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MO Team Algorithm Summary preface:Again, like the Chairman tree, Oi god Ben invented the algorithm.When I first heard the name of the algorithm, I guessed that the team was a great god Ben. Did not expect to know that he unexpectedly is our seniors! If you do not see the person, first smell his name!Until later CJ was designated as the PKUWC, CJ renovated the engine room and hung up the honor wall in order to provide a better environment. So I can see
2. Input/output optimizations are recommended. SourceSolutionSub-block + MO TeamVery good to think, at the beginning of the wrong question, did not write Mo team, directly divided into blocks +lower_bound and then found not a sample ...In fact, very good to think, the weight of the sub-block, with
Topic: Given an array of length n, m times ask for a MEX value within a rangeAngry write MO team 233The weights are divided into √n blocks, recording the number of occurrences of each weight and how many weights appear in each block.Modify O (1) to complete the query when you first scan to find the first block within the bloc
"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 te
Y (or root and an apple). Next there are m lines, each line containing four integers u, V, A and B, which means that the person wants to count the number of Apple U to Apple V, and this person thinks color A is color B. If a=b=0, it means that the person does not suffer from color blindness. OutputThe output is a total of M lines, each line contains only one integer, representing the number of colors the p
Topic link series searchThink of the pieces and run the team,Set $c[i]$ to the number of occurrences of the $i$ within the current maintenance interval,$g [i]$ is the number of occurrences of $i$ in the currently maintained interval,$BG [i]$ The number of occurrences, $bg the meaning of the i]$ is the number of occurr
Title Address: HDU 5145Mo team is really good magic. This complexity is only N*SQRT (n) ...Bare team tiles, first offline, then by the left end of the block, by the number of blocks as the first keyword sort, and then by the R value as the second keyword to sort. are small to large, can prove that such a complexity of only N*SQRT (n). Then the transfer between blocks is performed.The code is as follows:#inc
, fm/g); } } return 0;}This section of code space is too large to be AC, but very useful, and universal. If you do not want to see a direct look at the next article, more important and effective.This problem is not already used online algorithm a over, how to again?Don't forget! The range additive was used before!This time the algorithm does not use the interval to be additive, also does not have the interval to reduce the sex, how does it do?From one endpoint to the interval of each point
Title.#include "Mo Team Algorithm" "Weight-value block" bzoj3809 Gty's two-forced sister sequence
100% of data, n≤10000,m≤10000, no more than 1000 modifications, all the integers appearing in the input data are greater than or equal to 1 and no more than 10^6.2016.3.2 new data and two groups by Nano_ape Resources: 1.http://blog.csdn.net/a_crazy_czy/article/details/51544720 2.http://blog.csdn.net/werkeytom_ftd/article/details/51540342 Block of sequencesFor the query operation, the
number of a[i]^k, and update the answer on the LINE. The above is a different or processing part. Next is the MO team Algorithm. The MO team algorithm is a brute force algorithm used to solve interval problems. His principle is that if I know the value of [l,r], I can find the values of [l+/-1,r] and [l,r+/-1] at O1 time, then the values of query [l+/-k,r] and [
);printf("%lld\ n", sum[r]-sum[l-1]-(ss[r]-ss[l-1])*(L-1)); }}bool CMP (int x,int y){returnb[x]y];} BOOL Cmp1 (W a,w b) {returnbl[a.l]intMain () {scanf ("%lld%s", mod,s+1); N=strlen (s+1); block=sqrt(n);if(mod==2|| mod==5) {solve_2_5 ();return 0; } a[0]=1; for(intI=1; iTen*a[I1]%mod; for(inti=n;i>=1; i--) b[i]= (A[n-i]*(s[i]-' 0 ') +b[i+1])%mod; for(intI=1; iSort(c+1, c+n+1, CMP); for(intI=1; iif(b[c[i]]==b[c[i-1]]) H[c[i]]=tot;ElseH[c[i]]=++tot; for(
pairs of i and J for the first query is: (1, 2), ( 1, 4), (1, 5), ( 2,3 ), (3, 6), (5, 6), (6, 6). Not a single of these pairs are suitable for the second query. In the second, the sample XOR equals 1 for any subarrays of an odd length.Main topic:There is a series of N of length, M query (L,R) within how many pairs (i, j) make Ai ^ ai+1 ^ ... ^ aj = K;Problem Solving Ideas:The interval query here is offline and can be used in the legendary MO
First with the MO team algorithm to ensure the complexity of the transfer between inquiries, each transfer requires O (sqrt (m)) insertion and deletion, the weight of the block insertion/deletion is O (1).Then ask the time with the weight of the block query interval k small value, each time is O (sqrt (n)).So the total complexity is O (m* (sqrt (n) +sqrt (m)).The
any subarrays of an odd length.Test instructions Give you a sequence of length nM inquiries, every time you ask how many pairs of l,r between the sub-range are equal to KExercisesGodless, a team of Mo.#include using namespacestd;Const intN = 1e6+Ten, M = 5e6+Ten, INF = 1e9+7, mod = 1e9+7;intH[m];intSum[n],n,m,k,a[n],pos[n];Long LongAn[n],ans;structss{intL,r,id;} Q[n];intCMP (ss S1,ss S2) {if(POS[S1.L]==POS[S2.L])returns1.rElse returns1.lS2.L;}intCmp
]; }}q[n];structcmeow{intP,v,last;} Cq[n];intP,tim, ans[n];intC[m], now;intL=1, r=0, cur=0; inlinevoidAddintx) {now+= (++c[x]) = =1;} InlinevoidDelintx) {now-= (--c[x]) = =0;} InlinevoidChaintPintv) { if(Lr) Add (v), Del (A[p]); A[P]=v;}voidModui () { for(intI=1; i){ while(Cur, Cha (cq[cur].p, CQ[CUR].V); while(Cur>q[i].tim) Cha (cq[cur].p, cq[cur].last), cur--; while(R, add (A[r]); while(RGT;Q[I].R) del (A[r]), r--; while(L; while(LGT;Q[I].L) l--, add (A[l]); ans[Q[i].id]=Now ;
Title Link: Http://codeforces.com/problemset/problem/617/ETitle:Here you A1 A2 A3 An number, M inquiry: in [L, R] inside how much [L, R] makes Al xor al+1 xor AR is k.The following:in the subject only interval query has no interval modification, and the amount of data is small (10w ), so it can be solved in an offline way. Using the MO team algorithm to solve, you need O (1) Modification [L, r+1], [L, R-
Tags: here sele combine temp sap roo where SEQ itemSELECT * from Out_mo_res WHERE pegged_id= ' 001201271060 '; --5000175080/160_1-mfg0011SELECT * from V_out_mo;SELECT * FROM V_out_mo_item_1 out_mo_itemLeft join Sap_marc on substr (out_mo_item.loc_id, 0, 4) =sap_marc.werks and OUT_MO_ITEM.ITEM=SAP_MARC.MATNRWHERE productionordid= ' 5000175080/160_1-mfg0011 ';SELECT * from Out_mo WHERE productionordid= ' 5000175080/160_1-mfg0011 ';SELECT * from Out_mo_item WHERE productionordid= ' 5000175080/160_1
consists of n positive integers separated by a space, and the first I represents the size of the data that is numbered I.The third line is a positive 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.OutputQ lines, one positive integer per line, indicating the number of times the Mato need to be exchanged this day
Very interesting, you can carefully appreciate the problem, see White is Mo Team board. DescriptionSmall B has a large number of S, the length reached N bits; This number can be seen as a string, it may have a leading 0, for example 00009312345。 Small B also has a prime number p. Now, small B raises M queries, and
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