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content is less than 1G Pause>nul The second program, can display the current folder under the level directory, but do not cycle every time the size of the folder, the program is as follows: @echo off setlocal enabledelayedexpansion (for/d%%a in (c:\kugou\*) do ( set/a n=z=0 pushd "%%a" For%%b in (*.*.) Do ( set/a n+=1,z+=%%~zb ) set/a "z>>=20" popd [echo;%%a!n! M]) > Statistical results. txt Pause Through the above two programs, I integrated a bit, intend to

samples, each sample has p variables, which form the geographic data matrix of the NXP order.When P is large, it is troublesome to investigate the problem in the P-dimensional space. In order to overcome this difficulty, it is necessary to carry out dimensionality reduction treatment, that is, to replace the original more variable index with fewer comprehensive indexes, and to make these less comprehensive indexes reflect the information reflected by the original more variable index as much as

same as the column name of the result set of the first SELECT statement in the Union operator. The result set column name of another SELECT statement is ignored.Two different usages are union and unionall, the difference being that the union deletes duplicate rows from the result set. If you use UnionAll, all rows will be included and duplicate rows will not be deleted. The difference between union and UnionAll:Union Check RepeatUNION ALL do not checkFor example, select ' A ' union select ' A '

§1 the direction of the space line 1. Direction angle The angle between the straight OM and the three axis through the Origin o α,β,γ is called the direction angle of the line namely: Α=∠mox,β=∠moy,γ=∠moz 2. Direction cosine The cosine of the direction angle of the line is called the direction cosine. L=cosα=x/ρ, m=cosβ=y/ρ, n=cosγ=z/ρ which ， 3. Direction number If L is any line in space, and a straight line om,om from the Origin o to a straight line L, the coordinate of the point W is (P,q,r)

for (int i = 0; i TEMP[J] + = m[j][i] * V[i];memcpy (R, temp, sizeof (VEC));} Float Transformlength (Mat m, float R) {Return Sqrtf (m[0][0] * m[0][0] + m[0][1] * m[0][1] + m[0][2] * m[0][2]) * r;} Float sphere (Vec C, float R) {FLOAT dx = c[0]-sx, DY = c[1]-sy;float a = dx * dx + dy * dy;Return a } Float opunion (float z1, float z2) {Return z1 > Z2? Z1:Z2;} Floa

') , ' W ') r = Re.compile (' [\x80-\xff]+ ') m = r.findall (TXT) dict={} z1 = Re.compile (' [\x80-\xff]{2} ') z2 = Re.compile (' [\x80-\xff]{4} ') z3 = re.compile (' [\x80-\xff]{6} ') z4 = Re.compile (' [\x80-\xff]{8} ' For I in M: x = I.encode (' gb18030 ') i = Z1.findall (x) #i + = Z2.findall (x) #i + = Z2.findall (x[2 :] #i + + z3.findall (x) #i + +

Rethinking on the area division of the first type curved surface @ (Calculus) Some problems, look complex, but very good solution. In the same way, some problems look simple, but they are hard to do. Give an example of the calculation of the area of the first type of curvature. The solution of the first type curved area based on three things to do: projection generation calculation There is no logical order between three things, and whoever wants to do it first. The goal is to be converted into

Longest Common subsequence and public subsequence DescriptionSubsequence refers to the sequence obtained after several (0) elements are deleted from a sequence. To be accurate, if the given sequence X = (x1, x2, ·, xm), then the other sequence Z = (z1, z2, ·, zk ), the subsequence of X refers to the existence of a strictly incrementing subscript sequence (i1, i2, ·, ik) so that for all j = 1, 2 ,···, k has zj = xij. For example, sequence Z = (B, C, D,

#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