# Bzoj 2741 "Fotile simulation" L tiles + persistent trie tree

Source: Internet
Author: User

Main topic

A sequence is given to find the maximum continuous in [L, R] xor And.
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Ideas

First, divide the entire sequence into n   √ block, preprocessing the beginning of each block to the maximum continuous of each number xor And. We just have to deal with the prefix. xor And, after that, it can be done with a trie tree that can be persisted. So the right side of the question is the whole block. The rest of the left side of the casual violence can be over.

CODE
``#define _crt_secure_no_warnings#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 15000using namespace STD;intCNT, asks;intSrc[max];intblocks, size;intBelong[max];intSt[max], Ed[max];structtrie{Trie *son[2];intCNT;} Mempool[max * -], *c = Mempool; Trie *trie[max], **tree = Trie +1; Trie *newtrie (Trie *_, Trie *__,int___) {c->son[0] = _; c->son[1] = __; c->cnt = ___;returnC + +;} Trie *insert (Trie *consult,intCintDeep) {if(!deep)returnNewtrie (null, NULL, CONSULT-&GT;CNT +1);if(! (~c&deep))returnNewtrie (consult->son[0], Insert (consult->son[1], C, deep >>1), consult->cnt +1);Else        returnNewtrie (Insert (consult->son[0], C, deep >>1), consult->son[1], consult->cnt +1);}intGetmax (Trie *l, Trie *r,intXintDeep) {if(!deep)return 0;BOOLdir = x&deep?0:1;if(r->son[dir]->cnt-l->son[dir]->cnt)returnDeep + getmax (L->son[dir], R->son[dir], X,deep >>1);Else        returnGetmax (L->son[!dir], R->son[!dir], x, deep >>1);}intans[ -][max];intMain () {Cin>> CNT >> asks; for(inti =1; I <= CNT; ++i) {scanf("%d", &src[i]); Src[i] ^= src[i-1]; } size =sqrt(CNT +1);memset(St,-1,sizeof(ST)); for(inti =0; I <= CNT; ++i) {Belong[i] = i/size;if(St[belong[i]] = =-1) St[belong[i]] = i;    Ed[belong[i]] = i;    } blocks = belong[cnt]; tree[-1] = Newtrie (c, C,0); tree[0] = Insert (tree[-1],0,1<< -); for(inti =1; I <= CNT; ++i) Tree[i] = Insert (Tree[i-1], Src[i],1<< -); for(inti =0; I <= blocks; ++i) for(intj = St[i]; J <= CNT; ++J) Ans[i][j] = max (Getmax (tree[st[i)-1], Tree[j], src[j],1<< -), J? Ans[i][j-1] :0);intLast_ans =0; for(intX, y, i =1; I <= asks; ++i) {scanf("%d%d", &x, &y); x = ((Long Long) x + Last_ans)% cnt +1; y = ((Long Long) y + Last_ans)% cnt +1;if(x > Y) Swap (x, y); --x;intt = belong[x] +1; Last_ans = Ans[t][y]; for(intj = min (St[t]-1, y); J >= X; --J) Last_ans = max (Last_ans, Getmax (Tree[x-1], Tree[y], src[j],1<< -));printf("%d\n", Last_ans); }return 0;}``

Bzoj 2741 "Fotile simulation" L tiles + persistent trie tree

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