cansonic z2

*lon1/180.0f; double radlong2 = pi*lon2/180.0f; if (RADLAT1 // South if (radlat1 > 0) radlat1 = PI/2-Fabs (RADLAT1); // North if (radlong1 //west if (Radlat2 // South if (radlat2 > 0) radlat2 = PI/2-Fabs (RADLAT2); // North if (radlong2 //West //spherical coordinates x=r*cos (AG) sin (at), Y=r*sin (AG) *sin (at), Z=r*cos (at) //zero AG is up so reverse lat Double x1 = er * cos (radlong1) * sin (RADLAT1); Double y1 = er * sin (radlong1) * sin (RADLAT1); Double z1

} theInlineintLcaintXinty) + { - if(deep[x]Deep[y]) swap (x, y); $ intt=deep[x]-Deep[y]; $ for(intI= -; ~i;i--) - if(t (1Fa[x][i]; - if(x==y)returnx; the for(intI= -; ~i;i--) - if(fa[x][i]!=Fa[y][i])Wuyix=fa[x][i],y=Fa[y][i]; the returnfa[x][0]; - } Wu BOOLJudgeintAintBintCintd) - { About if(Max (b) return 1; $ if(Max (b) >max (C,D))return 0; - if(min (b) return 1; - if(min (b) >min (C,D))return 0; - if(a>=b)return 1; A return 0; + } the

Link to original topicidea: water problem, direct search1#include"Map"2#include"Queue"3#include"math.h"4#include"stdio.h"5#include"string.h"6#include"iostream"7#include"algorithm"8 #defineABS (x) x > 0? X:-X9 #defineMax (a) a > b? A:bTen #defineMin (a) a One A using namespacestd; - - intZ2,x2,y2,n; the intd[6][3]= {{0,0,1},{0,1,0},{0,0,-1},{0,-1,0},{1,0,0},{-1,0,0}}; - BOOLmap[ the][ the][ the],vis[ the][ the][ the]; - - structnode{ + intZz,xx,yy; - intStep; + }; A at voidBfs (intZi

OutputTAKTAKNIENIEHINT SourceAcknowledgement Claris provides questionschange the original image to a dual graph, and then check the set1#include 2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include"algorithm"Ten using namespacestd; OnetypedefLong LongLL; A Const intmax=1505; - intn,k; - intfa[max*MAX]; the int Get(intXinty) { - if(x>0 x0 yN) - return(X-1) * (n1)+y; - Else return(n1) * (n1)+1; + } - intGetfather (intx) { + if(fa[x]==x) A returnx;

just relatively prime t Riples).The InputThe input consists of a sequence of positive integers, one per line. Each integer in the input file would be less than or equal to 1,000,000. Input is terminated by End-of-file.The OutputFor each of the integer N in the input file print, the integers separated by a space. The first integer is the number of relatively prime triples (such, component of the triple is). The second number is the number of positive integers, and is not a part of any triple who

PortalThere are n points in the three-dimensional space, and the sum of the smallest distances is obtained by a point that makes it to the minimum of the Manhattan distance of the N-point.Point (X1,Y1,Z1) to (X2,Y2,Z2) The Manhattan distance is |x1-x2| + |y1-y2| + |z1-z2|. That is, the sum of the absolute values of the 3-dimensional coordinate difference.InputLine 1th: number of points N. (2 2-n + 1 lines:

. ascending = ascending; }public int compare (file O1, file O2) { long z1 = Util.getzxidfromname (o1.getname (), prefix); long z2 = Util.getzxidfromname (o2.getname (), prefix); int result = Z1 Z2?-1: (z1 > z2 1:0); return ascending? Result:-result; }}/*** Sort the list of files. Recency as determined by the version component* of the file name.** @p

traffic data of a famous Webshell management tool "kitchen knife" that remotely controls the "kitchen knife horse" and sends instructions: We can see that the "kitchen knife" uses base64 to encrypt the command sent to the "kitchen knife Horse". Through analysis, we can observe two key payload z1 and Z2. z1=Y21k z2=Y2QgL2QgIkQ6XHd3d1xxaHJkd3dcIiZ3aG9hbWkmZWNobyBbU10mY2QmZWNobyBbRV0%3D You can obtain the d

x = r * cosa, y = r * sina So x1 = x * cosb-y * sinb Also available: y1 = y * cosb + x * sinb This is the formula for z rotation. The formula of x rotation and y rotation can be introduced in the same way. Summary: Given Point: (x, y, z) point after rotation around the x axis (x1, y1, z1) point after rotation around the y axis (x2, y2, z2) point (x3, y3, z3) after rotating around the Z axis x (x remains unchanged) x1 = x y1 = y * cosb-z * sinb z1 = z

Preface: /V * C ,~ $ X, A3 P3 J2 R! A; m, K + u: P: Q. x 'C' g/L $ A + V2] 7 S) L I. Version :? /R0 Z. D5 C6 V1 C (^ ! S % W1 C6 ^ 3 P/Q 2. Basic compilation % \/A % o $ T: i7 g ! Y3 C2 X) M4 _ + u I 3. SQLite operations6 o/J '1 X % _/L1 N _ * { K$ P T1 z G9 G-Q (1) Basic Process '[6 R3 N) {; \-G + @0 ^ % B) A2 I9 K(2) SQL statement operations(P) H V! V8 m'1 {8 Q. m )@;'8 y-T1 K) N # M ([(3) binary; q; P0 J 'x. m "g K/n-H9 A8 L; Z' \ 5 u(4) Transaction Processing) I + H # Prepare a $ t! @ 7

% @. m) E: Z; r copy file 8 J _ Z' y # n gcp parameter source file target file: Z4 m'r2 _ + O # L $ y, F example: the natit of the desktop. copy kext to the driver directory 7 I "W4 S4 S, C, A, F CP-r/user name/desktop/natit. kext/system/library/extensions8 J. R ^ "] 3 M $ Q ? The R parameter indicates recursive operation on the directory. kext looks like a file on the GUI, but actually a folder .! W7 @/~ "Y $ }% {! E. Back up all the files in the driver directory to the desktop backup! M) \]

three wires with the same material, diameter, and length. A 2 V voltage is added to a bridge circuit consisting of R14, R20, R15, Z1, pt100, and wire resistance. Z1, Z2, Z3, D11, D12, d83, and various capacitors play a filtering and protection role in the circuit. When static analysis is performed, it can be ignored. Z1, Z2, and Z3 can be considered as short circuits, d11, D12, d83, and each capacitor can

/account.ring.gz.......Then, rebalance the ring file:Swift-ring-builder account. Builder create 18 3 1Swift-ring-builder account. Builder add z1-192.168.1.50: 6002/SDC 100Swift-ring-builder account. Builder add z2-192.168.1.51: 6002/SDC 100Swift-ring-builder account. Builder add z3-192.168.1.52: 6002/SDC 100Swift-ring-builder account. Builder add z4-192.168.1.54: 6002/SDC 100Swift-ring-builder account. Builder add z5-192.168.1.53: 6002/SDC 100Swift-ri

I should have eaten this question. At first, the equation was wrong and I didn't want to think about the problem of connecting columns. I am really confused. If I can't do it, I should repeat it, fuck, I gave N groups of X, Y, and Z and asked if there are two or more IDs. Yes, the value of ID % x is between the range [y, z, if yes, the output is cannot take off. Otherwise you know Based on the meaning of the question: A * X1 + Y1 B * X2 + y2 If there is a solution, there must be duplicates be

Numerical calculation refers to the algorithm in the field of numerical analysis. Numerical analysis is a special study of data problems related to numerical and approximate values, and numerical calculation plays a particularly important role in the study of numeric analysis.polynomial interpolation is a method of calculating the approximate value of a function . Where the function value is known only on a few points.The basis of this algorithm is to establish an interpolation polynomial pn (z)

# + #class Son (Father): - #def Big Health (self): the #print (' I want big Health care ') * #def Eat Chicken (self): $ ## Super (son,self). Eat chicken () #执行父类中的方法Panax Notoginseng #Print ("I'm not eating chickens.") I hate to eat chicken ") - #Father. Eating Chicken (self) the # + # A #A=son () the # + #A. Eat chicken () - $ $ - #str () - #class Bar: the #def __init__ (self,name,age,gender): - #" "Wuyi #constructor method, the class name plus attribute automatically executes the construct

In Week 5, the job requires supervised learning (suoervised learning) to recognize Arabic numerals through a neural network (NN) for multi-classification logistic regression (multi-class logistic REGRESSION). The main purpose of the job is to feel how to find the cost function in the NN and the derivative value of each parameter (THETA) in its hypothetical function (GRADIENT derivative) (using backpropaggation).The difficulty is not high, but the problem is that you have to get used to the Malab

Calculate the distance method according to two point longitude and latitude: -(double) LatitudeAndLongitudeWithLon1: (double) lon1 LAT1: (double) lat1 Lon2: (double) lon2 lat2: (double) lat2{Double er = 6378137; 6378700.0f;Ave. Radius = 6371.315 (Someone said more accurate is 6366.707)Equatorial radius = 6378.388nautical mile = 1.15078Double RADLAT1 = m_pi*lat1/180.0f;Double radlat2 = m_pi*lat2/180.0f;Double radlong1 = m_pi*lon1/180.0f;Double radlong2 = m_pi*lon2/180.0f;if (RADLAT1 if (radlat1

Total time limit: 1000ms memory limit: 65536kB The description gives the n points in the three-dimensional space (no more than 10), the distance between the N-point 22, and the distance between the two points and the distance from large to small. The input input includes two lines, the first row contains an integer n representing the number of points, and the second row contains the coordinates of each point (the coordinates are all integers). The coordinates of a point are 0 to 100, and there i

can write Where the configuration in Zookeeper1 is as follows #默认配置 ticktime=2000 initlimit=10 synclimit=5 #需要修改的配置 datadir=d:/mq/apache-activemq/ cluster/zkdata/z1/ clientport=2181 server.1=localhost:2888:3888 server.2=localhost:2889:3889 server.3=localhost:2890:3890 The configuration in Zookeeper2 is as follows #默认配置 ticktime=2000 initlimit=10 synclimit=5 #需要修改的配置 datadir=d:/mq/apache-activemq/ cluster/zkdata/z2/ clientport=2182 server.1=localhost