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paint brush default style */Public void setpaintdefaultstyle () {mypaint. setantialias (true); mypaint. setcolor (color. red); mypaint. setstyle (paint. style. stroke); mypaint. setstrokewidth (3);}/** circle * @ Param x coordinate * @ Param YY coordinate * @ Param radius circle radius */Public void drawcircle (float X, float y, float radius) {mycanvas. drawcircle (X, Y, radius, mypaint); invalidate ();} /*** draw a straight line * @ Param fromx start point X coordinate * @ Param fromy start po

) Similarly, the coordinates of the H point can be obtained. Coordinate C: To simplify the calculation, we set the N point to the AG point. In this way, the triangle CJF and EHF are obtained: Cx = ex-EF/2; The coordinates of the cpoint are: (ex-EF/2, mheight) Similarly, the J-point coordinate is obtained. The following derivation requires a lot of mathematical knowledge. If you do not remember the shoes, you can consciously review them ~~ The function of a straight line is: Y =

Source: hdoj 1700 Mathematical Proof] If R is the circle radius, A, B, and C are the angles of triangles, A/Sina = B/sinb = C/sinc = 2R is obtained from the sine theorem.Therefore, A + B + C = 2R (SINA + sinb + sinc). When r determines that the circumference A + B + C is the largest, the sum is the maximum of SINA + sinb + sinc.Set y = SINA + sinb + sinc, first fixed angle,Y = SINA + sinb + sinc = SINA + 2sin (B + C)/2) Cos (B-C)/2)= SINA + 2cos (A/2) Cos (B-C)/2)Because a is a fixed value, onl

in the string, otherwise split will not search for the corresponding character and the entireThe string is returned as a unique list element, which returns the original string.Set y [Split 7/4/1776 "/"]Puts "We celebrate on the [lindex $y 1] ' th day of the [lindex $y 0] ' th month\n 'Split is the split character that separates a string, followed by "/".Example⑤: Learning from foreachforeach is the traversal function, and the foreach command/control structure iterates through the list, dropping

Title Link:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1511The main topic: In a 8*8 board, give you a starting position and an end location, but also give you a trap location, ask you from the beginning to bypass the trap to the end of the shortest distance. (You can walk up and down or you can walk diagonally)Problem-solving ideas: can be directly with the search, can also use mathematical knowledge to solve, we have learned before, two points directly between the shortest, so when the tra

can only be used for sequences or data frames, and a one-dimensional array is a without this method.Here you customize a function that summarizes all of these statistical description metrics together:def stats(x): return pd.Series([x.count(), x.min(), x.idxmin(), x.quantile(.25), x.median(), x.quantile(.75),x.mean(), x.max(), x.idxmax(), x.mad(), x.var(), x.std(), x.skew(), x.kurt()],index=[‘Count‘, ‘Min‘, ‘Whicn_Min‘, ‘Q1‘, ‘Median‘, ‘Q3‘, ‘Mean‘, ‘Max‘,‘Which_Max‘, ‘Mad‘, ‘Var‘, ‘Std‘, ‘

queue is to create a temporary directory node/queue/x (i), and then each member gets/queue All directory nodes of the directory, that is, X (i). Determine if the value of I is already the number of members, and if it is less than the number of members waiting for/queue/start to appear, create/queue/start if it is already equal.Product flowchartApplication examplesIcon Explanation APP1,APP2,APP3,APP4 is a 4 independent business system. ZK1,ZK2,ZK3 is the 3 connection point of the zo

intMainvoid) One { A inti,j,k,p,q; - intx1,x2,x3; -scanf"%d%d",p,q); the { - for(i=0; i3*n; i++) - { -quan[i]=1; +dian[i]=i; - } + intCnt=0; A while(q--) at { -scanf" %d%d%d",x1,x2,x3); - intxx1,yy1; - intXx2,yy2; - intXx3,yy3; - for(xx1=x2;dian[xx1]!=xx1;) inxx1=dian[xx1]; - for(yy1=

title:http://acm.hdu.edu.cn/showproblem.php?pid=1496Test instructionsConsider equations has the following form:A*x1^2+b*x2^2+c*x3^2+d*x4^2=0 A, B, C, D is integers from the interval [ -50,50] and any of the them cannot is 0.It is consider a solution a system (X1,X2,X3,X4) that verifies the equation, Xi is an integer from [ -100,100] and Xi! = 0, any i∈{1,2,3,4}.Determine how many solutions satisfy the given

Triangle Area time limit:MS | Memory limit:65535 KB Difficulty:2 Describe give you three points, representing the three vertices of a triangle, now your task is to find out the area of the triangle. Input each row is a set of test data, with 6 integer x1,y1,x2,y2,x3,y3 representing the horizontal ordinate of three points, respectively. (Coordinate values from 0 to 10000) Input 0 0

Research discuzx3 source code analysis sequence nbsp; someone on the Internet said that discuz source code analysis from. include/common. inc. php file, but I downloaded discuz nbsp; x3, and none of them found this file... Nbsp; nbs research discuz x3 source code analysis sequence Some people on the Internet say that the discuz source code analysis starts from the. include/common. inc. php file, but I

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# Data AcquisitionX1 = round (runif (100, min = 80, max = 100 ))X2 = round (rnorm (100, mean = 80, SD = 7 ))X3 = round (rnorm (100, mean = 80, SD = 18 ))X3 [which (X3> 100)] = 100Num = seq (2005138101, length = 100)X = data. Frame (Num, x1, x2, X3)Write. Table (x, "grade.txt ")# Data AnalysisY = read. Table ("grade.txt

// The radius can be easily obtained using the cosine and sine formula! # Include "iostream" # include "cmath" # include "cstdio" using namespace STD; const double Pi = 3.141592653589793; int main () {double X1, Y1, X2, Y2, x3, Y3; double A2, B2, C2, R; while (CIN> x1> Y1> X2> Y2> X3> Y3) {a2 = (x3-x2) * (x3-x2) + (y3-

both sidesDealwithleft (temp1,temp2,result,list);Dealwithleft (temp2,temp1,result,list);//Pay attention to each time you want to bring results with you to store the result set return result;} private static void Dealwithleft (point p1,point p2,linkedlist result,linkedlist list) {Recursive processing p1,p2 the left point of the ray compositionIterator it = List.iterator (); Find the highest point on the left PmaxPoint Pmax = null;int max = 0;while (It.hasnext ()){Point next = (point) it.next ();

The code is as follows:1 varCanvas=document.getElementById ('my'), CTX=Canvas.getcontext ('2d');2 　　 SetInterval (function(){3 　　　　 Ctx.clearrect (0,0, -, -);4 　　　　 Ctx.save ();5 Ctx.translate ( $, $);6 　　　　 varci= -, Pi=Math.PI/CI, x1=Ten0, y1=0, X2=0, y2=0, X3=0, Y3=0;9 　　　　 Ctx.beginpath ();Ten 　　　　 for(varI=ci*2; I>0; I--){ One 　　 Ctx.rotate (PI); A 　　　　　　 Ctx.moveto (x1,y1); - 　　　　　　 y2=X1*Math.sin (PI); - 　　　　　　 X2=X1*Math.Cos (PI); t

calculate the frequency of each sentence by counting the sentences appearing in the dataset, the number of occurrences C (x), and the total number of sentences in the dataset. X∈s,p (x) = C (x)/n indicates the frequency of X, apparently σp (x) = 1.In summary, we can find a few questions:1) The above language model only exists theoretically, when the training data set is infinitely large, the frequency of the data set can be infinitely close to the actual probability in the grammar;2) for most s

column K in B is summed in the above calculation to obtain the data in line K of column C, and the code is as follows: void arymul1 (int a[4][5], int b[5][3], int c[4][3]) { int i, j, K; int temp[3] = {0}; for (i = 0; i This algorithm can easily go to the multiplication algorithm of sparse matrices. PS: implementation of Strathearn algorithmStrathearn method, a method proposed by V. Strathearn in 1969. We first discuss the method of calculating the second-order matrix.For sec

size is not reduced. The size is 35x35, the number of channels increases, and the number of 4 branch channels and 64 + 64 + 96 + 32 = 256, the final output tensor size is 35x35x256. 1st Inception Module group: 2nd Inception modules, named Mixed_5c. Step 1: padding model SAME. Four branches, and the fourth branch is connected to 64 output channels with 1x1 convolution. The size of the output tensor is 35x35x288. 1st Inception Module group: 3rd Inception modules, named Mixed_5d. The size of the o

correspondence relationship. The number of bar codes that meet the conditions is the number of k tuples. That is, the number of integer solutions of the equation Σ (I = 1 .. k) xi = n, 1 ≤ xi ≤ m. The easiest way to think of this is to search for algorithm 1: Since the value range of xi has been determined, we can leave all xi values empty and then check how many groups of solutions meet the requirements of Σ (I = 1 .. k) xi = n. The program is easy to write, but the complexity is very high. Fo