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left-side function partial_left (f) {var args = arguments; return function () {var a = array (args, 1); a =. concat (array (arguments); return f. apply (this, a) ;};}// pass the real parameter of the function to function partial_right (f) {var args = arguments; return function () {var a = array (arguments); a =. concat (array (args, 1); return f. apply (this, ); };} // This function real parameter is used as a template. // The undefined value in the real parameter list is filled with the actual

There is a statistical requirement that requires IP address classification statistics. Mysql providesInet_aton () and inet_ntoa ()Function process processing.YesView official manual. INET_ATON (expr)Returns an integer that represents the network address of a string. The address can be 4 or 8 bits. Mysql> select inet_ntoa (3507806248 );+ ----------------------- +| Inet_ntoa (3507806248) |+ ----------------------- +| 209.20.224.40 |+ ----------------------- +1 row in set (0.00 sec) The number is a

: filetype: Get type; Is_file: Determine if it is a normal file; Is_link: Determine if it is a link. There are several functions for obtaining properties: file_exists: Determine whether a file or directory exists, FileSize: Get file Size, is_readable, is_writable, is_executable: Readable, writable, executable; Filectime, Filemtime, fileatime: Get the file creation time (create), modify time (modify), Access time (access), return time stamp; stat: Get some basic information about the file, Return

adjust the position of the CD two points, and use principle 1 to calculate the hybrid bad plus the hybrid DAC (after the position is adjusted, the Hybrid CAD is used ). The problem is when we need to adjust the CD position. In principle 1, there is no intersection between the points in the polygon and the points on the polygon and the boundary of the polygon, the graph has an intersection. Therefore, if the first Vertex a is connected to any vertex D on the map that goes out of the BC two point

is the largest common factor of N and REM (m%n) ..., then, only the maximum common factor of N and REM is the maximum common factor of M and N to allow this recursion to go on. So the question is, why is the largest common factor of N and REM the largest common factor of M and n?Idea: Suppose m > N, and even if M rem = m%n-M-zn (z = m/n), when Rem=0,n is the largest common factor of both; Rem>0,rem=xa-zya, obviously rem%a = 0The largest common factor of M and N is the largest common factor of N

then we ask for the Fresnel value ,Float Fresnel = saturate (Pow (1-dot (Viewdir, Halfvec), 5.0));The Fresnel value can be changed dynamically by our external _fresnelvalFresnel + = _fresnelval * (1-fresnel);Edge Illuminationfloat rim = saturate (Pow (1-dot (Viewdir, s.blurrednormals), _rimpower)) * Fresnel;First reflection High light calculation ,Dot product of the linear and half-width vectorsfloat specb

There is a statistical need for IP classification statistics, MySQL provides Inet_aton () and Inet_ntoa () function process processing.The official manual can be viewed in detail. Inet_aton (expr)Gives a point address representation of a network address as a string that returns an integer representing the numeric value of the address. The address can be a 4-or 8-bit address. Mysql> Select Inet_ntoa (3507806248);+-----------------------+| Inet_ntoa (3507806248) |+-----------------------+| 209.2

position of the vector to hold the core points, boundary points, noise pointsVectorVectorVector4 Defining vectors to preserve the resulting clustersvectorIv. Source Code#include #include #include #include #include using namespace Std;#define EPS 3//eps as neighborhood radius value#define MINPTS 3//Neighborhood density threshold#define N 20//datasets contain n objectsDouble point[n][2];//to save all data pointsVectorVectorVectorvectorvectorInitialize n coordinate pointsvoid init (int n) {Srand (

;Double sum = 0;For (I = 1; pow (n, I)/fac (I)> 0.000001; I + = 2){Sum + = pow (n, I)/fac (I) * k;K =-1 * k;}Return sum;}// Cos () function definitionDouble cos (double m){Int I;Int k = 1;Double sum = 0;For (I = 0; pow (m, I)/fac (I)> 0.000001; I + = 2){Sum + = pow (m, I)/fac (I) * k;K =-1 * k;}Return sum;} /* Start th

+ s. b> s. c s. a + s. c> s. B s. B + s. c> s. a) // The condition for forming a triangle is that the sum of any two sides is greater than the third side. { Double r = (s. a + s. B + s. c)/2; Area = sqrt (r * (r-s.a) * (r-s. B) * (r-s.c); // use the Helen formula to calculate the area Printf ("the Triangle area is % g", area ); Return area; } Else { Printf ("cannot form a triangle! "); Return area; } } // ================================================ ========== /** * 7. Determine wheth

many, but many English abbreviations I only know what it is, impetuous mentality has made me do not know what its full name is. For example, "Wi-Fi", I believe most of the mobile phone users can not leave it, but do not know what its full name, a bit of English-based people (such as i) probably can guess "wi" is the abbreviation of "Wireless", but almost do not care about "Fi" is how to come. Degree Niang later know, originally "fi" Is "Fidelity" (fidelity) abbreviation, so "Wi-Fi" full name is

There is a statistical requirement that requires IP address classification statistics. Mysql providesInet_aton () and inet_ntoa ()Function process processing.YesView official manual. INET_ATON (expr)Returns an integer that represents the network address of a string. The address can be 4 or 8 bits. Mysql> select inet_ntoa (3507806248 );+ ----------------------- +| Inet_ntoa (3507806248) |+ ----------------------- +| 209.20.224.40 |+ ----------------------- +1 row in set (0.00 sec) The number is a

greater than 6, it is carried;3. when the number to be modified is equal to 5, it depends on the number before 5. If it is an odd number, it is carried. If it is an even number, it is rounded off, that is, the number at the end of the repair is an even number; if there is not any number after 5 as "0", then whether the front of 5 is an odd or even number, it should be carried. For example, use the above rules to retain three valid numbers for the following data: 9.8249 = 9.82, 9.82671 = 9.83 9.

buffer to the receive ring of the port. If it is added more, the buffer added to a port receive ring cannot be shared by other ports, resulting in waste. If it is added less, it may cause instant buffer depletion and packet loss. It's hard to balance. The simplest thing is to waste more memory. In octeon, pip applies for free buffer only after receiving the message. In Pip, only a small amount of free is prefetch. Buffer, so free buffer can wait for all ports and all hardware and software units

';Default:return ' unknown type ';}} FileSize returns the data in bytes, if it is a large file number or large, you can deal with the number first, the code is as follows Copy the Code code as follows:Working with file sizesfunction GetSize ($path = ", $size =-1) {if ($path!== null $size = =-1) {//only path is calculated size, or it can be processed only by the number$size = FileSize ($path);}if ($size >= pow (2, 40)) {Return round ($size/

have not passedPublic:Void addcity (int city );Int tabu [iCityCount];Void Clear ();Void UpdateResult ();Double m_dLength;Double m_dShortest;Void move ();Ant ();Void move2last ();}; Void ant: move2last (){Int I;For (I = 0; I If (AllowedCity [I] = 1 ){Addcity (I );Break;}} Void ant: Clear (){M_dLength = 0;Int I;For (I = 0; I Prob [I] = 0;AllowedCity [I] = 1;}I = tabu [iCityCount-1];M_iCityCount = 0;Addcity (I );} Ant: ant (){M_dLength = m_dShortest = 0;M_iCityCount = 0;Int I;For (I = 0; I Allowed

Algorithm 1: time complexity O (N) double pow(double x, long n){ if(n==0) return 1; /*when n=0 whatever the x is,the pow is 1*/ if(n==1) return x; /*when n=1 the result is x itself*/ else { double tempa=1.0000; while(n--) tempa=tempa*x; return (tempa); /*every time make n-- then multiply a xto get the result*/ }} Algorithm 2: time complexity O (log (n )) double

This article mainly introduces about PHP according to latitude and longitude sorting, according to latitude and longitude to filter the distance segment, has a certain reference value, now share to everyone, the need for friends can refer to SQL statements: Select Location.* from (select *,round (6378.138*2*asin (sqrt () (Pow () (Sin (36.668530*pi ()/180-px_lat*pi ()/180)/2), 2) +cos (36.668530*pi ()/180) *cos (Px_lat*pi ()/180) *

capitalized, False is returned. the previously used format () function and evaluate the string length len () function all belong to the string function library, help (str. replace) can be used to query specific function usage. This function is used to replace the old string with the new string. # Determine whether the string is in lowercase >>> str1 = "Eastmount" >>> str1.islower () False # string replacement >>>> str2 = 'adfababdfab '>>> str2.replace (' AB ', 'AB') 'adfababdfab '# string leng

The largest argument. in [+]: Max (' Hsdhsjd ', ' 5687 ', ' a ') #最长的一个Out []: ' HSDHSJD '3. Get length Len ()in [+]: s= ' 1234 ' in [[]: Len (s) #取字符创的长度Out []: 4help (len) to the built-in function Len in module __builtin__:len (...) Len (object), Integer #处理对象, returns the integer return the number of items of a sequence or collection. in [+]: Len ([up]) #取列表的元素个数, 2 out[31]: 2 in [Max]: Len ((' A ',)) #取元祖的元素个数, 1 out[30]: 1In [+]: Len ({' A ': 3, ' d ': 4}) #len () Statistical dictionaries