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Hi, here is the implementation of Triple DES on Java card. /* Package handson_crypto_des; import javacard. Framework. *; import javacard. Security. *; import javacardx. crypto. *;/***** class handson * */ Public Class Handsoncryptodes Extends Javacard. Framework. Applet { // Globals Secret ey secret ey; cipher ciphercbc; Final Short Dataoffset = (Short ) Iso7816.offset _ CDATA; Static Byt

The meaning of covarianceReproduced in: http://bbs.mathchina.com/cgi-bin/topic.cgi?forum=5topic=14444 (thanks to the original author)In probability theory, two random variables X and Y are related to each other in roughly the following 3 cases:When the joint distribution of x, Y is like that, we can see that there are roughly: the larger the X, the greater the X, the smaller the Y, the less the case we call " positive correlation ".When the joint distribution of x, Y is like that, we can see tha

; o.root.ondragend = new function ();nbsp; O.ROOT.O Ndrag = new Function ();nbsp;},nbsp; nbsp; Start:function (e) nbsp; {nbsp; var o = drag.obj = this;nbsp; E = drag.fi XE (e);nbsp; var y = parseint (o.vmode? o.root.style.top:o.root.style.bottom);nbsp; var x = parseint (o.hmode? o.root.s Tyle.left:o.root.style.right);nbsp; O.root.ondragstart (x, y);nbsp; nbsp; o.lastmousex = e.clientx;nbsp; O.lastmousey = e.clienty;nbsp; nbsp; if (o.hmode) {nbsp; if (O.minx!= null) O.minmousex = E.clientx-x + o.

. minmousey =-o. Maxy + E. clienty + Y;} Document. onmousemove = drag. Drag;Document. onmouseup = drag. end; Return false;}, Drag: function (E){E = drag. fixe (E );VaR o = drag. OBJ; VaR ey = E. clienty;VaR EX = E. clientx;Var y = parseint (O. vmode? O. Root. style. Top: O. Root. style. Bottom );VaR x = parseint (O. hmode? O. Root. style. Left: O. Root. style. Right );VaR NX, NY; If (O. minx! = NULL) EX = O. hmode? Math. Max (ex, O. minmousex): Math.

N points, M edge, the edge is divided into a, b two classes, to construct a minimum spanning tree, and the number of a side of the tree is K.We can control the number of a edges in the tree by adding a weight dx to all a edges. Obviously, when the DX increases, the A-side number KK will decrease.Two-minute DX,When kk>=k, increase the DX (i.e. l=mid+1) while updating the Ans=sum (MST)-mid*k;When kk1#include 2#include 3#include 4 using namespacestd;5 structnode6 {7 intu,v,w;8}

;o.xMapper = fXMapper ? fXMapper : null;o.yMapper = fYMapper ? fYMapper : null;o.root.onDragStart= new Function();o.root.onDragEnd= new Function();o.root.onDrag= new Function();},start : function(e){var o = Drag.obj = this;e = Drag.fixE(e);var y = parseInt(o.vmode ? o.root.style.top : o.root.style.bottom);var x = parseInt(o.hmode ? o.root.style.left : o.root.style.right );o.root.onDragStart(x, y);o.lastMouseX= e.clientX;o.lastMouseY= e.clientY;if (o.hmode) {if (o.minX != null)o.minMouseX= e.cli

A very classical problem, to be done by the method of division and treatment.Each time the chessboard is divided into 4 pieces, for a chessboard that does not have a blank lattice, an L-domino is used to construct the blank lattice.#include #include#include#include#include#include#defineLL Long Longusing namespacestd;intnum;intgrid[ -][ -];inlineBOOLJudgeintSxintSyintSizeintDxintdy) { return(sx1) (sy1);}voidDfsintSxintSyintSizeintDxintdy) { if(size==1)return; intlen=size/2; ints; //Top le

,bx];y=[ay,by];hold on; Plot (x,y, ' linewidth ', 2); hold off; cx=ax+ (Bx-ax)/3; cy=ay-d; dx=bx-(Bx-ax)/3; dy=by-d; ay=ay-d; by=by-d; Cantor (AX,AY,CX,CY); Cantor (Dx,dy,bx,by); End Run Cantor (0,5,5,5), the following legend appears: Recursive algorithm of Koch curve On a unit length line segment, the middle segment is replaced by a equilateral that removes the bottom edge, and then repeats the above operation on each line, so that the fractal curve Koch cu

$ (' img '). On ("Touchstart", function (E) {//e.preventdefault (); E.stoppropagation (); var El=this;var me=$ (This), $ ("#tip"). Text ("in Touchstart"); var t=e.touches[0]; Console.log ("touchstart!,mx,my= (", T.pagex, ",", T.pagey, "), el.x,y= (", El.offsetleft, ",", El.offsettop, ")"); Me.data ("MX", T.pagex); Me.data ("My", t.pagey);

Java source code: {code...} Source: www. cnblogs. commailin... Java source code: Import java. security. security; import javax. crypto. cipher; import javax. crypto. secretKey; import javax. crypto. spec. secretKeySpec; public class ThreeDES {private static final String Algorithm = "DESede"; // defines the encryption Algorithm. DES, DESede, Blowfish // keybyte can be used as the encryption key, the length is 24 bytes // src is the encrypted data buffer (source) public static byte [] encryptMode

InputT 1X yOutputHow many starting points can walk N (n>=0) step to (x, y), only from (x, y) (X,Y+LCM (x, Y))/(X+LCM (x, y), y)Standard solution: From (x,y0) to (x, y), then set X=AG,Y0=BG,G=GCD (X,Y0), there is y=bg+abg= (a+1) BG, because A, b coprime, A, (a+1) coprime, so A and (a+1) b coprime, so if you can go from (x,y0) to (X, Y), gcd (x,y0) =gcd (x, y), and then divide x and y gcd (x, y) by dividing it continuously (x+1)1#include 2#include 3#include Set>4#include 5 6 using namespacestd;7 8

My Game-segment Data In the context of the drawing from the Bezier curve generated line segments, with Drawnode to draw polygons, while a background has two mountains, a mountain has more than two Bezier curve saved, with nested data class:Bezier,linelayer,bglayerdata This is also done to facilitate reading data from a filePut the background data in the file for easy modification, you can also do a tool to quickly draw the background, the following is the background of the data saved XML file:1

multiple test cases. the first line of each test case contains three integers n, m, and T (1 Maze layout, with each line containing M characters. A character is one of the following: 'X': a block of wall, which the doggie cannot enter;'S ': the start point of the doggie;'D': the door; or'.': An empty block. The input is terminated with three 0's. This test case is not to be processed. Outputfor each test case, print in one line "yes" if the doggie can have ve, or "no" otherwise. Sample Input 4

// obtain the full name of the file from the text box, including the path string strfilepath = txupfile. text; // obtain the file length fileinfo Fi = new fileinfo (strfilepath); long lfilelength = Fi. length; // obtain the file name string strfilename = strfilepath. substring (strfilepath. lastindexof ("//") + 1); // create an httpwebrequest object, pass in the path name, and pass the file name and length as parameters to the server httpwebrequest objrequest = (httpwebrequest) httpwebrequest.

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

[Android development experience] a safer algorithm than DES encryption-3DES encryption algorithm, android3des Reprinted please indicate the source: http://blog.csdn.net/zhaokaiqiang1992 In the previous article, we discussed the DES algorithm and learned how to ensure the consistency of encryption and decryption results on different platforms. However, DES, as an encryption algorithm that has been around for a long time, is prone to brute-force cracking as the computer's computing power is enhanc

B. sailtime limit per test 1 second Memory limit per test 256 megabytes Input Standard Input Output Standard output The polar bears are going fishing. They plan to sail from (SX,SY) To(EX,EY) . However, the boat can only sail by wind. At each second, the wind blows in one of these directions ctions: east, south, west or north. Assume the boat is currently (X,Y) . If the wind blows to the east, the boat will move(X+ 1,Y). If the wind blows

Html: Css:@charset "Utf-8";Html,body{margin:0;width:100%;height:100%;overflow:hidden;}body{width:100%;Overflow:hidden;Background: #000;}#container {width:100%;height:100%;Overflow:hidden;}#wrap {width:100%;height:100%;}. page{position:relative;Float:left;width:100%;Overflow:hidden;Background-position:center;Background-repeat:no-repeat;}#music {position:absolute; width:44px; height:44px; right:30px; top:80px; Background:url (...) /img/music.png) No-repeat; z-index:999;}#p0 {background: #f00

/*************************************** * ********************* Function: // DES_EN (byte [] src, byte [] key) Description: // DES encryption algorithm Input: // byte [] src encryption source, byte [] key encryption key Output: // encrypted src Return: // byte [] ********************************* * ***********************/public static byte [] DES_EN (byte [] src, byte [] key) {try {SecretKey secret ey = new SecretKeySpec (key, "DES"); // generate ke

)--------------------------------------------------------------------------------73697499 563007521 4425075667654 4390076987782778878397844 510007876790079027934 The query results have only one column, which is the concatenation of the value of empno with the subsequent arithmetic value. Description a) using "| |" for data merging in select b) Query result error. Rows with an empty value of COMM no calculation results Q-quote delimiter Query statements