macbook 2002

Frontpage 2, J-bots 2002-javascript special effects Plug-insThe effects of JavaScript are exciting, but the complexity of JavaScript language is daunting for many people. Although FrontPage itself offers many components that allow you to add special effects to a Web page without having to know the JavaScript language, FrontPage itself builds only a handful of components and dynamic effects that are far from satisfying everyone's needs. Here I introduc

ERROR 2002 (HY000): Can ' t connect to local MySQL server through socket '/var/lib/mysql/mysql.sock ' (2)MY.CNF add[Client]Socket =/tmp/mysql.sock (tmp is the actual path)This article is from the "Network Engineering topic" Blog, please be sure to keep this source http://houzhibo.blog.51cto.com/3103241/1597814ERROR 2002 (HY000) Can ' t connect to local MySQL server through socket ' Varlibmysqlmysql.sock ' (

Workaround: ERROR 2002 (HY000): Can ' t connect to local MySQL server through socket '/var/lib/mysql/mysql.sock ' (2)Vim/etc/my.cnfIn the last new paragraph, because in the compilation of the installation, the path is written sock=/tmp/mysql.sock, but at startup, the system goes back to the default to find Var/lib/mysql/mysql.sock, so the new field to specify the actual path[MySQL]Socket=/tmp/mysql.This article is from the "Jurchens" blog, make sure t

Label:There was an error when connecting to the MySQL database with thinkphp: sqlstate[hy000] [2002] No such file or directoryThis is because PDO cannot connect to the database due to the Mysql.sock (or mysqld.sock) file path not specified in the php.ini configuration fileConfigure the following three items in the php.ini fileMysql.default_socket =/tmp/mysql.sockpdo_mysql.default_socket=/tmp/mysql.sockMysqli.default_socket =/tmp/mysql.sockThe database

Fault descriptionPrompt for error 2002 at startup (HY000)When you start the service, prompt the server quit without updating PID file.Workaround One:1. Because MySQL was uninstalled after reloading, view log files found that port 3306 was occupied.2. Check which process is occupied, end the process, and restart the service.# Lsof-i: 3306# kill-9 4420 #4420为占用3306端口的进程号# service Mysqld StartIf there are other ways, please comment to addWorkaround for e

Label:I did a MySQL multi-instance experiment today,Enter 3307 no problem.Enter 3306 : Enter : mysql-s/data/3306/mysql.sock error : ERROR 2002 (HY000): Can ' t connect to local MySQL server Solve:1.Kill 33934Kill 342282,I removed the password from the startup script MySQL.I put the startup script MySQL all chmod +x processing (before 700)3, restartMysqld_safe--defaults-file=/data/3306/my.cnf 2>1/dev/null Login : Enter : Mysql-s/data/33

Label:Foreword: At present the problem solves, but still do not know what reason causes, before the problem that installs Uwsgi, MySQL appears this question, which hero says this is how to return a matter?Text: Mysql Error 2002 bug solved under LinuxCheck the/ETC/RC.D/INIT.D/MYSQLD status first to see if MySQL has been started.If MySQL does not start, etc/init.d/mysqld start MySQLStartup fails, probably is/etc/my.comf file configuration problem, and t

Label: bzoj bzoj2002 multipart Questions and LCT version question: see http://blog.csdn.net/popoqqq/article/details/38849471 Today, I used a part-by-part method to rewrite this question. The part-by-part method is short. Divide the spring into √ n Blocks For each spring, we record the number of bounce times and point points starting from this spring until it is exceeded. Query and modify the position from block to point until the point is exceeded. In this way, both the modification and qu

Ball Size Calculation Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)Total submission (s): 81599 accepted submission (s): 33601The Problem description calculates the volume of the Ball Based on the input radius value. There are multiple groups of input data. Each group occupies one row. Each row contains a real number, indicating the radius of the ball. The size of the ball corresponding to the output. For each group of input data, a row is output, and the cal

(int i = 1; I //cout // }//cout LL ans =0; for(inti =1; I i) { for(intj = i +1; J j) {structNode t[8]; t[0].x = Arr[i].y-arr[j].y +arr[i].x; t[0].Y = arr[j].x-arr[i].x +arr[i].y; t[1].x = Arr[j].y-arr[i].y +arr[j].x; t[1].Y = arr[i].x-arr[j].x +arr[j].y;//cout //cout //for (int k = 0; k //cout // }//cout if(t[0].x 0].x = = arr[i].x t[0].y arr[i].y) {t[0].x = Arr[j].y-arr[i].y +arr[i].x; t[0].Y = arr[i].x-arr[j].x +arr[i].y; } if(t[1].x 1].x = = arr[j].x t

); - if(b -) F+=dfs (a,b+1, c,d,e,f) * ( --B)/( Wu-sum); + if(c -) F+=dfs (a,b,c+1, d,e,f) * ( --c)/( Wu-sum); - if(d -) F+=dfs (a,b,c,d+1, e,f) * ( --D)/( Wu-sum); + DoubleMd= -;if(e==4) { for(intI=0;i4; ++i) Md=min (DFS (A,B,C,D,I,F)/( Wu-sum), MD); f+=MD;} AMd= -;if(f==4) { for(intI=0;i4; ++i) Md=min (DFS (a,b,c,d,e,i)/( Wu-sum), MD); f+=MD;} atvis[a][b][c][d][e][f]=true; - returndp[a][b][c][d][e][f]=F; - } - intMain () - { -memset (Vis,false,sizeofvis); in inta,b,c,d;

, and the geometric properties of the squaresfind the coordinates of the other two points to determine whether in the set of points (hash, two, set ...). ) Can I use the hash. specific to all the points at the horizontal axis, ordinate from small to large sort. The number of statistics $tot$, the final answer is ${tot/2}$ (the symmetry of the square)so the question is, how do you know two points and ask for two other points? now give the formula: Record $ A (x_1,y_1) \ \ \ B (x_2,y_2) \ \ \overr

if((I-1) >=0) Dp[i][j] + = dp[i-1][j]; - if(J-1) >=0) Dp[i][j] + = dp[i][j-1]; the } + } A } thecout Endl; + } - $ intMain () $ { - -CIN >> n >> m >> x >>y; thememset (arr,0,sizeof(arr)); -Memset (DP,0,sizeof(DP));Wuyi //Remove the other horse's control points the for(inti =0; i i) - { Wu inttx = x +Avoidx[i]; - intTy = y +Avoidy[i]; About if(TX >=0) (TX 0) (Ty m)) $ { -Arr[tx][ty] =true; -

ItemsSample output Sample outputs5250It is easy to think of constructing a diagram, each node of the graph is an item, the right value is the price after the discount set the merchant to 0., find the shortest possible 1. Due to the level limit, the node must be 1 or less than the level of M, then for each of the upper and lower difference between the interval of M to find the shortest possible, and finally take the minimum value.Easy to understand, draw a very bad picture, we gather live to lo

seconds, the four-layer cycle of violence will definitely be timed out. So there is a way of thinking: First point sorting, double-loop enumeration before (N-2) points, in order to prevent repeated judgment, the second loop in the J to start from the i+1, two points after the search (N-J) point in the presence of a can and s[i],s[j] form the point of the square, so the second layer loop end condition is jKnown 2 points, write the coordinates that can form a square with these 2 points, such as (

Describe Lostmonkey on the ground along a straight line to put n devices, each device set the initial elastic coefficient ki, when the sheep reached the first device, it will bounce Ki step back to the first I+ki device, if there is no i+ki device, the sheep are bounced. The sheep wanted to know that when it started from the device I, it was bounced a few times. In order to make the game more interesting, Lostmonkey can modify the elastic coefficient of an elastic device, and the elasti

; - ElseA[i].bs= (a[a[i].num+i].bs+1), a[i].next=a[a[i].num+I].next; + } A } at inline ll read () - { -ll x=0, f=1;CharCh=GetChar (); - while(ch>'9'|| ch'0'){if(ch=='-') f=-1; ch=GetChar ();} - while(ch>='0'ch'9') {x=x*Ten+ch-'0'; ch=GetChar ();} - returnx*F; in } - intMain () { to inttot=0, M; +scanf"%d",n); -Mid= (int) sqrt (n) +1; the for(intI=1; iread (); * for(intI=1; i){ $ if(I>tot*mid) tot++;Panax Notoginsenga[i].block=tot; - } the

(inti=x;! IsRoot (i); i=Fa[i]) st[++top]=Fa[i]; for(inti=top;i;i--) pushdown (St[i]); while(!isroot (x)) { inty=fa[x],z=Fa[y]; if(!isroot (y)) { if(c[y][0]==x^c[z][0]==y) rotate (x); Elserotate (y); } rotate (x); }}voidAccessintx) { intt=0; while(x) {splay (x); c[x][1]=T; T=x;x=fa[x];}}voidRever (intx) {access (x); splay (x); Rev[x]^=1;}voidLinkintXinty) {rever (x); Fa[x]=y;splay (x);}voidCutintXinty) {rever (x); Access (y); splay (y); c[y][0]=fa[x]=0;}intMain () {scan

outputSample Output17Data range and TipsData Size HintAs described ExercisesRecursion.Using F "i,j" means the number of methods that go from the i,j to the i,j (if the point is controlled by a horse), the state transition equation is "f" = 0, "=f", "I-1,j" +f "I,j".var n,m,x,y,i,j:longint;A:array[-2..21,-2..21]of Boolean;F:array[-1..21,-1..21]of Int64;BeginREADLN (N,m,x,y);A[x,y]:=true;A[x-2,y-1]:=true;A[x-2,y+1]:=true;A[x-1,y-2]:=true;A[x-1,y+2]:=true;A[x+1,y-2]:=true;A[x+1,y+2]:=true;A[x+2,y

Title Description DescriptionKnown: sn= 1+1/2+1/3+ ... +1/n. Obviously for any one integer k, when n is large enough, SN is greater than K.An integer k (1Enter a description input DescriptionKeyboard input KOutputs description output DescriptionScreen Output nSample input to sample1Sample output sample outputs2Data size HintExercisesSimulation.Each time the cumulative 1 is divided by the value of n until the value is greater than S.var i,k,n:longint;S:real;BeginREADLN (k);While SBeginInc (N);