how to reset moto z

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

() { while(~SCANF ("%d%d",n,m)) { intTP =ceil (sqrt (n)); for(inti =1; I ) {scanf ("%d",A[i]); Pos[i]= (I-1)/TP;//The position of i-th} memset (num,0,sizeof(num)); for(inti =1; I ) {scanf ("%d%d",q[i].l,Q[I].R); Q[i].id=i; } sort (Q+1, q+m+1, CMP); intPL,PR; PL=1, PR =0; Ans=0; for(inti =1; I ){ if(Q[I].L = =Q[I].R) {Son[q[i].id]=0; Mom[q[i].id]=1; Continue; } Else{ if(PR Q[I].R) { for(intj = pr +1; J ) {Updata (J,1); } PR=Q[I].R; } Else{

problem_id=1551 "> Topic linksMain topic:The logarithm of the number in the interval less than or equal to 2 is calculated.Thinking Analysis:MO team algorithm.Then analyze it.Suppose you add a number. Then add it next to the number of the difference of 2.Conversely lower one. It is necessary to reduce the number of the difference between 2 and. Subtract yourself again this 1.#include WHU OJ 1551 Pairs (Mo team algorithm)

+1, b[i]=A[i]; Sort (b+1, b+n+1); intCnt=2; for(intI=2; ii)if(b[i]!=b[i-1]) b[cnt++]=B[i]; --CNT; for(intI=1; ii) a[i]=lower_bound (b +1, B +1+cnt,a[i])-(b +1); memset (b,0,sizeof(b)); scanf ("%d",q); for(intI=1; ii) scanf ("%d%d", AMP;RES[I].L,AMP;RES[I].R), res[i].id=i; Sort (Res+1, res+1+Q,CMP1); for(intI=1, l=1, r=0; ii) { for(; rr) Change (R+1,1); for(; r>res[i].r;--r) Change (R,-1); for(; ll) Change (L,-1); for(; l>res[i].l;--l) Change (l-1,1); Res[i].ans=sum; } Sort (Res+1, res+q+1

of the output should contain single positive integer-the p Ower of the i-th query subarray. %lld specificator to read or write 64-bit integers in C + +. It is a preferred to use cout stream (also). Examplesinput3 21 2 11 21 3Output36Input8 31 1 2 2 1 3 1 12 71 62 7Output202020NoteConsider the following array (see the second sample) and its [2, 7] Subarray (elements of the Subarray is colored): Then k1 = 3, k2 = 2, k3 = 1, so the power was equal to 32 1 + 2 2 2 + 12·3 =. AC Co

Same as BZOJ2038. Review it.1#include 2#include 3#include 4 using namespacestd;5 intBlock;6 structquery{7 intI,l,r;8 BOOL operatorConstQuery q)Const{9 if(L/block==q.l/block)returnrQ.R;Ten returnl/blockBlock; One } A}query[220000]; - intcnt[1111111]; - Long Longres,ans[220000]; the voidInsertLong Longx) { -res-=x*cnt[x]*Cnt[x]; -++Cnt[x]; -res+=x*cnt[x]*Cnt[x]; + } - voidRemoveLong Longx) { +res-=x*cnt[x]*Cnt[x]; A--Cnt[x]; atres+=x*cnt[x]*Cnt[x]; - } - inta[220000]; -

Shaped like Hdu 5381: Click to open linkThe function complexity of add (int x, int y) is O (|x-y|)Del similarly#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. MO Team Algorithm Template

consists of n positive integer ci, where CI denotes the color of the sock, and the same color is represented by the same number. The next m line, two positive integer l for each line, R indicates a query.OutputContains m rows, for each query the output fraction of a line is a/b indicating the probability of randomly extracting two socks of the same color from the range [l,r] of the query. If the probability is 0 then output 0/1, otherwise the output of A/b must be the simplest fraction. (See ex

http://www.lydsy.com/JudgeOnline/problem.php?id=1878Test instructions: ...Idea: It's a lot easier than the previous question. Number is small, open an array hash record occurrence number (remember the array to open 1e6), and then directly counted on the line.1#include 2#include 3#include 4#include 5 using namespacestd;6 #defineN 500107 structNode {8 intL, R, ans, id;9} p[n*4];Ten intmp[n* -], Num[n], ans, kuai; One A BOOLcmpConstNode a,ConstNode b) { - if(A.l/kuai = = B.l/kuai)returnA.R

This code is based on the MTK platform Android 5.1 as the analysis object, and Google native Aosp a little difference, please read the reader.This figure is mainly based on the Android source code to dial the phone process to draw, recorded the phone calls the main process:Reference Blog: http://blog.csdn.net/yihongyuelan/article/details/45098339Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.Android5.1 Telephony Process Analysis-call pr