BZOJ2038 [2009 Countries Training team] small Z socks (hose) (Mo team algorithm)

The magic of the MO team algorithm, to solve the problem can be offline without modification of the interval query:

• First, the original sequence is divided into blocks, √n block each block √n;
• Then the interval [l,r] of all queries is sorted, first by the block ordinal in which L is in ascending order, if the same as in ascending order by R;
• Finally, in order to find out the results of each query: Know [L,r] answer, and on this basis can be relatively fast in O (x) to get the adjacent interval [l+1,r], [L-1,r], [l,r-1], [l,r+1] answer, so that can be from [l,r] On the basis of the LR plus minus minus gets the answer of any one interval [l ', R '].

Seemingly violent, but the time complexity of doing so is O (x*n*√n)! Because:

• L is arranged according to the number of the block, the same piece within a maximum of √n times ++l or--l to reach the target, a piece of up to about √n times plus minus, a total of √n block, so L change the number of times also √n*√n*√n.
• R in the same piece is ascending, so the same piece up to n times ++r; the next block when R is assumed to reach the farthest in the last piece, that is up to N times--r back to the target, a total of √n blocks, so the R change number is at most (n+n) *√n.
• The cost of the new answer is X, so the time complexity is O (x*n*√n)!

This problem, set each interval [l,r] Each color of the socks number is $a,b,c,d,\dots$, each interval [l,r] The answer is $ (c_a^2+c_b^2+c_c^2+c_d^2+\cdots)/c_{r-l+1}^2$, expand to simplify:

$$ (A^2+b^2+c^2+d^2+\cdots-a-b-c-d-\cdots)/((r-l+1) * (r-l+1-1)) $$

$$ (a^2+b^2+c^2+d^2+\cdots-(r-l+1))/((r-l+1) * (r-l)) $$

where $ (a^2+b^2+c^2+d^2+\cdots) $ can be used as the team algorithm to process the interval answer, open the number of sets of records ABCD ... can be transferred to an adjacent interval in O (1).

In addition, the case of l=r interval.

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <algorithm>5 using namespacestd;6 #defineMAXN 555557 8 intBlock;9 structquery{Ten     intI,l,r; One     BOOL operator< (ConstQuery &q)Const{ A         if(L/block==q.l/block)returnr<Q.R; -         returnl/block<q.l/Block; -     } the }QUERY[MAXN]; -  - Long LonggcdLong LongALong Longb) { -     if(b==0)returnA; +     returnGCD (b,a%b); - } +  A intSEQ[MAXN]; at Long LongCNT[MAXN],ANSX[MAXN],ANSY[MAXN]; - voidInsertLong Long&res,inta) { -res-=cnt[a]*Cnt[a]; -++Cnt[a]; -res+=cnt[a]*Cnt[a]; - } in voidRemoveLong Long&res,inta) { -res-=cnt[a]*Cnt[a]; to--Cnt[a]; +res+=cnt[a]*Cnt[a]; - } the intMain () { *     intn,m; $scanf"%d%d",&n,&m);Panax Notoginsengblock=sqrt (n); -      for(intI=1; i<=n; ++i) scanf ("%d", seq+i); the      for(intI=0; i<m; ++i) { +Query[i].i=i; Ascanf"%d%d",&query[i].l,&QUERY[I].R); the     } +Sort (query,query+m); -     intL=1, r=1; $++cnt[seq[1]]; $     Long Longres=1; -      for(intI=0; i<m; ++i) { -         if(query[i].l==QUERY[I].R) { theansx[query[i].i]=0; ansy[query[i].i]=1; -             Continue;Wuyi         } the          while(l<QUERY[I].L) { - Remove (res,seq[l]); Wu++l; -         } About          while(l>QUERY[I].L) { $--l; - Insert (res,seq[l]); -         } -          while(r<QUERY[I].R) { A++R; + Insert (Res,seq[r]); the         } -          while(r>QUERY[I].R) { $ Remove (res,seq[r]); the--R; the         } the         Long Longa=res-(query[i].r-query[i].l+1), b= (QUERY[I].R-QUERY[I].L+1LL) * (QUERY[I].R-QUERY[I].L), c=gcd (b,a); theANSX[QUERY[I].I]=A/C; ansy[query[i].i]=b/C; -     } in      for(intI=0; i<m; ++i) printf ("%lld/%lld\n", Ansx[i],ansy[i]); the     return 0; the}

BZOJ2038 [2009 Countries Training team] small Z socks (hose) (Mo team algorithm)