straight ruler

The first type of code: less CodeThrough regular expression to obtain the real address of the Baidu network disk file, to achieve the effect of straight chain Save the following code as downbd.php Copy CodeThe code is as follows: $canshu =$_server["Query_string"]; if ($canshu = = "") { Die ("file does not exist"); } Else { $wangzhi = "Http://pan.baidu.com/share/link?". $canshu; $file =file_get_contents ($wangzhi); $pattern = '/a>/i '; Preg_match_al

To find the distance from the point to the straight line:Double dis (point p1,point p2){if (Fabs (p1.x-p2.x) {return Fabs (P2.X-PEGX);}Else{Double k= (P2.Y-P1.Y)/(p2.x-p1.x);Double b=p2.y-k*p2.x;return Fabs (k*pegx-pegy+b)/sqrt (k*k+1);//return distance}}Determine whether a polygon is convexIf the hole contains any protrusions (i.e), if there exist any of interior points in the hole so, if connected by a Lin e segment, that segment would cross one or

// histogram 1 const// straight-side Figure 2 int Method// contrast methods ); Method has Cv_comp_correl, Cv_comp_chisqr,cv_comp_intersect,cv_comp_bhattacharyya four kinds of methods, The corresponding formula is as follows: Practice: Comparing the histogram histogram of two images with different illumination conditions is mainly used to judge the matching degree of two images, we test the following two image histogram c

DescriptionThere are n linear l1,l2 on the Xoy Cartesian plane,... Ln, if the Y value is positive infinity down, you can see a sub-segment of Li, it is said that Li is visible, otherwise li is covered.For example, for straight lines: l1:y=x; L2:y=-x; L3:y=0Then L1 and L2 are visible, and L3 are covered.gives a line of N, expressed as a form of y=ax+b (| a|,| b|InputThe first behavior n (0 OutputFrom small to large output visible line number, 22 is sep

Through regular expression to obtain the real address of the Baidu network disk file, to achieve the effect of the straight chain, suitable for some use outside the chain website. /I '; Preg_match_all ($pattern, $file, $result); $tempurl =implode ("", $result [1]); $FILEURLT =str_replace ("", "" ", $tempurl); $fileurl =str_replace ("", "", $FILEURLT); Header ("Location: $fileurl"); }?> As the effect: http://www.bkjia.com/PHPjc/371957.html www.bk

Portal: Intersecting LinesTest instructions: Gives the n set of lines, each group of 2 straight lines, to find out whether the line intersects. If the collinear outputs line, the intersection enters point coordinates, otherwise the output is none.Analysis: Template naked title, directly on the template ...#include #include#includestring.h>#include#include#include#include#includeSet>#includestring>#includeusing namespacestd;Const DoubleEPS = 1e-8;Const

Introduction: This is a Data Structure Algorithm(PhP description) insert a detailed page of the sort straight binary sort in a half-fold manner, introducing PHP, related knowledge, skills, experience, and some PHP source code. Class = 'pingjiaf' frameborder = '0' src = 'HTTP: // biancheng.dnbc?info/pingjia.php? Id = 341542 'rolling = 'no'> 1 2 /**3 * Half-fold insert sorting straight binary sort4 *5 * pr

Question: give n straight lines, divide a plane into multiple parts, and then give m the coordinates of the server guard. In one part, only one server guard can protect this part, ask if the M server guard provided protects all parts. Solution: Calculate the positional relationship between each server guard and each line (above or below), expressed in, so each server guard has a 01 string, finally, the number of parts divided into the n

Andrew Stankevich Contest 4 -- Driving Straight, stankevichcontest Question connection Question:Enter m and n, followed by M-1 rows and 2 * n-1 columns. ''Indicates that it cannot pass, '+' can pass, '-', '|' only allows one direction to pass (in the same direction as the symbol ), calculate the shortest path from the lower left to the upper right corner and output the result: when outputting the result, try to maintain the previous motion directio

POJ 1269 Intersecting Lines (determining the intersection of straight Lines) Address: POJ 1269 You can directly set the template... I really don't want to write the template myself... It stinks and grows .... However, you must first determine whether there is a straight line vertical X axis. The Code is as follows: #include #include #include #include #include #i

IOS 2D plotting (Quartz 2D) path (point, straight line, dotted line, curve, arc, elliptical, rectangle) A path can contain one or more shapes and subpaths. quartz provides many convenient shapes that can be called directly. For example, point, line, Arc, Curves, Ellipse, and Rectangle ). You can perform stroke or fill on these paths, or use path to intercept a region (clip ). For example, you can intercept a circular area. If you still have no ide

/* Obviously, if A and B are on the same side of a straight line, min (PA + Pb) = DIS (a, B); otherwise, obtain B's symmetric point on the straight line B ', min (PA + Pb) = DIS (A, B ') */# include

Work encountered in the flow of the project, need to draw a straight line with arrows, after groping, solve;(1) two points (P1,P2) determine a line, with one end of the line (assuming P2) as the origin, set an angle(2) the vector P2P1 (p) is obtained at the origin of the P2, the vector p rotates theta angle to get the vector P1, the vector p rotates-theta angle to get the vector P2(3) Stretch vector to length, translate variable to end of line(4) Now

The example in this article tells you how to draw a straight line in PHP. Share to everyone for your reference. The implementation methods are as follows: Copy Code code as follows: 1. Create Canvas $im = Imagecreatetruecolor (300,200);//Create a new true color image, the default background is black, and return the image identifier. Another function, imagecreate, is deprecated. 2, to draw the desired image $red = Imagecolorallocate ($im, 25

§ 1 Introduction of complex elements, intersection of quadratic curves and straight lines IComplex elements on the plane A Cartesian coordinate system {O; I, j} is created on the plane. The concept of vertices on the plane is expanded as follows: any pair of ordered plural numbers (x, y) it is the coordinate of a point p on the plane. If x and y are all real numbers, p is called the real point. Otherwise, p is called the virtual point, and the real po

Poj3819 Coverage (percentage of the cross between a straight line and a circle), poj3819coverage Question: I will give you a straight line and several circles, and find out the proportion of the length of the circle and the straight line to the whole line. Solution: calculate the percentage of the Cross-circle between the circle and the

1. Mathematical Analysis 1) draw a straight line I thought it would be easy to draw a straight line. Just take a straight line formula and traverse X to find Y. Isn't that true. Take the 2D line as an example, because the 3D line only introduces a zcoordinate. Key problem: the straight lines we have learned in mathemat

EventsVtkinteractoreventrecorder *recorder = Vtkinteractoreventrecorder::new ();Recorder->setinteractor (Iren);Recorder->readfrominputstringon ();Recorder->setinputstring (Tsweventlog);Test on OFF mechanismipw->setenabled (0);Spline->enabledoff ();ipw->setenabled (1);Spline->enabledon ();Test Set Get Handle positionsDouble pos[3];int i;For (I=0;i{Spline->gethandleposition (I,pos);Spline->sethandleposition (I,pos);}Test Closed on OffSpline->closedon ();Spline->closedoff ();Render the imageIren->

This rough question is still considered to be the intersection of a line segment and a line segment, but the subject requires that the line segment is completely within the rectangle. Therefore, it can be changed to the intersection problem between the straight line of the target line segment a and each line segment of the rectangle, and then check whether the line segment a is within the rectangle range. [Cpp]# Include Using namespace std;Struct Poi

= -SinA; SinD= -CosC; Cosd=SinC; Break; default: } //Cross point to draw line linesDrawLine (Point, [Point[0]+c*cosd, Point[1]+c*sind], {color: ' Red '}); DrawLine (Point, [point[0]+c* (-COSD), point[1]+c* (-sind)], {color: ' Red '}); varPoint_v_1 = point[1]-line[1][1]; varpoint_h_1 = point[0]-line[1][0]; varpoint_dist_1 = Math.sqrt (Math.pow (point_v_1,2) +math.pow (point_h_1,2));//point to line up a little distance varPoint_v_0 = point[1]-line[0][1]; varPoint_h_0 = point[0]-line[0][0