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1#include 2 3 using namespacestd;4 Const intN =1 -;5 structnode6 {7 intL,r,id;8 }q[n];9 Ten intAns[n]; One Long LongAnswer[n]; A Long LongFlag[n]; - intA[n]; - BOOLCMP (node X,node y) the { - if(ans[x.l]==ANS[Y.L]) - returnx.rY.R; - returnans[x.l]ANS[Y.L]; + } - intn,m,k; + intL =1, R =0; A Long Longans=0; at voidAddintx) - { -ans+=flag[a[x]^K]; -flag[a[x]]++; - } - voidDelintx) in { -flag[a[x]]--; toans-=flag[a[x]^K]; + } - intMain () the { * $scanf"%d%d%d",n,m,k);Panax Not

1 /*2 Test Instructions:3 Ideas:4 Time:5 */6#include 7 using namespacestd;8 9typedefLong LongLL;Ten Const intmaxn=100005; One ConstLL MOD7 = 1e9+7; A - structQuery - { the intL,r; - intidx; - }Q[MAXN]; - + intA[MAXN]; - intBelong[maxn],qsize,num; +LL flag[1 -],ans; A LL ANS[MAXN]; at intn,m,k; - - intCMP (Query oa,query OB) - { - if(belong[oa.l]==BELONG[OB.L]) - returnoa.rOB.R; in returnbelong[oa.l]BELONG[OB.L]; - } to + voidADD (intx) - { theans+=flag[a[x]^K]; *flag[

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.Test instructions: There are n number and M inquiry, each inquiry will have a L and R, indicating the interval of inquiry,Ask how many consecutive sub-ranges are in this interval or KSuppose we now have a prefix XOR or array sum[], now we ask for the XOR value of the interval [l,r],Using

"BZOJ3529" [Sdoi2014] Tables descriptionThere is a table of NXM, and the number of rows J (1 InputThe input contains multiple sets of data.The first line of input, an integer q, represents the number of data groups in the test point, and the next Q row, three integers per line n,m,a (|a| OutputFor each set of data, the output is an integer that represents the value of the answer modulus 2^31.Sample Input24 4 310 10 5Sample Output20148HINT1 The first thing we need to know is the approximate numb

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51nod 1244 The sum of the function of the MO link to original topic the https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1244 topic is to let you calculate the Metos function. Classic du sieve u∗i=e u*i=e m (n) =1−∑i=2nm (⌊ni⌋) m (n) =1-\sum_{i=2}^nm (\big\lfloor\frac{n}{i}\big\rfloor) #include

The sum of the functions of the MO-colorimetric Base time limit: 3 seconds Space limit: 131072 KB Score: 320 Difficulty: 7-level algorithm problemhttp://www.51nod.com/onlineJudge/questionCode.html#!problemId=1244 Möbius function, presented by German mathematician and astronomer Möbius. Mertens (Mertens) first uses μ (n) (Miu (n)) as the token of the Möbius function. It is defined as follows:If a number contains a square factor, then Miu (n) = 0. For

Title Address: HDU 4638First wrote a team, MO team can water. Very simple MO team, not much to say.The code is as follows:#include #include #include #include #include #include #include #include #include #include using namespace STD;#define LL Long Long#define PI ACOs ( -1.0)#pragma COMMENT (linker, "/stack:1024000000")Const intMod=1e9+7;Const intinf=0x3f3f3f3f;Const Doubleeqs=1e-9;Const intmaxn=100000+Ten;i

Learn a very basic block and Mo team algorithm, because not very will write Manhattan distance smallest spanning tree, so wrote a block version of (in four cases, probably the meaning of it) ...cogs1775| | bzoj2038 Little Z's socksTopic: Static interval Query the number of different kinds of elements.Idea: With Mo team Sweep, and then numerator denominator multiply 2, you will find that the denominator is c

), the upper operating time is not reduced.The MO team algorithm is a good way to solve the above problems. The MO team algorithm thinks that after the left endpoint is sorted, the left end of the terminal movement must be few, but the right endpoint moves much more often. If you reduce the number of right end moves appropriately, even a little bit more left end-point moves can be cost-effective in total co

equivalent to the selection of B in a different item number of selection scheme."Data size and conventions"30% of the data are n,m≤5000;60% of the data are n,m≤25000;100% of the data in N,m≤50000,1≤l HINT SourceCopyright owner: Mo TaoIdea: Mo team algorithm, from the interval L, R transfer to L1, R1 need to pay the price |l-l1| +| r-r1|, block-after-offline query algorithm complexity o (n * sqrt (n))Attent

Tag: MAT ret Indicates practice struct update color LLD prefixTest instructionsA sequence of length n, an M inquiry, and a K. Each query asks how many consecutive sub-sequences from the L to R in the sequence are different or up to K.Analysis:Because know beforehand this question can use Mo team to write, just use this question practice Mo team.Preprocessing each prefix XOR or.Then, after the team is sorted

output " No", otherwise output "Yes".Exerciseshash+stl-map+-Mo Team algorithmSince the length of the inquiry sequence is given and does not exceed 20, we can pre-preprocess all successive sub-sequences with K in the original sequence and store them in a hash method, which is discretized with the map.For each query to hash it out, if it does not appear in the original sequence of the direct answer is 0, otherwise with the

Http://codeforces.com/problemset/problem/86/DThe MO team algorithm is to adjust the order of the query, then the brute force solution.Each time the interval solution of the interval (l+1,r), (L-1,r), (l,r+1), (l,r-1) can be obtained through the existing interval solution ans (l,r).Adjustment mode http://blog.csdn.net/bossup/article/details/39236275This problem is more simple than that, the query is k^2*z, very clear is the

https://www.lydsy.com/JudgeOnline/problem.php?id=2120In the title there are two different methods.With modified Mo team and ordinary Mo team more than a modification operation, the impact is not big, but notice the details do not appear zz error.The number of revisions to the problem can be written in the team, but it is probably not possible to modify the numbers or the extreme data.1#include 2#include 3#i

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