sap srm

)==0) ans =ABS(x) +ABS(y);ElseAns =ABS(x) +ABS(y) +2; }Else if(x 0 y 0) {if((x1)==1|| (y1)==1) ans =ABS(x) +ABS(y) +2;ElseAns =ABS(x) +ABS(y) +4; }Else if(X >0 y 0) {if((x1)==1|| (y1)==0) ans =ABS(x) +ABS(y);ElseAns =ABS(x) +ABS(y) +2; }Else if(X 0 y >0) {if((x1)==0|| (y1)==1) ans =ABS(x) +ABS(y);ElseAns =ABS(x) +ABS(y) +2; }returnAns }};The first method of code#include #include #include #include using namespace STD;structdata{intx, y;intIdintD;} d[100010];intnd queueQintTin

;>1; Merge_sort (A,l,mid); Merge_sort (A,MID,R); inti = l, j = Mid, k =l,p; while(I R) { if(Times++,a[i]>=a[j]) t[k] = a[j++]; ElseT[k] = a[i++]; K++; } if(i = = mid) for(P = j; p A[p]; Else for(p = i; p A[p]; for(k = l;k t[k];}classbearsortsdiv2{ Public: DoubleGetprobability (Vector int>seq) { for(inti =0; I ) {A[seq[i]-1] =i; } Times=0; Merge_sort (A,0, Seq.size ());//log (ans);d ouble ans = returnTimes*onecmp; }}bear;/*int main () {freopen ("In.txt", "R",

, found that there is a better wording, do not need to pre-processing what, direct O (n) sweep, with two variables sentenced to the line.Best Code#include using namespace STD;classniceorugly{ Public:BOOLIsvol (Charc) {returnc = =' A '|| c = =' E '|| c = =' I '|| c = =' O '|| c = =' U '; }stringDescribestrings) {BOOLCan1 =0;intNV, NC; NV = NC =0; for(inti =0; I 0; i++) {if(S[i] = ='? ') nv++, nc++;Else if(Isvol (S[i])) nv++, NC =0;ElseNV =0, nc++;if(NV >=3|| NC >=5) Can1 =1; }BOOLCan2 =1; NV = NC

Topic descriptionGiven a sequence of size n (nExercisesThe number of n as a binary, if the number of n is required and equal to the number of N or value, then for each bit of n number, there may be a maximum of 1, because more than one will produce a carry.Because number I if the current bit is placed 0, but the current bit of r[i] is 1, then the position can be placed at will, there is no limit, so we can use dp[i][j] to indicate in the first bit, the current state is J of the number of conform

Almost the same title. For more information, see: http://community.topcoder.com/stat?c=problem_statementpm=13211rd=15857Thinking:1 sequence2 calculation of the current number of buildings I, all possible minimum building changes3 each time the calculation i+1. Total possible minimum number of building changes4 minimum number of changes when you can get i+1 at the same timeAnd the procedures are not complicated.#include Copyright notice: The author Heart Jing, Jing Space address: http://blog.csd

what the Scrum Team would work on NE xt And,? Ensuring the development Team understands items in the Product Backlog to the level needed.3. The Development TeamThe development Team consists of professionals who does the work of delivering a potentially releasable Increment of ' done ' Product at the end of each Sprint. Only members of the development Team create the Increment.ArtifactsOrder Product Backlog: high-level requirements sorted by priority. Sprint Order Sprint Backlog: A checklist of

-------always wanted to play SRM, but it felt TopCoder used too much trouble. The topic is English, but there is nothing to do or a dozen. But just registered the number can only play DIV2, anyway, I am so weak is only suitable for DIV2.T1:Main topic:The 8*8 's chessboard is judged by whether there are only 1 pieces per row in each column.Puzzle: I don't know how to write a puzzle. T2:Main topic:There are e e,em a em,m m,mh a mh,h H. is less tha

Test instructions[Topic link] This How to send a link ah .....There \ (n\) Zhang Fuka in a queue, each spell card has two attributes, Rank \ (li\) and damage \ (di\).You can do any operation, each one of the following:Move the team's first spell card to the end of the team.Use the team's first spell card to damage the enemy ( d_i\) and discard the l_i\ Zhang Fuka (including the spell card you used). If the queue is insufficient \ (l_i\) Zhang Fuka Then you cannot use it.The maximum value of the

A: eliminate two adjacent equal characters. I really don't like STL .. #include B: Give a tree with N nodes, and find the maximum m value (n First, Floyd finds the distance between all vertices. Then, the solution is enumerated in a way similar to the State compressed DP. Then compare the maximum value. #include C: Exercise your thinking. Reference others' practices: Use an array similar to a suffix. It's just that the array sets are really drunk. My solution is to find the minimum string that

Http://community.topcoder.com/tcI saw a document about how to use topcoder's SRM for exercises and competitions, and then I learned a little bit about it. I still want to explain it to me !!First of all, registration is very tedious. A bunch of information needs to be filled in, and it is all in English, but the English level is not so good, so I had to look up the dictionary while filling in the information. For details, refer to the registration and

Topcoder SRM 638 DIV 2 (great miracle), topcodersrm Water is a brute force attack. A great miracle. Problem Statement There is a narrow passage. Inside the passage there are some wolves. You are given a vector SizeThat contains the sizes of those wolves, from left to right.The passage is so narrow that some pairs of wolves cannot pass by each other. More precisely, two adjacent wolves may swap places if and only if the sum of th

SRM 631 DIV1SRM 631 DIV1 A: A maximum of two steps are required. The two rows in the middle are black and one line is white. In this case, you only need to consider the first step, and enumerate a row to dye the satisfied situation, just brute force. B: greedy. A record records the current location of a cat and the current location of more than one cat, and then sorts the positions from left to right. If the current location can be moved to more than

SRM 452 div2Problems 1000 question: http://www.topcoder.com/stat? C = problem_statement PM = 10572 RD = 13906 answer: If there exist loop formed by required edges or node with degree> 2, then the answer is 0. if there are no loops and no nodes with degree> 2, then the graph will consist of S separate nodes and C chains. we need to count the number of ways to concatenate them to form a single path. we have (S + C )! Ways to order them, and for each c

The brain hole is too big. Simple things are complicated and deserve div2; A: water, basic judgment a [I] B: I don't know how others do it. My number is 100 * n. I can't think too much. I forget that it is a continuous series. We can enumerate the tolerances and find the number of tolerances. C: I think of map, but I am too confused to write it in just a few minutes .. But it is not necessarily written .. It feels good to come in to DP .. We can find that the product is actually less .. Then, j

Here you can swap any two adjacent wolves. Thus, all 3! = 6 permutations are possible. 2) {1, 2, 3} 4 Returns: 3 You can get {1, 2, 3}, {2, 1, 3} a

= =0){//found one node whose in-order degree is zeroVisit[i] =true; VectorBOOL> tmp (N,false); for(j =0; J j) { if(Conn[i][j]) {Tmp[j]=true; CONN[I][J]=false;//remove its out-order edge.}} CNT+=Dfscount (conn); for(j =0; J j) { if(Tmp[j]) conn[i][j] =true; } Visit[i]=false; } } returnCNT; }};DFS contains a lot of repetitive calculations, after all, recursion.The solution of the dynamic programming can be referred to the answer to the questio

each N and for each permutation p of 1 through N the probability that p appears as the output of L Imak ' s program is strictly positive. Constraints - sortedsequence would contain exactly N elements. - N would be between 1 and inclusive. - Elements of Sortedsequence would be between 1 and N, inclusive. - Elements of sortedsequence would be pairwise distinct. Main topic:In the merge sort process,

Label: blog ar OS for SP Div 2014 Log Code 2333... Due to the small number of TC entries and the constant FST, I dropped to div2. It's okay to go back to div1 .. 250 Question 500 Question .. Just extend the BFS directly. Pay attention to the heavy judgment. I also launched it with Kanto .. 1000 This question is understood as incorrect .. I told you why someone else's code looks wrong. However, it is very easy to answer a question. Which leaf nodes are burned by binary e

flipped to the substring whose start length is k in position J in string B. steps int n = a.size(); memset(dp, 0x3f, sizeof(dp)); for(int j = 0; j 1000 The formula is simple. N * (1/n + 1/(n-1) + 1/(n-2) +... + 1/(n-k + 1 )) The key issue is coming. N and K are huge. Then we found that this is a harmonic series sum. When the number is large, only the approximate formula is used. Try it. (1/n + 1/(n-1) + 1/(n-2) +... + 1/(n-k + 1) is approximately equal to log (n + 1) + R R i

Question: Give a vertex (x, y), give some step delta1, delta2... deltan: Can I reach the (x, y) point after completing n steps in strict accordance with the step (0, 0. Solution: in fact, it is to determine whether these line segments and (0, 0)-(x, y) can constitute a multilateral (angle ?) You only need to judge whether the longest side is less than half of the length of all sides. Code: #include View code Topcoder SRM 633 div.2 500 jumping