at t independence mo

Test instructions: Chinese question.Analysis: The very famous MO team algorithm, the first expression of this probability, should be the molecule: C (x1, 2) + C (x2, 2) + C (X3, 2) + ... + c (xn, 2) Denominator: C (n, 2), and then the expression of fractions, [X1 (x1-1) +x2 (x2-1) +...+XN (xn-1)]/(n (n-1)) then the simplification is obtained (Sigma (XI*XI)-N)/(n (n-1)), then the operation of each interval, offline, so that the sequence is divided into

Title Link hh NecklaceThis problem can be directly on the chairman of the tree template#include Of course, you can do it with the MO team algorithm.#include Bzoj 1878 [Sdoi2009]hh's Necklace (Chairman tree or MO team algorithm)

Facet links/*code correctness is not guaranteed. in the DFS sequence do mo teamwhen a point is not another point of the LCA, it is necessary to add their LCA contribution*/#include #include #include #include #define GC () GetChar ()Const intn=4e5+5, m=1e5+5;intn,m,size,enum,h[n],nxt[n1],to[n1],dep[n],fa[n],top[n],son[n],sz[n];intnow,ans[m],seq[n1],in[n],out[n],id,a[n],tm[n];BOOLVis[n];structques{intL,r,lca,id,x,y;BOOL operatorConstQues a)Const{returnl

Topic Link Tree and QueriesThe title gives the color of a tree and each node. Every time you ask VJ, KJYou need to answer the number of colors that meet the conditions in a subtree rooted in VJ,Condition: The number of nodes with this color is at least KJ.(Mo team can actually live)Turn the DFS sequence first, and this becomes the interval query.Then apply the team directly, find the t[],t[k in each query state] to indicate the number of colors that h

)) \\=\sum\limits_{x} (s_x (1,r_1) · S_x (1,r_2)-s_x (1,l_1-1) · S_x (1,r_2)--s_x (1,r_1) · S_x (1,l_2-1) +s_x (1,l_1-1) · S_x (1,l_2-1)) \\=q (r_1,r_2)-Q (l_1-1,r_2)-Q (r_1,l_2-1) +q (l_1-1,l_2-1) $which$Q (A, B) =\sum\limits_{x}s_x (1,a) · S_x (1,B) $You can then split each query into 4, using the MO team algorithm to calculate the contribution to each answer separately.Note that when one of the $a$ or $b$ is 0 o'clock, it needs to be filtered out,

HDU-6333Test instructions: There are n different apples, you can use up to M, ask how many kinds of ways, multiple sets of data, groups and n,m are 1e5, so the table can not play.Idea: This problem to use the combination of the properties of the number of S (n,m) from N to the maximum number of M-method total, is obviously C (n,0), C (n,1) ... C (N,m) and.Obviously s (n,m+1) = S (n, m) + C (n,m+1);Another equation is less obvious, s (n+1,m) = 2 * S (n,m)-C (N,M);I also understand under the guida

DescribeGiven a sequence of n (n≤50000), M (m≤50000) asks for the minimum number of times the sequence does not fall by the adjacent element 22 interchange in the [L, R] interval.Analysis The first conversion to a reverse order of the problem, the minimum number of exchanges is the number of reverse pairs. The latter proves to be not rigorous or even wrong, but it can be used as an inspiration and reference: for an element x in a sequence, there is a C for the element that is smaller than it

Mo Team Classics. The number of times the number of open arrays maintained a[i] occurred.1#include 2#include 3#include 4 using namespacestd;5 intnum,ch[ A],f,c;6InlinevoidRintx) {7C=0; f=1;8 for(;c'0'|| C>'9'; C=getchar ())if(c=='-') f=-1;9 for(x=0; c>='0'c'9'; C=getchar ()) (x*=Ten) + = (c'0');Tenx*=F; One } AInlinevoidPintx) { - if(xTen) Putchar (x+'0'); - Else{P (x/Ten);p Utchar (xTen+'0');} the } - intans,anss[50001]; - intn,m,k,

The main topic: a total of M inquiry, each query given a range [L. R], the value of Sigma (C (i) ^2), where I is the value from 1 to K, where C (i) represents the number I in [L. The number of repetitions in R].Idea: Mo team walk up.CODE:#include Bzoj 3781 Small b query Mo team algorithm

Briefly:The MO team algorithm is an algorithm for offline (online) Maintenance of intervals, trees, or other structures based on basic algorithms such as brute force maintenance, tree arrays, chunking, minimum Manhattan distance spanning trees, and a simple, easy-to-read and short-written algorithm for blending. In many cases, this algorithm can easily cut down some complex and difficult-to-write data structure problems.Example: Bzoj2038descriptionAs

"title" #58. "WC2013" Candy Park"Test Instructions" given n points of the tree, m kinds of candies, each point has candy ci. Given n number of WI and M number VI, the value of the nth tasting of the first candy is V (i) *w (j). Q Times ask for a link on the value of each point and or modify a point of candy ci. N,m,q"Algorithm" tree chunking + with modified MO team algorithmReference:WC 2013 Candy Park vfleakingFirst the tree is divided, referring to

"Test Instructions" given the arrangement of the length n, the longest continuous range of the interval [l,r] of M inquiry. n"Algorithm" Mo team algorithmThe "puzzle" considers the amount of information that the team maintains to add a number: Set up[x] to indicate the maximum length of the value x upward, and down[x] to indicate the maximum length of the value x extending down.When you add a number x, up[x]=up[x+1]+1,down[x]=down[x-1]+1. Make t=up[x]

The main topic: give some socks in the order, each time asked how many of the same color in the socks pair.Idea: Mo team algorithm is really a magic algorithm. First, the brute force enumeration is the time complexity of O (n^2), which is certainly not possible. If the interval is guaranteed not to overlap, then the total time transfer complexity can be reduced to O (n). Unfortunately, there is no such guarantee in the title. Thus, the mysterious team

Title DescriptionGive a number string and ask multiple times how many sub-intervals correspond to multiples of p. where P is prime.InputThe first line is an integer: P. The second line is a string: S. The third line is an integer: M. Next m line, two integers per line fr,to, represents a query to the substring of S s[fr...to]. Note: The leftmost number of S has a position ordinal of 1; For example, S is 213567, then s[1] is 2,s[1 ... 3] is 213.N,mOutputOutput m lines, one integer per line, and l

Topic Links:https://www.lydsy.com/JudgeOnline/problem.php?id=3289Ideas:To find the interval of the minimum number of times to change the interval into a non-descending sequence is actually to find the number of interval reverse, the sample interpretation of the problem can not be seen, the sample should be4 and 2 Exchange and 3, so that is the minimum number of exchanges into a non-descending sequence, from the example we can see in fact this is the process of reverse order, but this problem isT

integer q, indicating that Mato will look at the data for a few days. After the Q line of two positive integers per line L, R, indicating Mato this day to see the [l,r] interval of the file. N,q Q lines, one positive integer per line, indicating the number of times the Mato need to be exchanged this day.Sample Input41 4 2 321 22 4Sample Output02Sample interpretation: The first day, Mato do not need to exchangeThe next day, Mato can move the number 2nd 2 times to the end. as long as the team an

http://www.lydsy.com/JudgeOnline/problem.php?id=3289Very naked Mo team ...After the interval is offline and sorted by block, the query is done.In this case, if you know $[l, r]$, consider how to transfer $[l, r+1]$, find is $a[r+1]$ color in this interval ranking, then $r-l+1-rank $ is the number of times that need to move.So in the case because only need bare ranking, so you can consider using bit, that is, after the discrete.And then it's okay.#incl

Mo Team Example.MO Team Study: https://www.cnblogs.com/Paul-Guderian/p/6933799.htmlThe subject molecule is Sigma (C (sum[a[i]],2)) and the denominator is Sigma (l-r+1,2); Maintain molecules and can.MO Team Scope of application: offline, interval, transfer to the next block O (1).1#include 2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 9 Const intn=50010;TentypedefLong LongLL; One intN,m,sq,a[n]; A LL Ans,gcd,sum[n]; - structn

The main topic: given n points, each point has a color, m inquiry, each asked a range randomly selected two points the same probability of the same numberOtz-mo Team algorithm ...The specifics of this, whether it's a block or a Manhattan minimum spanning tree, I'm not going to talk about it.The practice of this problem is to record a CNT array representing how many of each color in the current intervalAdd a point with a color of x on ans+=cnt[x] and c

plus 1temp+= (Long Long) num[a[r]]*Num[a[r]]; } while(l>M[i]. L) {L--; Temp-=(Long Long) num[a[l]]*Num[a[l]]; NUM[A[L]]++;//A[l] The number plus 1temp+= (Long Long) num[a[l]]*Num[a[l]]; } while(r>M[i]. R) {Temp-=(Long Long) num[a[r]]*Num[a[r]]; NUM[A[R]]--;//A[l] The number minus 1temp+= (Long Long) num[a[r]]*Num[a[r]]; R--; } while(lM[i]. L) {Temp-=(Long Long) num[a[l]]*Num[a[l]]; NUM[A[L]]--;//A[r] The number minus 1temp+= (Long Long) num[a[l]]*Num[a[l]]; L++; } ans[m[