motorola z2

#define PI 3.1415926double LantitudeLongitudeDist(double lon1,double lat1, double lon2,double lat2){double er = 6378137; // 6378700.0f;//ave. radius = 6371.315 (someone said more accurate is 6366.707)//equatorial radius = 6378.388//nautical mile = 1.15078double radlat1 = PI*lat1/180.0f;double radlat2 = PI*lat2/180.0f;//now long.double radlong1 = PI*lon1/180.0f;double radlong2 = PI*lon2/180.0f;if( radlat1 if( radlat1 > 0 ) radlat1 = PI/2 - fabs(radlat1);// northif( radlong1 if( radlat2 if( radla

Hdu_3624 This question requires that the volume of the part that covers three or more times be obtained. If we discretize the zcoordinate, if a part of each layer Z is overwritten by three or more times, it is equivalent to that the projection of this layer Z on the XY plane is covered by three or more times, therefore, for each layer of Z, We can first find the area covered by the projection three or more times, then multiply the height of the layer Z, that is, the layer Z is covered by thr

Intersection of cubesTime Limit: 500 ms memory limit: 32768kb 64bit Io format: % LLD % LlUSubmitStatusPracticeLightoj 1211DescriptionYou are given n cubes, each cube is described by two points in 3D space: (x1, Y1, Z1) being one corner of the cube and (X2, Y2, Z2) being the opposite corner. assume that the sides of each of the cubes are parallel to the axis. your task is to find the volume of their intersection.InputInput starts with an integer T (≤

);// Compute the angleFloat fdot = d3dxvec3dot ( m_vstart, m_vend );Float angle = ACO (fdot );// AxisD3dxvector3 vaxis;D3dxvec3cross ( vaxis, m_vstart, m_vend );M_qcurrent = d3dxquaternion (vaxis. X, vaxis. Y, vaxis. Z, angle );}Void getrotationmatrix (d3dxmatrix * pmatrot){D3dxmatrixrotationquaternion (pmatrot, m_qcurrent );};PRIVATE:Void maptosphere (int x, int y, d3dxvector3 * PVEC){Int nradius = (m_rect.width ()> m_rect.height ())? M_rect.width (): m_rect.height ();// Translate to [0...

quaternion: getmatrix () const{Float X2 = x * X;Float y2 = y * Y;Float Z2 = z * z;Float xy = x * Y;Float xz = x * z;Float YZ = y * z;Float wx = W * X;Float WY = W * Y;Float WZ = W * z; // This calculation wocould be a lot more complicated for non-unit length Quaternions// Note: the constructor of matrix4 expects the matrix in column-Major format like expected// OpenGLReturn matrix4 (1.0f-2.0f * (y2 + Z2),

We know that the coordinates of the triangle vertices are (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), (x4, y4, z4 ), you can use either of the following two methods to calculate the volume of a single triangle: 1. Use the hybrid product of Vectors Set the three-way amount passing through a vertexA,B,CWhich of the following is the volume of the expected triangle? | (A×B)·C|/6. Assume that (x1, y1, z1) is a four-dimensional vertex A = (x2-x1, y2-y1,

Gaussian noise # code: myhaspl@myhaspl.comimport cv2import numpy as npfn = "test2.jpg" myimg = cv2.imread (fn) img = cv2.cvtColor (myimg, cv2.COLOR _ BGR2GRAY) param = 30 # grayscale range grayscale = 256 w = img. shape [1] h = img. shape [0] newimg = np. zeros (h, w), np. uint8) for x in xrange (0, h): for y in xrange (0, w, 2): r1 = np. random. random_sample () r2 = np. random. random_sample () z1 = param * np. cos (2 * np. pi * r2) * np. sqrt (-2) * np. log (r1)

from the previous operation, and then insert only the newly opened node to record the modified node, because one insertion only affects log2n nodes, so the total space complexity is O (nlog2n), At the same time, the time complexity is O (nlog2n) because the point at which a pointer is inserted is LOG2N, and the complexity of each insertion time is as long as it is two points apart. The subject can not be discretized, but I am more counseling so it is still a bit of separation of space.Code:1#in

1#include 2 voidSortint*PA,int*PB) {//A , B is arranged in order from large to small (using pointers to achieve "bidirectional" delivery)3 intT;4 if(*papb) {5t=*pa;*pa=*pb;*pb=T;6 }7 }8 intGcd_1 (intAintb) {//seeking greatest common divisor with the method of dividing9 intC;Ten if(b!=0){ One while(a%b!=0){ AC=b; -b=a%b; -A=C; the } - } - returnb; - } + intGcd_2 (intAintb) {//the recursive method of dividing the Euclidean - intC; + if(b==0) c=A

class object is eating"); - } the } - - classZiextendsfu{ - intx = 20; + PublicZi (String name) { - Super(name);//calling the parent class with a method for constructing a parameter +System.out.println ("Zi class with parameter construction method"); A } at PublicZi () {//the JVM defaults to calling the parent class's parameterless construction method -System.out.println ("Method of constructing Zi class without parameters"); - } - Public voidprint () { -

), two switches share a ZigBee node "Z1" (equivalent to two keys on the development Board, which should not be difficult to understand), they together control the right three light bulbs (also on the same node-"Z2"). Z1 's Key1 controls the lamp1 of Z2, while Key2 controls Z2 lamp2 with Lamp3.There are several concepts that have not been seen: Cluster ID, Binding

This semester has been to follow up on the Coursera Machina learning public class, the teacher Andrew Ng is one of the founders of Coursera, machine learning aspects of Daniel. This course is a choice for those who want to understand and master machine learning. This course covers some of the basic concepts and methods of machine learning, and the programming of this course plays a huge role in mastering these concepts and methods.Course Address https://www.coursera.org/learn/machine-learningThe

C # array intersection, difference set, and Union set, C # array intersection, difference set, and Union set At work, this knowledge is often used to check which fields are required and which are forbidden. 1 using System; 2 using System.Collections.Generic; 3 using System.Linq; 4 using System.Text; 5 6 namespace Collection 7 { 8 class Program 9 {10 static void Main(string[] args)11 {12 int[] x = { 6,7,8,9,10};13 int[] y = { 4,5,6,7,8};14 15

("%.2f+%.2fi", a-c,b-d); if(s== ' * ') System.out.printf ("%.2f+%.2fi", a*c-b*d,a*d+b*c); if(s== '/') System.out.printf ("%.2f+%.2fi", (a*c+b*d)/(C*c+d*d), (b*c-a*d)/(c*c+d*d)); System.out.println (); }}-----------------------About plural--------------------------------------The addition of complex numbers is carried out according to the following rules : set z1=a+bi,z2=c+di is any two complex number,Then their and is (A+BI) + (C+di) = (a+c) + (b+

public static string Mypf_function () {var lee = 0; float D1 = 0, d2 = 0; var ret = "Multipatch does not intersect!! "; Space two multipatch p1[0] = new Polytopepoint (0, 0, 1); P1[1] = new Polytopepoint (0, 0, 2); P1[2] = new Polytopepoint (2, 2, 2); P1[3] = new Polytopepoint (2, 2, 1); P2[0] = new Polytopepoint (2, 1, 0); P2[1] = new Polytopepoint (0, 1, 0); P2[2] = new Polytopepoint (0, 1, 3);

From collections import Defaultdictfrom collections Import ORDEREDDICTD = Defaultdict (list) d[' a '].append (1) d[' a ']. Append (2) d[' B '].append (5) print (d) #di = Dict (list) #di [' B '].append (one) #di [' B '].append (a) #di [' C '].append () #print (di) ds = Defaultdict (set) ds[' B '].add (one) ds[' B '].add (All) ds[' C '].add (in) print (ds) do = Ordereddict () do[' foo '] = 1do[' Far '] = 2do[' coo '] = 5for key in Do:print (key, Do[key]) # Dictionary of operations prices = {' IBM

// calculate the distance between two coordinatesDouble Distancewithcoord (double lon1,double lat1,double lon2,double lat2){Double er = 6378137;Double RADLAT1 = pi*lat1/180.0f;Double radlat2 = pi*lat2/180.0f;Double radlong1 = pi*lon1/180.0f;Double radlong2 = pi*lon2/180.0f;if (RADLAT1 if (radlat1 > 0) radlat1 = PI/2-Fabs (RADLAT1);//Northif (radlong1 if (Radlat2 if (Radlat2 > 0) radlat2 = PI/2-Fabs (RADLAT2);//Northif (Radlong2 Double x1 = er * cos (radlong1) * sin (RADLAT1);Double y1 = er * sin

lengthen the display. The w parameter can be added a few more, up to three can be added to display longer process information. 3. Kill The kill command can terminate the process and then connect the process number. 4. Free Free displays the memory usage of the system. -B,-k, and-m parameters indicate memory usage in bytes, kilobytes, and megabytes. 5. Vmstat Run the vmstat 2 command to display a line of system information every two seconds, including CPU usage, memory usage, and disk IO. With i

[Cpp]// Calculate the ry + least Spanning Tree// Precision// If the distance between the two units is small and the sum of the radius, the distance is 0.// Conversely, the sum of distance-radius# Include # Include # Include # Include Using namespace std;Const int MAXN = 101;Const int MAXM = 500000;Int n;Int M;Double cost;Double eps = 0.0000001;Struct edge{Int u;Int v;Double w;} Edges [MAXM];Struct unit{Double x;Double y;Double z;Double r;} Units [MAXN];Int cmp (const void * a, const void * B){Ed

After Google began to release HCE support, a huge response was triggered. Get rid of the dependence of SWP-SIM on operators, get rid of the dependence of eSE on mobile phone manufacturers, its impact on the industrial ecology is self-evident. The prospects and application discussions based on HCE on the Internet are not satisfactory, but there is little technical information. Now we will reveal the HCE through a specific implementation and give a technical understanding. First, the required envi