题目大意：给定一个序列，提供下列操作：

1.在数组结尾插入一个数

2.给定l,r,x,求一个l<=p<=r,使x^a[p]^a[p+1]^...^a[n]最大

首先我们能够维护前缀和 然后就是使x^sum[n]^sum[p-1]最大

x^sum[n]为定值，于是用Trie树贪心就可以

考虑到l-1<=p-1<=r-1,我们不能对于每一个询问都建一棵Trie树，可是我们能够对于Trie数维护前缀和，建立可持久化Trie树

每一个区间[l,r]的Trie树为tree[r]-tree[l-1]

注意0要插入一个数字0。所以把-1作为空节点。然后把数组向前推进一位就可以

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 600600
using namespace std;
struct Trie{int cnt;Trie *son[2];
}*tree[M],node[14000000];
int n,m,tot,sum[M];
inline int getc() {static const int L = 1 << 15;static char buf[L], *S = buf, *T = buf;if (S == T) {T = (S = buf) + fread(buf, 1, L, stdin);if (S == T)return EOF;}return *S++;
}
inline int getint() {int c;while(!isdigit(c = getc()));int tmp = c - '0';while(isdigit(c = getc()))tmp = (tmp << 1) + (tmp << 3) + c - '0';return tmp;
}
inline int getch() {int c;while((c = getc()) != 'A' && c != 'Q');return c;
}
inline Trie* New_Node(int _,Trie*__,Trie*___)
{node[tot].cnt=_;node[tot].son[0]=__;node[tot].son[1]=___;return &node[tot++];
}
Trie* Build_Tree(Trie *p,int x,int pos)
{if(!pos)return New_Node(p->cnt+1,tree[0],tree[0]);if( (x&pos)==0 )return New_Node(p->cnt+1,Build_Tree(p->son[0],x,pos>>1),p->son[1]);elsereturn New_Node(p->cnt+1,p->son[0],Build_Tree(p->son[1],x,pos>>1));
}
int Get_Ans(Trie *l,Trie *r,int x,int pos)
{int num=x&pos?

1:0; if(!pos) return 0; if(r->son[!num]->cnt-l->son[!num]->cnt) return pos + Get_Ans(l->son[!num],r->son[!num],x,pos>>1); else return Get_Ans(l->son[num],r->son[num],x,pos>>1); } int main() { int i,x,l,r; char p[10]; tree[0]=New_Node(0,0x0,0x0); tree[0]->son[0]=tree[0]->son[1]=tree[0]; tree[1]=Build_Tree(tree[0],0,1<<25); cin>>n>>m; for(i=1;i<=n;i++) { x=getint(); sum[i]=sum[i-1]^x; tree[i+1]=Build_Tree(tree[i],sum[i],1<<25); } for(i=1;i<=m;i++) { p[0]=getch(); if(p[0]=='A') { x=getint(); sum[n+1]=sum[n]^x; tree[n+2]=Build_Tree(tree[n+1],sum[n+1],1<<25); ++n; } else { l=getint(); r=getint(); x=getint(); x^=sum[n];--l;--r; printf("%d\n", Get_Ans(tree[l],tree[r+1],x,1<<25) ); } } }