BZOJ2741: "Fotile simulation Game" L

2741: "Fotile simulator" L time limit:15 Sec Memory limit:162 MB
submit:1170 solved:303
[Submit] [Status] DescriptionFotile gets a long sequence of N, a, in order to save the Earth, he wants to know the largest contiguous XOR within certain intervals. that is, for a query, you need to ask for Max (Ai xor ai+1 xor ai+2 ... xor Aj), where l<=i<=j<=r. to reflect the online operation, for an inquiry (x, y):L = Min ((( X+lastans) mod n) +1, ((Y+lastans) mod n) +1 ).
r = Max (((X+lastans) mod n) +1, ((Y+lastans) mod n) +1). where Lastans was the answer to the last question, starting at 0. InputThe first line is two integers n and M. The second line has n positive integers, where the number of I is AI and there are extra spaces. The two digits x, y for each line of the rear m row represent a pair of queries.
OutputA total of M lines, and a positive integer on line I indicates the result of the I query. Sample Input3 3
1 4 3
0 1
0 1
4 3


Sample Output5
7
7HINT

HINT

N=12000,m=6000,x,y,ai is within the range of signed Longint.

Source

by Seter

Exercises
Handling: by Seter

Divided into blocks of length L, where the first point of each block is called the key point, there is a total of k=n/l key points.

Processes an array of k*n that represents the answer to each point after each key point. This for a key point is can O (NLGN) out, with the general trie on it, this will not be able to first go to water Usaco again.

Then for a query (U,V), it can be decomposed into (u,v) and (X,v), where X is the first key after U.

Then (X,V) has been processed out, the rest is u ... X the number of O (L) in Max Xor (U,V).

The problem is converted to, for a period of time and X of the difference or maximum.

This thing can be easily chairtrie with the use of a. Chairtrie and Chairtree Almost, is the function of the trie, from high to low processing x one, if you can go in the opposite direction, go, you can ...

Some comments are written in the code.
Code:

1#include <cstdio>2 3#include <cstdlib>4 5#include <cmath>6 7#include <cstring>8 9#include <algorithm>Ten  One#include <iostream> A  -#include <vector> -  the#include <map> -  -#include <Set> -  +#include <queue> -  +#include <string> A  at #defineINF 1000000000 -  - #defineMAXN 150000+5 -  - #defineMAXM 5000000+5 - #defineMAXK 150 in  - #defineEPS 1e-10 to  + #definell Long Long -  the #definePA pair<int,int> *  $ #defineFor0 (i,n) for (int i=0;i<= (n); i++)Panax Notoginseng  - #defineFor1 (i,n) for (int i=1;i<= (n); i++) the  + #defineFor2 (i,x,y) for (int i= (x); i<= (y); i++) A  the #defineFor3 (i,x,y) for (int i= (x); i>= (y); i--) +  - #defineMoD 1000000007 $  $ using namespacestd; -  -InlineintRead () the  - {Wuyi  the     intx=0, f=1;CharCh=GetChar (); -  Wu      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} -  About      while(ch>='0'&&ch<='9') {x=Ten*x+ch-'0'; ch=GetChar ();} $  -     returnx*F; -  - } A intn,m,q,tot,rt[maxn],id[maxm],t[maxm][2],A[MAXN],B[MAXK][MAXN]; +InlinevoidInsertintPreintXintk) the { -     intnow=rt[k]=++tot;id[tot]=K; $For3 (I, -,0) the     { the         intJ= (x>>i) &1; thet[now][j^1]=t[pre][j^1];//the null pointer points to the original thet[now][j]=++tot;id[tot]=K; -now=tot; inPre=T[pre][j]; the     } the } AboutInlineintAskintLintRintx) the { the     intans=0, now=Rt[r]; theFor3 (I, -,0) +     { -         intJ= (x>>i) &1)^1; the         if(id[t[now][j]]>=l) ans|=1<<i;Elsej^=1;//The following nodes have IDs that are smaller than L, so change the directionBayinow=T[now][j]; the     } the     returnans; - } -  the intMain () the  the { the  -Freopen ("Input.txt","R", stdin); the  theFreopen ("output.txt","W", stdout); the 94N=read (); q=read (); theFor1 (i,n) a[i]=a[i-1]^read (); theid[0]=-1; theInsert (rt[0],a[0],0);//Insertion 098For1 (i,n) Insert (rt[i-1],a[i],i);//Insert Prefix XOR value About     intLEN=SQRT (n); m=n/len+ (n%len!=0); -For0 (i,m-1)101For2 (j,i*len+1, N)102B[i][j]=max (b[i][j-1],ask (i*len,j-1, A[j]));//Chunking , using a[j] to query the maximum value in I*len-j-1103     intans=0;104      while(q--) the     {106         intx= (LL) read () + (ll) ans)%n+1, y= (LL) read () + (ll) ans)%n+1;107         if(x>y) Swap (x, y); x--;108         intbx=x/len+ (x%len!=0);109Ans=bx*len<y?b[bx][y]:0;//chunks of the answer have been given theFor2 (J,x,min (bx*len,y))111Ans=max (Ans,ask (x,y,a[j));//query Max with point violence within a small block theprintf"%d\n", ans);113     } the  the     return 0; the 117}

View Code

BZOJ2741: "Fotile simulation Game" L