BZOJ3261: Maximum XOR and

3261: Maximum xor and time limit:10 Sec Memory limit:512 MB
submit:543 solved:237
[Submit] [Status] Description     

Given a non-negative integer sequence {A}, the initial length is N.
There are two operation types for m operations:

1, a x: Adds an operation that represents adding a number x to the end of the sequence, n+1 the length of the sequence.
2, Q l R x: Ask the operation, you need to find a position p, meet l<=p<=r, make:

A[P] XOR a[p+1] xor ... xor a[n] xor x maximum, the maximum output is what.

Input

The first line contains two integers N, M, meaning as shown in the problem description.
The second line contains n non-negative integers that represent the initial sequence a.

Next m line, each line describes an operation, the format is described in title face.

Output

Assuming that the ask operation has a T, the output should have a T line, and an integer per line indicates the answer to the question.

Sample Input5 5
2 6 4) 3 6
A 1
Q 3 5 4
A 4
Q 5 7 0
Q 3 6 6
For Test point 1-2,n,m<=5.

For Test point 3-7,n,m<=80000.
For Test point 8-10,n,m<=300000.

Where test points 1, 3, 5, 7, 9 guarantee no modification operation.
For 100% of data, 0<=a[i]<=10^7. Sample Output4
5
6
HINT

For 100% data, 0<=a[i]<=10^7 .

Exercises

Magical and sustainable trie!

The prefix and the X-xor-or-l-1-r-1 trie tree can be queried.

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 1000000+5 -  - #defineMAXM 20000000+5 -  in #defineEPS 1e-10 -  to #definell Long Long +  - #definePA pair<int,int> the  * #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 #defineFor3 (i,x,y) for (int i= (x); i>= (y); i--) the  + #defineMoD 1000000007 -  $ using namespacestd; $  -InlineintRead () -  the { - Wuyi     intx=0, f=1;CharCh=GetChar (); the  -      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} Wu  -      while(ch>='0'&&ch<='9') {x=Ten*x+ch-'0'; ch=GetChar ();} About  $     returnx*F; -  - } - intn,m,tot,rt[maxn],id[maxm],t[maxm][2]; AInlinevoidInsertintPreintXintk) + { the     intnow=rt[k]=++tot;id[tot]=K; -For3 (I, -,0) $     { the         intJ= (x>>i) &1; thet[now][j^1]=t[pre][j^1]; thet[now][j]=++tot;id[tot]=k;now=tot; thePre=T[pre][j]; -     } in } theInlineintQueryintLintRintx) the { About     intans=0, now=Rt[r]; theFor3 (I, -,0) the     { the         intJ= (x>>i) &1)^1; +         if(id[t[now][j]]>=l) ans|=1<<i;Elsej^=1; -now=T[now][j]; the     }Bayi     returnans; the } the  - intMain () -  the { the  theFreopen ("Input.txt","R", stdin); the  -Freopen ("output.txt","W", stdout); the  theN=read (); M=read ();intsum=0; theid[0]=-1;94Insert (rt[0],0,0); the For1 (i,n) the     { thesum^=read ();98Insert (rt[i-1],sum,i); About     } -      while(m--)101     {102         CharCh=GetChar ();103          while(ch!='A'&&ch!='Q') ch=GetChar ();104         if(ch=='A') {sum^=read (); Insert (rt[n],sum,n+1); n++;} the         Else106         {107             intL=read (), R=read (), x=read ();108printf"%d\n", Query (l1, R-1, sum^x));109         } the     }111  the     return 0;113  the}

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BZOJ3261: Maximum XOR and