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代写赚钱平台_网络兼职日结_赚钱资讯

0 人参与  2019-09-20 23:03  分类 : 网络兼职日结  点这评论

代写赚钱平台今天是中国一年一度的传统七夕佳节,这也是万千情侣撒狗粮的专属日子,在《逆水寒》里,情缘们已经开始结伴过七夕了。七夕这一天对单身青年来说,就像在柠檬树下吃柠檬一样,对于没有情缘的呢,今天也是个找寻伴的日子,所以很多找寻情缘的内容也应运而生:

就像上文中动媒体平台介绍的那样,软文代写的需求主要分为几个方面:最简单的就是评论这种会打字的人就能写,不过评论一写就是几十上百条,所以如果你想写的不重样其实很难,但是如果只是凑数,那么也是挺好写的;下来是SEO优化的文章,这种稿件不太要求逻辑,也不太在乎有创意,字数大概就600左右,需要注意的就是符合关键词的布局和原创。

动媒体小编接下来介绍的就是新闻稿,这种一般要依据某一个事件来撰写,软文代写人员用词要严谨,结构要清晰,表达的要有条理;再下来就是KOL文,这种可能是发微信公众号也可能是发今日头条、百家号等自媒体平台,这种稿件一般就是要有创意,写的生动活泼,能激发读者的传播欲;最后一种就是那种商业评论文、调查报告这一类的,这种在动媒体平台小编看来最难,因为它不仅要对行业熟悉还要有一定的商业知识,要有很多的数据证明,还要查阅很多的资料,只有这样软文代写人员才可以写出重要的内容。

日常工作中,并不是每个人都具备一定的写作能力,写好一篇演讲稿需要一定的文笔能力、知识积累能力以及思维能力,然而演讲稿又是公开演讲的过程,意味着更加重视演讲稿的写作内容,这也是很多人想到代写演讲稿,写作水平高、写作能力强以及擅长写演讲稿的团队,只要支付一些写作酬劳,就能获取一篇优秀的演讲稿,何乐而不代写呢?并且演讲过程的阵阵掌声,相信多少钱都是有价值的。【阿甘文案】专业代写各类演讲稿总结等。

2、专业写作团队

function TEST_CODE2()R = 10 / 1000;H = 20 / 1000;%number of elements in z directionN_z=5;%number of elements in r directionN_r=10;%number of total elementsN_E=(N_z)*(N_r);%number of nodes in z directionNN_z=(N_z)+1;%number of nodes in r directionNN_r=(N_r)+1;% number of total nodesNN=NN_z*NN_r;%r = zeros(NN, 1);theta = zeros(NN, 1);z = zeros(NN, 1);IEN = zeros(N_E*4, 1);%a=(2*pi*R)/(N_r);b=H/(N_z);c=(2*pi)/(N_r);N_z_index = 1;z_value = 0;%r_value = 0;theta_value = 0;for i = 1:1:NN% r(i) = r_value;z(i) = z_value;theta(i) = theta_value;if N_z_index >= NN_zN_z_index = 1;z_value = 0;%r_value = r_value + a;theta_value = theta_value + c;elsez_value = z_value + b;N_z_index = N_z_index + 1;endendIEN_count = 1;IEN_index = 1;for i = 1:1:N_Etemp = 4*(i-1);IEN(4*(i-1) + 1) = IEN_index;IEN(4*(i-1) + 2) = IEN_index + NN_z;IEN(4*(i-1) + 3) = IEN_index + NN_z + 1;IEN(4*(i-1) + 4) = IEN_index + 1;if IEN_count == N_zIEN_index = IEN_index + 2;IEN_count = 1;elseIEN_index = IEN_index + 1;IEN_count = IEN_count + 1;endendh = zeros(N_E, 1);h_theta = zeros(N_E, 1);mu=1;c=2;e=3;for i = 1:1:N_EF1 = 0;k = 0;F2 = 0;F = 0;for xi=(-1/sqrt(3)):(2/sqrt(3)):(1/sqrt(3))for eta=(-1/sqrt(3)):(2/sqrt(3)):(1/sqrt(3))NodeNum1 = IEN(4*(i-1) + 1);NodeNum2 = IEN(4*(i-1) + 2);NodeNum3 = IEN(4*(i-1) + 3);NodeNum4 = IEN(4*(i-1) + 4);theta1 = theta(NodeNum1);theta2 = theta(NodeNum2);theta3 = theta(NodeNum3);theta4 = theta(NodeNum4);z1 = z(NodeNum1);z2 = z(NodeNum2);z3 = z(NodeNum3);z4 = z(NodeNum4);h1=c+e*cos(theta1);h2=c+e*cos(theta2);h3=c+e*cos(theta3);h4=c+e*cos(theta4);h = (h1+h2+h3+h4)/4;h1_theta = e*cos(theta1)-e*sin(theta1);h2_theta = e*cos(theta2)-e*sin(theta2);h3_theta = e*cos(theta3)-e*sin(theta3);h4_theta = e*cos(theta4)-e*sin(theta4);h_theta = (h1_theta+h2_theta+h3_theta+h4_theta)/4N1=0.25*(1-xi)*(1-eta);N2=0.25*(1+xi)*(1-eta);N3=0.25*(1+xi)*(1+eta);N4=0.25*(1-xi)*(1+eta);N=[N1,0,N2,0,N3,0,N4,0;0,N1,0,N2,0,N3,0,N4];dN1_xi=0.25*(-1+eta);dN1_eta=0.25*(-1+xi);dN2_xi=0.25*(1-eta);dN2_eta=0.25*(-1-xi);dN3_xi=0.25*(1+eta);dN3_eta=0.25*(1+xi);dN4_xi=0.25*(-1-eta);dN4_eta=0.25*(-1-xi);dN_xi = zeros(2,8);dN_xi(1,1) = dN1_xi;dN_xi(1,2) = 0.0;dN_xi(1,3) = dN2_xi;dN_xi(1,1) = 0.0;dN_xi(1,1) = dN3_xi;dN_xi(1,1) = 0.0;dN_xi(1,1) = dN4_xi;dN_xi(1,8) = 0.0;dN_xi(2,1) = 0.0;dN_xi(2,2) = dN1_xi;dN_xi(2,3) = 0.0;dN_xi(2,1) = dN2_xi;dN_xi(2,1) = 0.0;dN_xi(2,1) = dN3_xi;dN_xi(2,1) = 0.0;dN_xi(2,8) = dN4_xi;dN_eta = zeros(2,8);dN_eta(1,1) = dN1_eta;dN_eta(1,2) = 0.0;dN_eta(1,3) = dN2_eta;dN_eta(1,1) = 0.0;dN_eta(1,1) = dN3_eta;dN_eta(1,1) = 0.0;dN_eta(1,1) = dN4_eta;dN_eta(1,8) = 0.0;dN_eta(2,1) = 0.0;dN_eta(2,2) = dN1_eta;dN_eta(2,3) = 0.0;dN_eta(2,1) = dN2_eta;dN_eta(2,1) = 0.0;dN_eta(2,1) = dN3_eta;dN_eta(2,1) = 0.0;dN_eta(2,8) = dN4_eta;J11=-1*(1-eta)*theta1+(1-eta)*theta2+(1+eta)*theta3-(1+eta)*theta4;J12=-1*(1-eta)*z1+(1-eta)*z2+(1+eta)*z3-(1+eta)*z4;J21=-1*(1-xi)*theta1-(1+xi)*theta2+(1+xi)*theta3+(1-xi)*theta4;J22=-1*(1-xi)*z1-(1+xi)*z2+(1+xi)*z3+(1-xi)*z4;J = zeros(2,2);J = 0.25*[J11, J12;J21,J22];detJ=J11*J22-J21*J12;%invJ=[J22,-1*J12;% -1*J21,J11]/detJ;dN_theta=(J22*dN_xi-J12*dN_eta)/detJ;dN_z=(-1*J12*dN_xi+J11*dN_eta)/detJ;F1_xi_eta=(1/R)*(dN_theta)*(dN_theta);F2_xi_eta=(dN_z)*(dN_z);%h=c+e*cos(theta);% Gauss integral method% initializek = k + detJ*(h(i)^3)*(F1_xi_eta)*(F2_xi_eta)/(12*mu);V_theta1 = R*(theta1)/2;V_theta2 = R*(theta2)/2;V_theta3 = R*(theta3)/2;V_theta4 = R*(theta4)/2;V_theta = (V_theta1+V_theta2+V_theta3+V_theta4)/4V_z = (e)/2;F1 = F1+detJ*(h(i))*((F1_xi_eta)*(V_theta)+(F2_xi_eta)*(V_z));F2 = F2+detJ*(h_theta(i))*N;F = F1-F2;endP = F/k;endendy = StiffnessAssemble(K,k,NodeNum1,NodeNum2,NodeNum3,NodeNum4);end% change local to global%nodes i,j,m,nfunction y=StiffnessAssemble(K,k,i,j,m,n)K = zeros(NN*2,NN*2);K(2*i-1,2*i-1)= K(2*i-1,2*i-1) + k(1,1);K(2*i-1,2*1)=K(2*i-1,2*i)+k(1,2);K(2*i-1,2*j-1)=K(2*i-1,2*j-1)+k(1,3);K(2*i-1,2*j)=K(2*i-1,2*j)+k(1,4);K(2*i-1,2*m-1)=K(2*i-1,2*m-1)+k(1,5);K(2*i-1,2*m)=K(2*i-1,2*m)+k(1,6);K(2*i-1,2*n-1)=K(2*i-1,2*n-1)+k(1,7);K(2*i-1,2*n)=K(2*i-1,2*n)+k(1,8);K(2*i,2*i-1)=K(2*i,2*i-1)+k(2,1);K(2*i,2*i)=K(2*i,2*i)+k(2,2);K(2*i,2*j-1)=K(2*i,2*j-1)+k(2,3);K(2*i,2*j)=K(2*i,2*j)+k(2,4);K(2*i,2*m-1)=K(2*i,2*m-1)+k(2,5);K(2*i,2*m)=K(2*i,2*m)+k(2,6);K(2*i,2*n-1)=K(2*i,2*n-1)+k(2,7);K(2*i,2*n)=K(2*i,2*n)+k(2,8);K(2*j-1,2*i-1)=K(2*j-1,2*i-1)+k(3,1);K(2*j-1,2*i)=K(2*j-1,2*i)+k(3,2);K(2*j-1,2*j-1)=K(2*j-1,2*j-1)+k(3,3);K(2*j-1,2*j)=K(2*j-1,2*j)+k(3,4);K(2*j-1,2*m-1)=K(2*j-1,2*m-1)+k(3,5);K(2*j-1,2*m)=K(2*j-1,2*m)+k(3,6);K(2*j-1,2*n-1)=K(2*j-1,2*n-1)+k(3,7);K(2*j-1,2*n)=K(2*j-1,2*n)+k(3,8);K(2*j,2*i-1)=K(2*j,2*i-1)+k(4,1);K(2*j,2*i)=K(2*j,2*i)+k(4,2);K(2*j,2*j-1)=K(2*j,2*j-1)+k(4,3);K(2*j,2*j)=K(2*j,2*j)+k(4,4);K(2*j,2*m-1)=K(2*j,2*m-1)+k(4,5);K(2*j,2*m)=K(2*j,2*m)+k(4,6);K(2*j,2*n-1)=K(2*j,2*n-1)+k(4,7);K(2*j,2*n)=K(2*j,2*n)+k(4,8);K(2*m-1,2*i-1)=K(2*m-1,2*i-1)+k(5,1);K(2*m-1,2*i)=K(2*m-1,2*i)+k(5,2);K(2*m-1,2*j-1)=K(2*m-1,2*j-1)+k(5,3);K(2*m-1,2*j)=K(2*m-1,2*j)+k(5,4);K(2*m-1,2*m-1)=K(2*m-1,2*m-1)+k(5,5);K(2*m-1,2*m)=K(2*m-1,2*m)+k(5,6);K(2*m-1,2*n-1)=K(2*m-1,2*n-1)+k(5,7);K(2*m-1,2*n)=K(2*m-1,2*n)+k(5,8);K(2*m,2*i-1)=K(2*m,2*i-1)+k(6,1);K(2*m,2*i)=K(2*m,2*i)+k(6,2);K(2*m,2*j-1)=K(2*m,2*j-1)+k(6,3);K(2*m,2*j)=K(2*m,2*j)+k(6,4);K(2*m,2*m-1)=K(2*m,2*m-1)+k(6,5);K(2*m,2*m)=K(2*m,2*m)+k(6,6);K(2*m,2*n-1)=K(2*m,2*n-1)+k(6,7);K(2*m,2*n)=K(2*m,2*n)+k(6,8);K(2*n-1,2*i-1)=K(2*n-1,2*i-1)+k(7,1);K(2*n-1,2*i)=K(2*n-1,2*i)+k(7,2);K(2*n-1,2*j-1)=K(2*n-1,2*j-1)+k(7,3);K(2*n-1,2*j)=K(2*n-1,2*j)+k(7,4);K(2*n-1,2*m-1)=K(2*n-1,2*m-1)+k(7,5);K(2*n-1,2*m)=K(2*n-1,2*m)+k(7,6);K(2*n-1,2*n-1)=K(2*n-1,2*n-1)+k(7,7);K(2*n-1,2*n)=K(2*n-1,2*n)+k(7,8);K(2*n,2*i-1)=K(2*n,2*i-1)+k(8,1);K(2*n,2*i)=K(2*n,2*i)+k(8,2);K(2*n,2*j-1)=K(2*n,2*j-1)+k(8,3);K(2*n,2*j)=K(2*n,2*j)+k(8,4);K(2*n,2*m-1)=K(2*n,2*m-1)+k(8,5);K(2*n,2*m)=K(2*n,2*m)+k(8,6);K(2*n,2*n-1)=K(2*n,2*n-1)+k(8,7);K(2*n,2*n)=K(2*n,2*n)+k(8,8);y=K;end& 本团队核心人员组成主要包括BAT一线工程师,精通德英语!我们主要业务范围是代做编程大作业、课程设计等等。我们的方向领域:window编程 数值算法 AI人工智能 金融统计 计量分析 大数据 网络编程 WEB编程 通讯编程 游戏编程多媒体linux 外挂编程 程序API图像处理 嵌入式/单片机 数据库编程 控制台 进程与线程 网络安全 汇编语言 硬件编程 软件设计 工程标准规等。其中代写编程、代写程序、代写留学生程序作业语言或工具包括但不限于以下范围:C/C++/C#代写Java代写IT代写Python代写辅导编程作业Matlab代写Haskell代写Processing代写Linux环境搭建Rust代写Data Structure Assginment 数据结构代写MIPS代写Machine Learning 作业 代写Oracle/SQL/PostgreSQL/Pig 数据库代写/代做/辅导Web开发、网站开发、网站作业ASP.NET网站开发Finance Insurace Statistics统计、回归、迭代Prolog代写Computer Computational method代做因为专业,所以值得信赖。如有需要,请加QQ:99515681 或邮箱:99515681@qq.com 微信:codehelp QQ:99515681 或邮箱:99515681@qq.com 微信:codehelp

SIT221 Data Structures and Algorithms Trimester 2, 2019 Practical Task 6.2 (Credit Task) Submission deadline: 10:00am Monday, September 2 Discussion deadline: 10:00am Saturday, September 14 General Instructions In this task, answer all the following questions and complement each answer with a detailed explanation. 1. After hearing about a Binary Max‐Heap data structure, Johnny English decides to implement it for his supper‐duper algorithm. However, as he was not paying enough attention in class, his implementation of the DownHeap procedure (also widely known in the literature as Max‐Heapify) is not absolutely correct. He gave us the following pseudocode of his implementation: Function JE_DownHeap(index x):If ( H[Left(x)].key > H[x].key ){Swap elements H[x] and H[Left(x)]JE_DownHeap(Left(x))}ElseIf ( H[Right(x)].key > H[x].key ){Swap elements H[x] and H[Right(x)]JE_DownHeap(Right(x))}Your task is to give an example of a max‐heap for which the above JE_DownHeap procedure fails. More specifically, you need to propose a heap where the root 1 is the only node breaking the essential Max‐Heap Property so that running the above procedure JE_DownHeap(1) does NOT result in a valid max‐heap. Use your instance to clearly illustrate how such heap may look like. You must draw the heap before and after running JE_DownHeap(1). Note that your instance can be made quite small. 2. Assume now that Johnny English has eventually fixed the aforementioned problem and developed a correct implementation of a max‐heap that provides all the standard operations such as: ‐ DeleteMax, Insert(K), IncreaseKey(X, K), in a logarithmic log time; and ‐ Max, in a constant 1 time, where is the number of stored elements. Johnny English wants to add a new capability to this max‐heap .Specifically, he would like to implement a DeleteElement(X) operation that removes element H[x] from the max‐heap consisting of elements in a logarithmic log time. Note that the H[x] does not need to be the maximum‐key element in .Also, naturally, has to maintain the Max‐Heap Property after deletion of H[x]. Your task is to help Johnny English by providing a pseudocode of such a DeleteElement(X) operation with a brief analysis of its running time as function of elements. 3. Johnny English knows that the maximum element of a binary max‐heap can be obtained by means of Maxoperation in a constant 1 time. He however wants you to develop an efficient method (in pseudocode) that finds the maximum element in log time, where 1 Remember that both insertion and removal in a binary max‐heap of size takes a log time. SIT221 Data Structures and Algorithms Trimester 2, 2019 Further Notes You may find exploring the content of chapter 9.3 of the course book “Data Structures and Algorithms in Java” by Michael T. Goodrich, Roberto Tamassia, and Michael H. Goldwasser (2014) useful to solve these tasks. You may access the book on‐line for free from the reading list application in CloudDeakin available in Resources Additional Course Resources Resources on Algorithms and Data Structures Course Book: Data structures and algorithms in Java. Marking Process and Discussion To get this task completed, you must finish the following steps strictly on time: Submit your answers to the task via OnTrack submission system. You may submit a hand‐written and then scanned document, but ensure that the text is very clear to read. Note that this is a theoretical task, thus we do not expect you to write any program code. Meet with your marking tutor to explain your solutions. When the solutions are hand‐written, do not forget to bring them with you. Cloud students must record a short video explaining their work and use a sort of white‐board, e.g. a graphical editor, or a scanned document with the answers. Answer all additional (theoretical) questions that your tutor may ask you. Questions are likely to cover lecture notes, so attending (or watching) lectures should help you with this compulsory interview part. Please, come prepared so that the class time is used efficiently and fairly for all the students in it. You should start your interview as soon as possible as if your answers are wrong, you may have to pass another interview, still before the deadline. Use available attempts properly. Note that we will not check your solution after the submission deadline and will not discuss it after the discussion deadline. If you fail one of the deadlines, you fail this task that may impact your performance and your final grade in the unit. Unless extended for all students, the deadlines are strict to guarantee smooth and on‐time work through the unit. Remember that this is your responsibility to keep track of your progress in the unit that includes checking which tasks have been marked as completed in the OnTrack system by your marking tutor, and which are still to be finalised. When marking you at the end of the unit, we will solely rely on the records of the OnTrack?system and feedback provided by your tutor about your overall progress and quality of your solutions. 本团队核心人员组成主要包括BAT一线工程师,精通德英语!我们主要业务范围是代做编程大作业、课程设计等等。我们的方向领域:window编程 数值算法 AI人工智能 金融统计 计量分析 大数据 网络编程 WEB编程 通讯编程 游戏编程多媒体linux 外挂编程 程序API图像处理 嵌入式/单片机 数据库编程 控制台 进程与线程 网络安全 汇编语言 硬件编程 软件设计 工程标准规等。其中代写编程、代写程序、代写留学生程序作业语言或工具包括但不限于以下范围:C/C++/C#代写Java代写IT代写Python代写辅导编程作业Matlab代写Haskell代写Processing代写Linux环境搭建Rust代写Data Structure Assginment 数据结构代写MIPS代写Machine Learning 作业 代写Oracle/SQL/PostgreSQL/Pig 数据库代写/代做/辅导Web开发、网站开发、网站作业ASP.NET网站开发Finance Insurace Statistics统计、回归、迭代Prolog代写Computer Computational method代做因为专业,所以值得信赖。如有需要,请加QQ:99515681 或邮箱:99515681@qq.com 微信:codehelp QQ:99515681 或邮箱:99515681@qq.com 微信:codehelp

⑱伯希和根据他的研究兴趣和经验,把选取文书的重点放在汉文以外的少数语种的语言资料和佛典以外的汉文资料上,至于佛典,则选那些有题记的写卷。

非洲肯尼亚尼耶利(Nyeri)的一名大学毕业生玛莉·姆布瓜(Mary Mbugua)最近正因缴纳学费和承担房租经历经济难关。她努力出去找工作,还尝试起卖保险,但她一单也没卖出去,只能每天坐在酒吧前台里发愁。

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