hackerrank for work

If coin Order matters, that's, each sequence are unique, the DP function is a simple enough-to-make it 1D DP. But the key is this order doesn ' T matter, so we need to add one more state:ending coin. And for each DP advance step, we only put >= coins.#include #include#include#includeusing namespacestd;typedef unsignedLong LongULL;intMain () {unsigned n, m; CIN>> N >>m; VectorV (m); for(Unsigned i =0; I ) Cin>>V[i]; Std::sort (V.begin (), V.end ()); //dp[val][ending Coin]Vector1, Vector0)); d

Test instructions: Give you a positive integer n, ask how many positive integers you exist on a, B (aThe approximate number of all the numbers within N is preprocessed, then the violent enumeration A, the violent enumeration of X, and then enumerates all the n-ax, the sentence weight, the statistical answer can be.#include "Enumeration approximate" Hackerrank-week of Code 26-satisfactory Pairs

I made a few questions in HackerRank, so let's talk about it... Question: Give n and k, then give n numbers, and find the number of k pairs in the array. Where [N /* * Author: illuz

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

insert and delete the elements multiple times, so we cannot sort them every time, this method will inevitably time out. Here I use the STL et in STL (set cannot be used because there may be duplicate elements). set can be inserted and deleted in the time complexity of O (logn, in addition, the order will be maintained during the insertion and deletion process, so that we do not need additional sorting work, which undoubtedly saves time. However, set

The knowledge-based elderly people are also implementing the "Distribution by work" principle, so you can work harder to get more. God will be bored if there is nothing to do. What is the secret of learning? Don't trust the secret. If you are a sprout seedling, a good book is a bit of rain. How do you know that there are mountains outside the mountains, open your eyes, and know that there are days outside t

When I visited the website, I saw an old article titled "The Day Programmer vs. the night Programmer", which was written by Mitch Denny, because Mitch Denny served as CTO for a company, and the article has been issued to attract hot and Post. The article is translated by netizens, in order to let us see more clearly, some places to modify.Years ago I believed that there were two kinds of programmers in the world, no matter what technology they used, I would call it:1. Day Programmers Daytime wor

The knowledge of the elderly is also in the implementation of the "distribution according to work" principle, you have more work to get more. If there's nothing to do, God will be bored. What is the secret of learning? Don't believe in the secret. If you are a budding seedling, then a good book is a drop of luscious rain. Stand on its laurels, know the mountains outside the mountains, open Horizons, Fang kn

GitHub Address: Https://github.com/cheesezh/python_design_patternsTopicUse code to simulate the work of the day, Good morning state, noon want to sleep, the afternoon gradually recover, overtime bitter torment.Basic version-Function editionhour = 0work_finished = Falsedef write_program(): if hour 当前时间: 9 点, 上午工作，精神百倍当前时间: 10 点, 上午工作，精神百倍当前时间: 12 点, 饿了，午饭，犯困，午休当前时间: 13 点, 下午状态还可以，继续努力当前时间: 14 点, 下午状态还可以，继续努力当前时间: 17 点, 收工，下班当前时间: 19 点, 收工，下班当前时间: 22