hackerrank competitors

Analysis on how to help your SEO by analyzing your competitors Competition among search engine organic search results is like competition for real estate in a favorable area. Because a good ranking in search engines means a good traffic, which means more traffic, more conversion, and more sales. High Ranking is not an easy task, especially for a highly competitive industry. Even if you hire the best seoer and have the best SEO strategy, you cannot al

SEO optimization is mainly to make their own site ranking beyond their competitors, but a lot of people do not realize this core point, just blindly to do outside the chain, to update the site, but the time is not obvious effect, this is behind closed doors, if we can carefully analyze the top of the site, analysis of their website , find the clues to their SEO optimization, you will be able to find a way to defeat

code should be safe, of course, any program is a loophole, but for the vulnerabilities have been found, we should be in the procedure to prevent. 5, Content Web site structure and procedures are completed, then you have to prepare to walk the lake, that is, the site online. After the online update 2~3 article original article, these articles are the basic skills of your test, his experiment, you still have no confidence to continue lakes, if there is, please continue to look down. Ii. imitat

Kinda similar with another palindrome DP from Leetcode. Feel it, it is a bottom-up dp-palindrome subsequence.str =input () Slen=Len (str) DP= [[0 forXinchRange (Slen)] forXinchrange (Slen)] forCleninchRange (1, Slen + 1): forIinchRange (0, Slen-clen + 1): ifClen = = 1: Dp[i][i]= 1elifClen = = 2: Dp[i][i+ 1] = 2ifStr[i] = = Str[i+1]Else1Else: ifStr[i] = = Str[i + clen-1]: Dp[i][i+ Clen-1] = dp[i + 1][i + clen-2] + 2Else: Dp[i][i+ clen-1] = max (dp[i + 1][i + clen-2], dp[

Greedy-though Simple, but fun!#include #includeusing namespacestd;intMain () {intN1; intN2; intN3; CIN>> N1 >> n2 >>N3; Long LongL1 =0, L2 =0, L3 =0; Vectorint>H1 (N1); for(intH1_i =0; H1_i ) {cin>>H1[h1_i]; L1+=H1[h1_i]; } Vectorint>H2 (N2); for(intH2_i =0; H2_i ) {cin>>H2[h2_i]; L2+=H2[h2_i]; } Vectorint>H3 (N3); for(intH3_i =0; H3_i ) {cin>>H3[h3_i]; L3+=H3[h3_i]; } intI1 =0, I2 =0, i3 =0; while(! (L1 = = L2 L2 = =L3)) { //which stack is highest now? intMaxi =1; intM

(1, CC, 1): - # Down ins[(rr-1 + i) * m + (j + i)] = (rr-1 + i) * m + (j-1 +i) - forJinchXrange (Rr-2,-1,-1): to # Right +s[(j + i) * m + (cc-1 + i)] = (j + 1 + i) * m + (cc-1 +i) - forJinchXrange (Cc-2,-1,-1): the #Top *S[i * m + (j + i)] = i * m + (j + 1 +)i) $i + = 1Panax Notoginseng returns - the defMultiply (A, b): +n =Len (a) Ac = [] the forIinchxrange (n): + c.append (A[b[i]) - returnC $ $ defDisplacepow (A, k): -

DescriptionHave \ (n\) skills, each time through a wizard, a skill can be turned into another skill, ask most how many different skills.SolNetwork flow.First of all, the very naive I started, the skill will be split, the middle plus a list of wizards, \ (s\) to the initial skills of the number of edge capacity, the corresponding point between the capacity of the inf\, and then from the point to \ (t\) to the edge, the capacity is \ (1\), the wizard from the left side of a point to the right even

Please note input constraints. String length won't exceed, which means, we can use relatively naive representation\calculation for Anagrams:sorti Ng.#include #include#include#include#include#includeusing namespacestd;intCalcstringstr) { intRET =0; size_t Len=str.length (); Unordered_mapstring,int>Rec; for(size_t Ilen =1; Ilen ) { for(size_t i =0; I ) { stringCurr =str.substr (i, Ilen); Std::sort (Curr.begin (), Curr.end ()); Rec[curr]++; }

Original title AddressLeetcode also have this problem, directly sweep over the line, even the array is not open, feel like a certain kind of moving into the thought in the inside, to not be a moving titleCode:1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 8 intMain () {9 /*Enter your code here. Read input from STDIN. Print output to STDOUT*/ Ten intN; One intCurr_score, Prev_score, curr_cnt; A intdir, Len; - intres =0; - thePrev_score =-1; -curr_c

Typical tree recursion works. The key Is:any node can be treated as the root, since it was about edge and not on node. This simplifies things a lot.#include #include#include#include#include#includeusing namespaceStd;typedefLong Longlong64;structnode{Node (): Val (0{} Node (unsigned v): Val (v) {} long64 val; Unordered_setint>Inx;}; LONG64 ret= std::numeric_limits:: Max (); long64 Go (Vectorintroot, long64 tsum) {Noden =Nodes[root]; long64 Currsum=N.val; for(intI:n.inx) {nodes[i].inx.eras

Simplied a dfs\bfs with minor modification.#include #include#include#include#include#includeusing namespacestd;typedef VectorChar>>Matrix;typedef pairint,int>Pos;BOOLIsway (Matrix m,intIintj) { returnM[j][i] = ='.'|| M[j][i] = ='*';}BOOLGo (Matrix m, pos curr, POS DST,Const intKintN) { if(Curr = =DST) { returnn = =K; } inth =m.size (); intW = m[0].size (); inti =Curr.first; intj =Curr.second; VectorCands; if(I >0 Isway (M, I-1, J)) {Cands.push_back (Pos (i-1, J)); } if(I 1 Isw

]; -START_INDEX_A[I][J] = start_index_a[i +1][j]; theSTART_INDEX_B[I][J] = start_index_b[i +1][j]; +}Else { ALCD[I][J] = lcd[i][j +1]; theSTART_INDEX_A[I][J] = start_index_a[i][j +1]; +START_INDEX_B[I][J] = start_index_b[i][j +1]; - } $ } $ } - - inti = start_index_a[0][0]; the intj = start_index_b[0][0]; - while(I >=0 J >=0) {Wuyicout " "; the intII = Start_index_a[i +1][j +1]; - intJJ = Start_index_b[i +1][j +1]; Wui =II; -j =JJ; About

) Constraints1 S-1000 XX∈ ar Sample Input 71 3 9 8 2 7 5 Sample output 1 ExplanationInsertion Sort will take 9 "shifts" to sort the array. quicksort will take 8 "swaps" to sort it, as shown in the dimo-below. 9-8 = 1, the output. Question: the difference in the number of moves between quick sorting and inserted sorting elements in sorting. For insert sorting, count the number of moving elements; for fast sorting, count the number of switching elements (including switching between yourself and

Static intGetnumberofprimes (intN) {intn = n+1;//To include 0 as the first number for easy index operations later Final BooleanA[] =New Boolean[n];//Initialized to NULL by default.Arrays.fill (A,true); a[0] =false; a[1] =false; for(inti = 2; i i) {if(A[i]) { for(intj = i + i; J i) a[j]=false; } } intCount = 0; for(inti = 2;ii) {if(A[i]) {//System.out.print (i+ ",");++count; } } returncount; }[Hackerrank] P

A natural thought is brutal-force. But as you may have already thought of, there must be a smarter one. And yes there is. Think like this: XOR gives you all different bits, if you cocould imagine the binary representation of L ^ r, it can be represented as: 1xxxxx... what is asked is the 'maximum 'value of L ^ r-and that means 1111111... so this maximum value only depends on which is the highest 1-it is irrelevant with all intermediate values. so now the question turns to be: Which is the highe

Original address As a qualified seoer, the analysis of the competitor website must be mastered, reasonable analysis of the weak part of the competitor SEO optimization and advantages, and then according to our own web site to develop corresponding optimization strategy to enhance the site keyword rankings. But there are still a lot of friends for the analysis of competitors do not have clear ideas, the following rhino cloud for everyone to comb unde

To optimize the success of SEO, analysis of competitors is an essential link, do not bother, it can let you optimize the optimization of SEO. As the saying goes, there are mountains outside the mountains, there are people outside. Not every seoer is the so-called master, each seoer has its own shortcomings, but also should learn from the strengths of others. And we received a new list of SEO, we can not ignore the strength of

The boss went out to recruit people and threw a document before leaving: project implementation 9 Yin Zhenjing, surpassing competitors-18 Practical Skills of software project managers The document does not contain the author's name, and I have not checked the source. Let's see it first. Here we will record the good sentences and personal insights in the document. This includes the author's and my thoughts. Comments as you like ------------------

These days have a lot of friends asked the new Chen, why the site ranking dropped? Why are they being attacked by others? Of course, this is undoubtedly related to the content of your site, the external chain and the user experience has a very big relationship, just. New Chen in this feel, also need to study the competitor's site, to do: no one I have. People have my gifted, just have this, talent will your site really do up; so compare your competitors