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，成功获取QQ号码：615947283 Use queue FIFO to get QQ number, according to the characteristics of the topic, just can use queue to process, queue simple and convenient, and better understand. #include struct queue {int *data;int head;int tail;}; int main(){int num, i;printf("请输出要破译的QQ号码长度：");scanf("%d", num);struct queue q;q.data = (int *)malloc(sizeof(int)*(num*2-1));//总共需要的数组长度为num*2-1 q.head = 0;q.tail = 0;for(i=1;i Say something. This idea comes from "

are also defined in the header file.Why static variables of a class cannot be initialized in place: the so-called in-place initialization is similar to the case:Class A{static char msg[] = "aha";};This is not allowed because the class declaration is usually in the header file, and if this is allowed, it is actually equivalent to defining a non-const variable in the header file.In C + +, the header file defines what a const object would be like:Not us

.(5) noun/pronoun binage + Word SegmentationI saw a cat running ss the road. I saw a cat crossing the road. 6. Slang: slang refers to the composition of the noun or pronoun after which the noun or pronoun is used to describe the former, which is close to the post-attribute.For example:We students shocould study hard ')We all are students ') 7. independent components: Sometimes there are some components in the sentence that are not associated with the syntax of the sentence, called the Independen

; BL = bp.length; Start walking down the tree looking for a discrepancy for (var i = 0; i A lot, isn't it? Have the wood to feel the moment is I pit? Aha, in fact, as long as you continue to debug breakpoints set, you will find that ~ ~ ~ No more than this pit! They're all loops! 1.8.3 inside in the first function inside the rotation, the hands are sour and did not see out, 1.11.0 the whole cycle, there are the number of parameters s

-(void) Printdata{Prepare for printing, create a reference to SharedprintcontrollerUiprintinteractioncontroller *printer = [Uiprintinteractioncontroller Sharedprintcontroller];Printer.delegate = self;Configure printing informationUiprintinfo *pinfo = [Uiprintinfo printinfo];Pinfo.outputtype = uiprintinfooutputgeneral;//printable text, graphics, imagesPinfo.jobname = @ "Print for Xiaodui";//optional attribute for identifying print jobs in the print centerPinfo.duplex = uiprintinfoduplexlongedge;/

Android Studio is really handy. But the build speed is extremely slow!!!！I went online to find a lot of build acceleration settings, but not fit into my 1.3RC version ...。。 Heart plugs. No resistance, endure super invincible slow speed obstinately after a week. Of course, almost every day of the week to find the set ...。。Until. Today.I'm really happy and broken!！！。！！！！。！！！！！。！！！！！！！！！。！！！。！。！！！！！！！！！。！！！。！。！！！！！ aha haha haha hahaBy chance, I got into

through all vertices of the 1~n number. In a nutshell: From the I vertex to the J vertex only passes the shortest distance from the front K points. In fact, this is a "dynamic planning" of the idea, about this idea we will be in the "Aha!" The algorithm 2--the great thinking shines in a detailed discussion. The complete code for this algorithm is given below: #include int main () { int e[10][10],k,i,j,n,m,t1,t2,t3; int inf=99999999; //U

");Session.save (student);Action.commit ();}catch (Hibernateexception e) {if (action!=null) {Action.roolback ();}E.printstacktrace ();}finally{if (session!=null) {Session.close ();}}Delete Student Information:try{Configuration context = new configuration (). Configure ();Sessionfactory sessionfactory = Context.buildsessionfactory ();Session session = Sessionfactory.opensession ();Transaction action = Session.starttransaction ();Student Student = new Student (1);Session.delete (student);Action.co

= Session.starttransaction ();Student Student = new Student (1, "hah");Session.save (student);Action.commit ();}catch (Hibernateexception e) {if (action!=null) {Action.roolback ();}E.printstacktrace ();}finally{if (session!=null) {Session.close ();}}To delete Student information:try{Configuration context = new configuration (). Configure ();Sessionfactory sessionfactory = Context.buildsessionfactory ();Session session = Sessionfactory.opensession ();Transaction action = Session.starttransaction

handle_message (int new_fd);Server.cpp file, the Epoll model is implemented here, as follows:#include "local.h" #include "utils.h" using namespace std;//store the client socket descriptor listlistTester.cpp file, simulate the high concurrency of the server, open 10,000 clients to connect to the server, as follows:#include "local.h" #include "utils.h" using namespace Std;char message[buf_size]; Accept server information listI will not give the results of the implementation of the program, but t

clients to connect to the server, as follows: #include "local.h" #include "utils.h" #include "head.h" #define Epoll_test_size 8000 using namespace std; Char Message[buf_size]; Accept Server Information List I will not give the results of the implementation of the program, but the following is the code author himself test, you can see, concurrency 10000 no pressure AHA single client goes to connect to the server, client.cpp file, as follo

, it also occupies a total of 50 character widths, so it can be written on the same lineInput Example 2:5 5 5 5 5 5 10 10 10 10 10 10 10 10 10 10 10 10 5 5 5 5 5 5 5 5HahahahahahahahaOutput Example 2:2 20Example Illustration 2:"5 5 5 5 5 5 10 10 10 10 10 10 10 10 10 10 10 10 5 5 5 5 5 5 5 5" Specifies the width of each character"Hahahahahahahaha" is the input string sSince the h width is 10,a width of 5, the ' Hahahahahahah ' occupies 100 character widths that can be written on the first line, a

#可变类型 Immutable typesStus = [' aaa ', ' BBB ', ' CCC ', ' ddd ', ' eee ']Words = ' aha haha '# take out every element of the list Print (i)# remove every character in a stringFor j in Words: Print (j)# slices, a way to take a list of valuesPrint (Stus[1:3]) #顾头不顾尾, takes only the element with the subscript 1-2, does not remove the element labeled 3, in list formPrint (Stus[:3]) #开头下标写0和不写一样Print (stus[-1]) #-1 represents the last elementPrint

completely handle it. These reasons have contributed to this short blog post.I plan to post all the following operations: Use Latex to write the original blog post, generate PDF documents; Convert the PDF to a high-definition PNG-formatted image; Combine images from multiple PNG formats into a single large picture; Upload the final big picture directly to the post editor Aha, it's done! OK, what if I convert the PDF docu

In the first blog, I introduced some of the basics of Rxjava and also introduced the map () operator. Of course, if you do not intend to use the Rxjava I am not surprised, after all, the touch of this point. After reading this blog, I am sure you want to immediately use Rxjava in your project, this blog will introduce many Rxjava in the operator, Rxjava's power comes from its defined operators.First, let's look at an example:Preparation assumptions I have such a method:This method returns a list

the specified character ValueOf () Specifies the data type variable to string 1, toUpperCase (), toLowerCase ()Converts all the letters in a string string to case and does not affect the original stringString str = "123abcABC aha"; // Convert all letters in a string to lowercase String lowerstr = str.tolowercase (); // capitalize all letters in a string String upperstr = str.touppercase (); System.out.println ("original string:" +str); System.out.

This is the most understandable algorithm in the world, hehe, from AHA algorithm.#include using namespacestd;inta[Ten],book[Ten],n;intTotal ;voidDfsintstep)//step said to stand in front of the first few boxes {if(step==n+1)//If you stand in front of the n+1 box, it means that the top N boxes have been put in place. {//Output an arrangement (poker number in 1-n box) for(intI=1; i) coutA[i]; coutEndl; Total++; return; Return to the previous step (where

code implementation is very simple, the core code only five elements: for (k=1;k for (i=1;i for (j=1;j if (E[i][j]>e[i][k]+e[k][j]) E[I][J]=E[I][K]+E[K][J]; Copy CodeThe basic idea of this code is that it is only allowed to go through the number 1th vertices at first, and then only the 1 and 2nd vertices are allowed to relay ... The shortest distance between any two points is allowed through all vertices of the 1~n number. In a nutshell: From the I vertex to the J

=Ha.next to ifLenA LenB: +Distance = LenB-LenA - forIinchRange (0,distance): theHB =Hb.next * $intersection =NonePanax Notoginseng whileHA is notNone andHb is notNone: - ifHA = =HB: the returnHA + Else: AHA =Ha.next theHB =Hb.next + returnIntersectionIdeas :1. First record the length of the two linked list2. Double pointers to two lists point to the long list fir

will be used as a reference count in a certain way. As mentioned above, naming a design pattern is valuable, and providing a name that the kernel developer uses will be useful for reviewers.Andrew Morton says:I would be very careful to say: "Aha, it used kref. I understand the reference counting mechanism, I know it's good to debug, I know it will track common errors." That's more than "Oh, this implements my own reference count, I need to check for