List of Comments

Looks like I should also use VPS for all my sites? But I have already paid a lot for my current shared web hosting, and it is working very well.

interresting article, thanks

Fortunately, my shared web hosting is really powerful to handle heavy traffic. But I did use dedicated IP for my site, because I have to use the IP for my SSL cert.

You are right, it is unnecessary checks that can be removed. post updated.

Why we need this step: if (st.size() >= 1) { // the first reversed word does not have leading white space s = st.top(); st.pop(); }

Create TCP server in Nodejs

I was struggling with this problem before I found your solution via googling. Can you please take some time to elaborate more on your solution? I can't see how you come up with the algorithm. Thanks.

another improvement can be done -1) try divison by 2-3 and 5 -2)switch the step by 2 and 4 alternatively to avoid finding other multiples of 3 exemple 2-3-5 (+2)=7 (+4)=11 (+2)=13 (+4)=17 I give you the code i wrote in C language /*méthode qui renvoie VRAI si la valeur entrée est un nombre premier */ BOOL estPremier (UINT x){ //déclaration des variables BOOL estPremier = VRAI;//on considère que le nombre est premier jusqu'à ce qu'on démontre le contraire UINT racine , cpt = 11; //début du code racine=sqrt(x)+1;//pour eviter du calcul superflu, on s'arrete a la racine du nombre (eratostene)) if (x>1){ if ((x==2)||(x==3)||(x==5||(x==7))){ return estPremier; }else{ if((x%2==0)||(x%3==0)||(x%5==0||(x%7==0))){ estPremier=FAUX; } } }else{ estPremier=FAUX; } while ((estPremier==VRAI)&& (cpt<=racine)){ if(x%cpt==0){ estPremier=FAUX; } cpt=cpt+2; if(x%cpt==0){ estPremier=FAUX; } cpt=cpt+4; } return estPremier; }

Nice Article really enjoyed programming niche cool

For a *.COM program i like to replace the last "int 20"-instruction with a simple "ret"-instruction, because DOS push a word of zero to our stack and additional it place an "int 20"-instruction to the top of our PSP just before it start to exexcute our *.COM program. Dirk

Here is more simply: /* * * * Get the directory size * @param directory $directory * @return integer */ function get_dir_size($directory) { $size = 0; foreach (new RecursiveIteratorIterator(new RecursiveDirectoryIterator($directory)) as $file) { $size+=$file->getSize(); } return $size; }

It would be interesting to produce combinations using a similar method!

I'd rather use binary search at least 🙂

findstr is a tool, you can google it.

I could not use a variable with path like "C:/program/" because findstr in doc i could not find what me help there is one way to fix this?

Here's more: //Declaration: const fn = 'Records.dat'; type TDatas = record ID:string[10]; Comment:string[25]; end; var Stream: TFileStream; //Form Loads: Read Multiple Records: var StreamSize: Integer; Begin if FileExists(fn) then Stream := TFileStream.Create(fn, fmOpenReadWrite) else begin Stream := TFileStream.Create(fn, fmCreate); end; StreamSize:= Stream.Size; Stream.Position := 0; while Stream.Position < StreamSize do begin Stream.ReadBuffer(Datas, SizeOf(TDatas)); Listbox1.items.add(Datas.ID); Listbox2.items.add(Datas.Comment); end; Stream.Free; End; //Write Records: Stream := TFileStream.Create(fn, fmOpenWrite); Stream.Position := Stream.Size; with Datas do begin ID := '1to10'; Comment := 'Say to GOD be the Glory Always'; end; Stream.WriteBuffer(Datas, SizeOf(TDatas)); Stream.Free;

For the adjacent list to represent the map I use a dict like this: {1: [2, 3, 4], 2: [1, 3, 4], 3: [4, 1, 2], 4: [3, 1, 2]} For this you can just define map like this map = dict() and construct the map by mapping each int element to a list(the node that can be reached from this int) when reading input.

You can time it using microtime easily. In the post, it has a quick comparison.

I've read many blogs made by professionals and they suggest the same thing. Echo over print. Could you also give a sample code using print for better comparison?

python genetaor.

lol... interesting! quite strange why C# is designed in this way. BTW, why not use some socialization plugins to enable logging in by open ID?

Linux is a open-source free project   Linux is *an open-source free project

acm.zju.edu.cn supports php~

好奇葩的算法。。