dbx 286

How to make the browser and focus of a mobile phone do not automatically zoom in on the page when entering a text box-silence 286 Method 1: Disable page Scaling Add the following code to the head label: Description of content Width = device-width // The width is equal to the device screen width initial-scale = 1.0 // The initial zoom ratio minimum-scale = 1.0 // The minimum zoom ratio maximum-scale = 1.0 // The maximum scale user-scala

Title Link: Http://codeforces.com/contest/505/problem/DMain topic:In one figure, there are n vertices, giving the m pair number (U,V) to indicate that the vertex u and vertex v must be connected directly or indirectly, so that you can construct a graph with a minimum number of edges required for the output. Analysis:Without a doubt, n vertices, we can use up to n edges, so that n vertices form a ring to satisfy all the situation (any two points are connected), but this is not necessarily the lea

Codeforces Round #286 (Div. 1) C, D, C: The number of steps in the question seems to be many. In fact, the maximum number of steps increases by about 250, because the number of moving steps is 1 + 2 + 3 + .. n, so there will be only sqrt (n) steps, so dp [I] [j] indicates that the value of step j is increased at the I position, and then transferred. D: in fact, for a Unicom block, a maximum of n sides and a minimum of n-1 are required. The condition f

Idea: The value of the mass factor multiplicative is the demand#include #include#include#include#includestring.h>#includeusing namespaceStd;typedefLong LongLL;intPrime (LL N);intMain () {LL n,ans,m; while(~SCANF ("%lld",N)) {/*for (int i=2;i*/ans=1; if(Prime (n) | | | n==1) ans=N; Else { intlen=sqrt (n); for(intI=2; i) {m=n/i; if(n%i==0Prime (i)) ans*=i; if(M!=i m*i==n Prime (M)) ans*=m; }} printf ("%lld\n", ans); } return 0;}intPrime (LL N) {if(n==1) return 0; fo

[t].to=to; Edge[t].next=head[from]; head[from]=t;}intVis[n],loop[n];BOOLOkinline voidDfsintNow) {if(!ok)return; vis[now]=1; for(inti=head[now];i!=-1; i=edge[i].next) {if(!ok)return;intto=edge[i].to;if(!vis[to]) DFS (to);Else if(vis[to]==1) {ok=false;return; }Else if(vis[to]==-1)Continue; } vis[now]=-1;}intMain () {N=read (), M=read (); Init_nex (); Init_edge (); for(intI=1; iintA=read (), B=read (); Add_edge (A, b);intNex1=findnex (a);intNex2=findnex (b);if(NEX1!=NEX2) {nex[nex1

)? ≠? (aJ,?bJ,?CJ).The next line contains a integer- q (1?≤? Q≤?100), denoting the number of the queries.Then followsQLines, containing space-separated, integers- ui and vi (1?≤? u i,? v i? ≤? N ). It's Guaranteed that ui? ≠? v I .OutputFor each query, print the answer-a separate line.Sample Test (s) input4 51 2 11 2 22 3 12 3 32 4 331 23 41 4Output210Input5 71 5 12 5 13 5 14 5 11 2 22 3 23 4 251 55 12 51 51 4Output11112NoteLet ' s consider the first sample. The figure above shows the f

C: The number of steps in the topic seems to be many, in fact, the maximum growth of about 250 steps, because the number of steps to move 1 + 2 + 3 +. N, so there is probably only sqrt (n) step, so dp[i][j] is in the I position, growth is the value of J step, then transfer canD: This problem in fact for a unicom block, at most only need n edge, at least to n-1, then the condition of judgment, is this unicom block whether there is a ring, the use of topological sort to be sentencedCode:C:#include

,0,sizeof(CNT)); for(i=1; i) {scanf ("%d",j); CNT[J]++; } memset (DP,-1,sizeof(DP)); dp[d][ -] =Cnt[d]; Ans= dp[d][ -]; for(i=d;i30000; i++) { for(j=1; j -; j + +) { if(Dp[i][j] = =-1)Continue; intto = i+d+j- -; if(To 30000) {Dp[to][j]= Max (dp[to][j],dp[i][j]+Cnt[to]); Ans=Max (ans,dp[to][j]); } if(to+130000) {dp[to+1][j+1] = max (dp[to+1][j+1],dp[i][j]+cnt[to+1]); Ans= Max (ans,dp[to+1][j+1]); } if(To 30000 to-i >1) {dp[to-1][j-1] = ma

B. Mr. Kitayuta ' s colorful Graph time limit/test 1 second memory limit per test 256 megabytes input standard input out Put standard output Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the

ShusekiKingdomistheworldsleadingnationforinnovationandtechnology. Therearencitiesinthekingdom, numberedfrom1ton. ThankstoMr. Kitayutasresearch, role Shuseki Kingdom is the world's leading nation for innovation and technology. there are n cities in

Topic:You is given a m x n 2D grid initialized with these three possible values. -1-A wall or an obstacle. 0-Gate A. INF-Infinity means a empty room. We use the value of represent as assume that the 231 - 1 = 2147483647 distance to

Let's give you an undirected graph with each vertex degree of k (k is an even number) and ask if we can dye n edges in the graph so that each vertex has two colored edges. That is to say, is there a subgraph with n edges in the source image, so that

Test Instructions Link Test instructions: Give a graph that tells you the color of the edges between each point (with a heavy edge) and asks you how many sides are connected between any two points. Solution: and set Processing, F[x][c]: Represents

expected. I think it's normal, why? Because I believe most people are not the beta testers for the Delphi 2007 for Win32, so they have no chance to use it. In addition, many people do not understand the new DBX 4 ability, perhaps I can reveal some absolutely let delphiers happy things. The first is that the Delphi 2007 for Win32 starts at a faster rate and reacts faster when you enter the IDE. Delphi 2007 for Win32 not only in the speed of the boot

When we debug a more complex problem, it is difficult to locate the problem, you need to try to set breakpoints, multiple continue, view variable values, and so on, sometimes need to set some special debugging variables and so on. This process often requires you to restart the debugger several times and restart these debugging commands. If debugging does not support reverse debugging, or if the reverse debugging performance is poor, the use of history and debug scripts simplifies this process, w

# console program in visual studio.net N Add the following class libraries to the reference tab: L InterOP. AutoCAD. dll L acadexample. dll L objectdbx16 (in Solution Explorer, right-click the reference tab and select "add reference" in the pop-up menu ", in the "add reference" dialog box, select "AutoCAD/objectdbx common 16.0 Type Library" in the drop-down list box under the "com" tab) Enter the following code: 1: using system; 2: using AutoCAD; 3: Using

Object| Graphics never open a graphic to get information about a tile? The answer is yes. Here is a description of the implementation of the specific method. Requirements: n is programmed with C # N Read what I wrote "Two development of AutoCAD using C #" (In Csdn) Begin: n Create a new C # console program in Visual Studio.NET n Add the following class libraries to the Reference tab: L Interop. AutoCAD.dll L AcadExample.dll L ObjectDBX16 (In Solution Explorer, right-click the References

, "username"); $password = Odbc_result ($rs, "password"); echo "$username"; echo "$password"; } Odbc_close ($conn); echo "";? >Copy code 0x4: using DBX to communicate with database PHP itself has a DBX function built into it, and the DBX module is a Database abstraction layer (the "X" in DBX means the X database it can

and took the time to try Firedac, this would like to find some ways to let FIREDAC take data and dbx like fast, and ultimately failed, the DBX implementation is too fast, 3472 records (110 fields of the table), 0 milliseconds to grab over,FIREDAC is almost 20 milliseconds faster. But FIREDAC has captured the data tfddatstable, know the number of records, (better than DB