else{ //输出字符编码 this->OutputEncode(value); //重新调整哈夫曼树 this->BalanceNode(this->list[value]->parent); } //重组哈夫曼树 if(this->root->count>=this->max_count) this->RearrangeTree(); return 0; } void Huffman::BalanceNode(Hbtree *node) { Hbtree *parent,*child,*brother; int i,j; parent=node->parent; if(parent==NULL)return;//根节点无需调整 if(node->value==NOT_CHAR){//非叶子节点 child=node->child[LEFT]->count > node->child[RIGHT]->count ? node->child[LEFT] : node->child[RIGHT]; if(child->count > parent->count - node->count){ //失衡 i=!(node->index); j=child->index; node->count=parent->count - child->count; brother=parent->child[i]; node->child[j]=brother; brother->index=j; brother->parent=node; parent->child[i]=child; child->index=i; child->parent=parent; } } this->BalanceNode(parent); } //输出一个字符的编码 int Huffman::OutputEncode(int value) { int stack[CODE_FINISH+2],top=0; Hbtree *tmp=this->list[value]; //输出编码 if(value<=MAX_VALUE){//字符 while(tmp!=NULL){ stack[top++]=tmp->index; tmp->count++; tmp=tmp->parent; } }else{//控制码 while(tmp!=NULL){ stack[top++]=tmp->index; tmp=tmp->parent; } } top--; while(top>0){ this->OutputBit(stack[--top]); } return 0; } void Huffman::PrintNode(Hbtree *node,int level) { int i; if(node){ for(i=0;i<level*3;i++)printf(" "); printf("%p P:%p L:%p R:%p C:%d",node,node->parent,node->child[0],node->child[1],node->count); if(node->value!=NOT_CHAR)printf(" V:%d",node->value); printf(" "); this->PrintNode(node->child[LEFT],level+1); this->PrintNode(node->child[RIGHT],level+1); } } int Huffman::Encode(unsigned char *s, int len) { int i; for(i=0;i<len;i++)this->Encode(s[i]); return 0; } void Huffman::PrintTree() { this->PrintNode(this->root,0); } int Huffman::RecountNode(Hbtree *node) { if(node->value!=NOT_CHAR)return node->count; node->count= this->RecountNode(node->child[LEFT]) + this->RecountNode(node->child[RIGHT]); return node->count; } void Huffman::RearrangeTree() { int i,j,k; Hbtree *tmp,*tmp2; //所有非控制码的计数值右移shrink_factor位,并删除计数值为零的节点 for(k=0;k<=MAX_VALUE;k++){ if(this->list[k]!=NULL){ tmp=this->list[k]; tmp->count >>= this->shrink_factor; if(tmp->count ==0){ this->list[k]=NULL; tmp2=tmp->parent; i=tmp2->index; j=!(tmp->index); if(tmp2->parent!=NULL){ tmp2->parent->child[i]=tmp2->child[j]; tmp2->child[j]->parent=tmp2->parent; tmp2->child[j]->index=i; }else{ this->root=tmp2->child[j]; this->current=this->root; this->root->parent=NULL; } delete tmp; delete tmp2; } } } //重新计数 this->RecountNode(this->root); //重新调整平衡 for(i=0;i<=MAX_VALUE;i++){ if(this->list[i]!=NULL) this->BalanceNode(this->list[i]->parent); } } void Huffman::InsertNewNode(int value) { int i; Hbtree *tmp,*tmp2; //将字符加入哈夫曼树 tmp2=this->list[CODE_FINISH]; tmp=this->NewNode(NOT_CHAR, tmp2->index, tmp2->parent); tmp->child[LEFT]=tmp2; tmp2->index=LEFT; tmp2->parent=tmp; tmp2=this->NewNode(value,RIGHT,tmp); tmp->count=tmp->child[LEFT]->count+tmp->child[RIGHT]->count; i=tmp2->count; while((tmp=tmp->parent)!=NULL)tmp->count+=i; //从底向上调整哈夫曼树 this->BalanceNode(tmp2->parent); } int Huffman::Decode(unsigned char c) { this->Decode(c,7); return 0; } int Huffman::Decode(unsigned char *s,int len) { int i; for(i=0;i<len;i++)this->Decode(s[i]); return 0; } int Huffman::Decode(unsigned char c, int start) { int value=c, candidates[]={1,2,4,8,16,32,64,128}, i,j; Hbtree *tmp; if(this->finished)return 0; i=start; if(this->current==NULL){//转义状态下 while(this->remain >= 0 && i>=0){ if((candidates[i] & value) !=0){ this->literal |= candidates[this->remain]; } this->remain--; i--; } if(this->remain < 0){//字符输出完毕 //输出字符 this->OutputChar(this->literal); //将字符插入哈夫曼树 this->InsertNewNode(literal); //重组哈夫曼树 if(this->root->count>=this->max_count) this->RearrangeTree(); //设置环境 this->current=this->root; } }else{ j=((value & candidates[i])!=0)?1:0; tmp=this->current->child[j]; i--; while(tmp->value==NOT_CHAR && i>=0){ j=((value & candidates[i])!=0)?1:0; tmp=tmp->child[j]; i--; } if(tmp->value==NOT_CHAR){//中间节点 this->current=tmp; }else{ if(tmp->value<=MAX_VALUE){//编码内容 j=tmp->value; this->OutputChar((unsigned char)j); //修改计数器 tmp=this->list[j]; while(tmp!=NULL){ tmp->count++; tmp=tmp->parent; } //调整平衡度 this->BalanceNode(this->list[j]->parent); //重组哈夫曼树 if(this->root->count>=this->max_count) this->RearrangeTree(); //设置环境 this->current=this->root; }else{ if(tmp->value==CODE_ESCAPE){//转义码 this->current=NULL; this->remain=7; this->literal=0; }else{//结束码 this->finished=true; return 0; } } } } if(i>=0)this->Decode(c,i); return 0; } int Huffman::OutputChar(unsigned char c) { this->buffer[this->char_top++]=c; if(this->char_top>=BUFFER_SIZE){//输出缓冲区 this->output(this->buffer,BUFFER_SIZE); this->char_top=0; } return 0; } ========end of Huffman.cpp================== ========Huffman.h============================ // Huffman.h: interface for the Huffman class. // ////////////////////////////////////////////////////////////////////// #if !defined(NULL) #include <stdio.h> #endif #if !defined(AFX_HUFFMAN_H__B1F1A5A6_FB57_49B2_BB67_6D1764CC04AB__INCLUDED_) #define AFX_HUFFMAN_H__B1F1A5A6_FB57_49B2_BB67_6D1764CC04AB__INCLUDED_ #if _MSC_VER > 1000 #pragma once #endif // _MSC_VER > 1000 #define MAX_COUNT 65536 //最大计数值,大于此值时 #define MAX_VALUE 255 //编码的最大值 #define CODE_ESCAPE MAX_VALUE+1 //转义码 #define CODE_FINISH MAX_VALUE+2 //结束码 #define LIST_LENGTH MAX_VALUE+3 //编码列表长度 #define SHRINK_FACTOR 2 //减小的比例,通过右移位实现 #define LEFT 0 //左孩子索引 #define RIGHT 1 //右孩子索引 #define NOT_CHAR -1 //非字符 #define TOP_BIT 7 //字符最高位 #define BOTTOM_BIT 0 //字符最低位 #define BUFFER_SIZE 81920 //缓冲区大小 //输出函数 typedef bool (Output)(unsigned char *s,int len); //哈夫曼树的节点定义 typedef struct Hnode{ int count;//计数器 int index;//父节点的孩子索引(0--左孩子,1--右孩子) Hnode* child[2]; Hnode* parent; int value; }Hbtree; class Huffman
2023-01-02 16:24:521

请问中国研究生入学考试的英文缩写是什么啊?

National Entrance Test for MA/MS Candidates简称NETM
2023-01-02 16:24:574

c++投票选举系统代码

hi,能不能说一下你的程序有什么需求?Voter Candidate PersonSet 这些类要自己实现吗
2023-01-02 16:25:211

“金榜题名”怎么翻成英文啊,急用,要准确,金榜题名

pass the examination
2023-01-02 16:25:314

中译英,能人进!

" How you know the candidates? To understand to what extent? Please introduce the students moral, academic level, research ability, knowledge structure and foreign language level, as well as your examinee registers for my school doctor graduate student opinion"
2023-01-02 16:26:0011

In the election__ ,the candidates have to make a

在选举竞争中,候选人必须发表许多鼓励的演说赢得来自各种人的选票选A
2023-01-02 16:26:383

We interviewed a number of candidates but none of them impressed us.!!

不只看意义上的转折,这是but连接的并列句,相当于but两边各自是一个句子,各自有完整的主谓。
2023-01-02 16:26:561

处理紧张的有效方法 英语作文

Many students to a special examination on the tension, I think, first of all candidates in the test must establish a strong self-confidence, it is necessary that "there must be Tianshengwocai", the study is not the only way to become useful, good only become useful to learn a Only factor. The day before the test to calm down, so a buffer among us, so that more clear-headed examination will be able to play well. Parents will have to continue to give comfort to the children and not ratio of students still tense, the task of parents is to create a comfortable environment for a child more confidence. Second, the candidates can and parents, friends and family, such as consultants to give vent to their pressure, to seek support. Can also have a rhythm to do some things, such as listening to music to ease the mood, dance, painting, to regulate the use of color. If the pressure in particular, can play to the beat of music. Candidates proposed to do some aerobic exercise, or to carry out a deep breath. In addition, the efforts of students to exercise restraint until their basic emotional stability, that is, pay attention to mental relaxation, at the same time I like to think of things scattered about the brain and nervous to come back, the better the outcome will be. Finally, the students believe they have to in order to best deal with the state exam. I hope people will test a success你紧张是因为肾上腺素分泌激素,你只要一边想着这是肾上腺素分泌激素然后做深呼吸,慢慢的就不会紧张了。
2023-01-02 16:27:021

UE接收控制信息的流程

在每个配置有PDCCH monitoring的activated serving cell上的active DL BWP上,UE根据对应的search space sets在一个或多个CORESETS上监听一组PDCCH candidates,此处监听指的是使用被监听的DCI formats对每个PDCCH candidates尝试解码。 对于服务小区上配置给UE的每一个DL BWP, 基站最多可以给UE配置3个CORESETS,基站通过信令          PDCCH-Config/PDCCH-ConfigCommon中IE:ControlResourceSet为每个CORESET提供以下信息:对于CORESET 0,UE认为在该CORESET接收到的PDCCH中的DMRS port与以下提到的天线端口具有准共址关系(quasico-located): 由TCI state配置的一个或多个下行参考信号,该TCI state由MACE CE activation command指示,或者, 在最近的随机接入流程后如果没有MAC CE activation command为该CORESET指示一个TCI state,则在最近的不是由PDCCH order发起的non-contention based随机接入流程中由UE确定的一个SSB与CORESET 0中接收到的PDCCH中的DMRS port具有准共址关系。 对于除了CORESET 0之外的其他CORESET,如果基站给UE为该CORESET提供了一个单一的TCI state或者如果UE接收到了MAC CE activation command(该命令中包含了为该CORESET配置的TCI states中的一个),则UE认为在该CORESET接收到的PDCCH中的DMRS port与TCI state中配置的一个或多个下行参考信号准共址。 对于服务小区上配置给UE的每一个DL BWP, 基站可以给UE配置个搜索空间集合,基站通过信令 SearchSpace为每个搜索空间集合提供以下信息:在每个服务小区中,一个UE最多只能在一个搜索空间集合中同时监听4种长度的DCI formats对应的PDCCH candidates,这4种长度的DCI formats最多可以有3种长度的DCI format是使用C-RNTI加扰的。下表给出了在单服务小区下运行的UE在SCS配置下,一个DL BWP中的每个slot上可以监听的PDCCH candidates的最大数量:如果一个UE: 配置为在单个小区上运行或者在相同的频段上运行载波聚合,并且, 在一个或多个小区的active DL BWP(s)上具有相同或者不同QCL-TypeD属性的多个CORESET中的overlapping的PDCCH monitoring occasions上监听PDCCH candidates 则该UE在一个或多个小区的active DL BWP(s)只监听一个CORESET以及在这多个CORESET中任意一个与该CORESET有相同QCL-TypeD属性的CORESET上监听PDCCH: 该CORESET对应索引最小的小区(指的是同时有多个小区同时进行下行调度的场景,比如载波聚合)中最小索引的SS set,如果该SS set为CSS set,则该CORESET对应的是该CSS set;否则该CORESET对应索引最小的小区内索引最小的USS set。 最小索引的USS set从overlapping的PDCCH monitoring occasions中所有至少有一个PDCCH candidate的USS set中选出。 为了确定CORESET:1.需要找到一个与CSI-RS具有不同的QCL-TypeD属性的SS/PBCH block;2.在第一个小区中第一个和一个SSB相关联的CSI-RS和在第二个小区中第二个和一个SSB相关联的CSI-RS, 这两个CSI-RS被认为具有相同的QCL-TypeD属性。 对non-overlapping CCE的分配以及用于PDCCH monitoring的PDCCH candidates的分配是基于在一个或多个小区上的active DL BWP中与多个CORESETS相关联的所有搜索空间集合。 active TCI states的个数由多个CORESETS确定。 对于一个被调度的小区,在任何时候,一个UE最多只能接收16个格式为DCI format 1_0或者1_1对应的用C-RNTI, CS-RNTI或者MCS-RNTI加扰的PDCCH,这些PDCCH对应最多16个UE尚未接收到的PDSCH;一个UE最多只能接收16个格式为DCI 0_0或者0_1对应的用C-RNTI, CS-RNTI或者MCS-RNTI加扰的PDCCH,这些PDCCH对应最多16个UE尚未发送的PUSCH。 如果一个UE, 在一个active DL BWP上的CORESET中,该UE被配置了在一个CSS set中监听第一个DCI格式为0_0和1_0的PDCCH candidate,并且在一个USS set中监听第二个DCI格式为0_0和1_0的PDCCH candidate,并且, 和第一个PDCCH candidate相关联的DCI format0_0/1_0与和第二个PDCCH candidate相关联的DCI format0_0/1_0具有相同的长度,并且, UE在同一组CCE上接收到第一个PDCCH candidate和第二个PDCCH candidate,并且, 第一个PDCCH candidate和第二个PDCCH candidate具有相同的扰码,并且, 第一个PDCCH candidate的DCI format0_0/1_0和第二个PDCCH candidate的DCI format0_0/1_0都由C-RNTI,或者MCS-C-RNTI,或者CS-RNTI加扰。 则则该UE只解码与第一个PDCCH candidate相关的DCI format0_0/1_0。http://blog.sina.com.cn/s/blog_3c3393b00102yr21.html
2023-01-02 16:27:081

高分求翻译,满意立刻给分

1。你的专业是什么关系/被提名的候选人? 2。你们认识多久他/她和他/她的主要责任的公司? 3。你会如何描述候选人的综合能力和技巧吗? 4。如何测量的候选人的主要性能目标在过去,无论他/她达到目标? 5。请问如何他/她相处。 6。这位候选人的能力如何处理危急,沉重的负荷和紧密的截止日期呢? 7。这位候选人的经验如何在相互难相处的人吗? 8。什么是他/她的伟大力量(s)或限制(■)组织? 9。如果候选人在任何方面一直执教他/她的表演,如果真是这样的话,这个自然? 10。这个离职的原因吗? 11分。你会如何描述他/她的正直、诚实吗? 12。你愿意与这个候选人说一遍好吗?为什么? 13。你有任何附加的评论会帮我评估候选人吗?
2023-01-02 16:27:142

aday译中文是什么意思

aday译中文“一天”的意思。简明解释:东不拉可以造句1.Enough of fantasy, the workaday world awaited him. 天马行空地幻想了一番之后,等待他的是平淡无奇的现实世界。2.This is a workaday newspaper. 这是一份枯燥乏味(或:普通)的报纸.
2023-01-02 16:27:222

what sets you aprat from all the other candidates?这句话什么意思??

set sb apart from:使某人远离。。。candidate:候选人,应聘者本句可译为:你有什么长于其他应聘者之处?即:你有什么不同之处?祝你进步!望采纳啊!
2023-01-02 16:27:312

数独游戏棋九宫格数独棋怎么玩的

玩家需要根据9×9盘面上的已知数字,推理出所有剩余空格的数字,并满足每一行、每一列、每一个粗线宫(3*3)内的数字均含1-9,不重复。数独盘面是个九宫,每一宫又分为九个小格。在这八十一格中给出一定的已知数字和解题条件,利用逻辑和推理,在其他的空格上填入1-9的数字。使1-9每个数字在每一行、每一列和每一宫中都只出现一次,所以又称“九宫格”。
2023-01-02 16:27:444

谁能给我提供一些关于美国总统选举的英文资料,谢谢了

United States Presidential Election - From Wikipedia, the free encyclopedia Elections for President and Vice President of the United States are indirect elections in which voters cast ballots for a slate of electors of the U.S. Electoral College, who in turn directly elect the President and Vice President. They occur quadrennially (the count beginning with the year 1792) on Election Day, the first Tuesday after the first Monday of November.[1] The most recent election occurred on November 4, 2008, with the next one scheduled for November 6, 2012. The process is regulated by a combination of both federal and state laws. Each state is allocated a number of Electoral College electors equal to the number of its Senators and Representatives in the U.S. Congress.[2] Additionally, Washington, D.C. is given a number of electors equal to the number held by the smallest state.[3] U.S. territories are not represented in the Electoral College. Under the U.S. Constitution, each state legislature is allowed to designate a method of choosing electors.[2] Thus, the popular vote on Election Day is conducted by the various states and not directly by the federal government. Once chosen, the electors can vote for anyone, but – with rare exceptions like an unpledged elector or faithless elector – they vote for their designated candidates and their votes are certified by Congress in early January. The Congress is the final judge of the electors; the last serious dispute was in United States presidential election, 2000. The nomination process, including the primary elections and the nominating conventions, were never specified in the Constitution, and were instead developed by the states and the political parties. 你可以去Wikipedia网站搜索一下,
2023-01-02 16:28:312

戴姆勒-克莱斯勒AMAP面试经历时间表

戴姆勒-克莱斯勒AMAP面试经历(时间表) hr管这个叫challenging,exciting and interesting.. 先是时间安排。。。   7:30 am -   8:00 am    Arrival & sign-in   8:00 am -   9:00 am    Welcome & briefing   9:00 am -   1:05 pm    Assessment Center morning session   1:10 pm -   1:45 pm    Noon Break   1:45 am -   5:00 pm    Assessment Center afternoon session   5:10 pm -   5:30 pm    Closing plenary session   6:00 pm -   9:30 pm    Dinner   9:30 pm - 10:30 pm    Feedback to the candidates 虽然时间经过一定调整,但总而言之,还是比较紧张的。 candidates 一共12人,编号1-12,1-6为一个小组,7-12为另一个小组(我所在)。 考官,也就是observer一共7人,其中一个rotating HR manager(德方),两个中方的HR经 理,还有1中+3德or美 的sales&marketing的经理。编号为A-G 具体的时间安排比较复杂,每个人的都不同,大家分别在不同时刻进入不同方便寻找自己的 hr进行面试,然后进入下一阶段,每个candidate 都会在不同时间看到不同的hr,每个hr也 会在不同场合看到不同的candidate。只有午饭和晚饭是12个人聚在一起的`,此时7个hr也在 开会讨论我们的表现-_- 上午的session 包括一轮1vs1的interview和negotiation,共两个部分。 以我自己为例:9:00-9:30   interview1a  6candidates VS 6HR,不同房间(我所在)               9:45-10:15  interview1b  另一组其他6人面试               9:35-10:35   我准备negotiation,看paperwork,准备case study               10:35-11:05 Triadic Negotiation1 第1组,包括我  3candidates VS 7HR,不同房间,看运气,有可能一人搞定两个HR有可能搞定3个               11:15-11:45 Triadic Negotiation2 第2组,               11:55-12:25   Triadic Negotiation3 第3组,               12:35-1:05pmTriadic Negotiation4 第4组, 每个组都在上一组进行negotiation的时候接到paperwork的资料然后准备自己的case study,准备时间是一个小时,答辩时间半个小时。               进行完这两个部分应该是1:10pm,所有candidates 一起午饭,就是简单的 汉堡三明治。  但由于上午提前结束,不到1点我们就开始吃午饭了。               1:15重新集结,正式开始下午的session 下午的session包括teamwork,另一轮1vs1的interview和Group discussion三个部分 以我为例:    1:15-2:00 teamwork。 两个大组同时开始,内容后面再谈-_-               中间hr有事去开会,我们break了一下,大家坐一起聊天               3:10-3:40,其中6个人开始又一轮的1vs1 interview2了               3:50-4:20,另外一组开始  1vs1 interview2,(我所在)               4:35-5:40,Group discussion 12个人随机分成3组,同时进行,每组 2-3名面试官。               5:50-6:30pm, closing plenary session  1-12,A-G,together               7:00-10:30pm, observersconference 讨论去留问题               10:30pm-11:00, feedback to the candidates 这部分是1vs1,每个人都是 被早上第一轮interviewer给做feedback的。 日程安排大致如此,不知道是不是太乱。。。。内容随后。。。 ;
2023-01-02 16:28:391