v vpn

Simple collation of the next, there are all, good chaos ... TLD is the abbreviation of an algorithm, which is called tracking-learning-detection by the original author. Author website Link: http://personal.ee.surrey.ac.uk/Personal/Z.Kalal/ Paper: (1) tracking-learning-detection (2) Forward-backward Error Automatic Detection of Tracking failures (3) Online learning of robust object detectors during unstable tracking Source Code Resources: 1. Original author Zdenek Kalal Author's homepage: http://

Reprint Address: http://blog.chinaunix.net/uid-191945-id-2792153.html Address "width=320 height=40 type=audio/x-pn-realaudio-plugin controls=" ControlPanel "loop=" true "autostart=" true " Volume= "initfn=" Load-types "mime-types=" Mime.types ">address " width=310 height=35 type=audio/x-pn-realaudio-plugin loop= "true" autostart= "true" controls= "Controlpanel,statusbar" >address " width=250 height=25 type=

Beginner programming problems-General Linux technology-Linux programming and kernel information. The following is a detailed description. Recently, I developed an application project in linux. System A is connected to subsystems B and C through Ethernet. B is responsible for sending some commands to A. After A breaks down the commands, it notifies C to execute them. For A, there is A common processing scenario. 1. after receiving the instruction from B, A sends the instruction to C. after rec

http://poj.org/problem?id=2478Test instructions: Given a number x, the value of the Euler function of the number of Solution: The data range is relatively large, like the poj1248 of the same practice is not feasible.First of all, we need to understand several properties and inferences of Euler's function: (Today we have a long discussion with Konjak, a good friend.) ）Inference (i): Phi (P^k) = (p-1) *p^ (k-1) Prove: For N=p^k, a number of positive integers less than or equal to N, a

for(intk=0; kk) { theLL PS = hm[idx].state[k], pn =Hm[idx].num[k]; the intSL = Get_state_at (PS, j), Su = Get_state_at (PS, j+1); the if(SL = =0 Su = =0){ - if(!city[i][j]) {//extend the state to a non-. Place inHm[cur].push (Set_state_at (PS, J,0,0), PN); the}//the plug should point to a blank grid the Else if(city[i][j+1

Linear Fitting: for the form of Y = A * x + BA = (N * Σ Xi * Yi-Σ Xi * Σ Yi)/(n * Σ Xi * Xi-(Σ xi) 2)B = (Σ Xi * xi) * (Σ Yi)-(Σ xi) * (Σ Xi * Yi)/(n * Σ Xi * Xi-(Σ XI) 2)MATLAB built-in functions can be used to achieve:Fitting Function: Pn = polyfit (X, Y, n) returns the PN coefficient vector, descending order, and N is the order.Function: yy = polyval (Pn, x)

label: style blog HTTP color Io OS AR for strong POJ 1845 - Sumdiv ( 数论 + 唯一分解定理 + 快速幂取模 ) 这是一道数论的好题，需要较好的数学基础 题意: 给定A,B，求A^B的所有因数的和，再MOD 9901 分析： 这里用到了数论当中相当一部分知识 a. 唯一分解定理 任何一个整数都可以分解为若干个素数的幂的乘积的形式 A = ( p1 ^ q1 + p2 ^ q2 + ..... + pn ^ qn ) p为素数 A^B = ( p1 ^ (q1*B) + p2 ^ (q2*B) + ..... + pn ^ (qn*B)) b. 约数和公式 Sum( A ) = ( 1 + p1 + p1^2 + .... + p1^q1 ) * ( 1 + p2 + p2^2 + .... + p2^q2 ) * ...... *

Description B June recently heard of a very amazing nature, set P=1+i p=1+i, for each Gaussian integerX+yi x + yi, we can all find a non-negative integer set S satisfying ∑k∈spk=x+yi () \sum _{k∈s}{p^k} = X+yi ()And all you have to do is output the set of s in the case of input x and Y.If you do not understand the above topic, we formally give the following definition.is for a complex number (Complex number), if his real and imaginary parts are integers, we call them Gaussian integers.The S col

. serialization;Using system. runtime. serialization. formatters. Binary;Using system. runtime. serialization. formatters. Soap;Using system. runtime. serialization;Using system. IO;Namespace multiserialize{Public partial class form1: Form{// Declare a string variable to store user input values in the MethodString N, NN, P, M, FN, success;Public form1 (){Initializecomponent ();}Private void xmlbtn_click (Object sender, eventargs E){// Assign the user input value to the variableN = This. nm. text

wireless networks, pilot pollution becomes a problem that needs special attention and needs to be solved. 1. Definition of pilot pollution In CDMA wireless networks, pilot pollution is divided into pilot phase pollution and pilot intensity pollution. The pilot pollution we usually mention is usually guided by the Frequency Intensity pollution. 1.1 Pilot phase pollution In the forward channel of the CDMA wireless network system, different PN offsets a

This problem examines the formula theorems of three number theory:Integer unique decomposition theorem: a= (P1^K1) * (P2^K2) **...* (PN^KN) (P (i) is prime)factors and formulas (known as a= (P1^K1) * (P2^K2) **...* (PN^KN)), then the sum of all the factors of a is: sums = (1+P1+P1^2+...P1^K1) * (1+P2+P2^2+...P2^K2) *...* (1+PN+

carefully, you will find that the number of stones is relatively small. We can compress this status into an octal number, in this way, there can be a maximum of 8*8*8*8 states, few. The status has been initially completed, but we found that there is a problem not solved. How can we avoid the same color or quantity as the previous group? Let's further refine the status. Add two integers, pc and pn, and record the color of the stones in front and the n

Park algorithm CodeTraining sequence structure t=[\ (c\) \ (d\) \ (c^{*}\) \ (d^{*}\)], where C represents a sequence of symbols obtained by N/4 transformation of a complex pseudo-random sequence of length Pn,ifft\ (C (n) = D (n/4-n) \)Original explanation: The training symbol is produced by transmittingA real-valued PN sequence on the even frequencies, while zerosis used on the odd frequencies. This means,

relationship between them, so we need to change the idea, starting with the decomposition decomposition, assuming two numbers a and B: A = p1^e1 * p2^e2 * ... * pn^en, B = p1^d1 * P2^D2 * ... * pn^dn , b divided by a can be expressed as: b/a = (P1^D1 * p2^d2 * ... * pn^dn)/(P1^E1 * p2^e2 * ... * pn