询问转成前缀异或和异或总异或和异或询问值,然后可持久化trie即可
md好长时间不写都不会写了
注意l==0的边界问题,我们需要一个等于0的前缀异或和
#include<iostream>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<cstdlib>
#include<cstdio>
#include<map>
#include<bitset>
#include<set>
#include<stack>
#include<vector>
#include<queue>
using namespace std;
#define MAXN 600010
#define MAXM 20000010
#define ll long long
#define eps 1e-8
#define MOD 1000000007
#define INF 1000000000
int n,m;
int S=0;
int rt[MAXN];
int son[MAXM][2],siz[MAXM];
int tot;
void change(int &x,int xx,int d,int y){
x=++tot;
memcpy(son[x],son[xx],sizeof(son[x]));
siz[x]=siz[xx]+1;
if(d==-1){
return ;
}
change(son[x][y>>d&1],son[xx][y>>d&1],d-1,y);
}
int ask(int x,int xx,int d,int y){
if(d==-1){
return 0;
}
bool f=y>>d&1^1;
if(siz[son[xx][f]]-siz[son[x][f]]){
return (1<<d)+ask(son[x][f],son[xx][f],d-1,y);
}else{
return ask(son[x][f^1],son[xx][f^1],d-1,y);
}
}
int main(){
int i,l,r,x;
char o[2];
scanf("%d%d",&n,&m);
n++;
change(rt[1],rt[0],30,0);
for(i=2;i<=n;i++){
scanf("%d",&x);
S^=x;
change(rt[i],rt[i-1],30,S);
}
while(m--){
scanf("%s",o);
if(o[0]=='A'){
scanf("%d",&x);
n++;
S^=x;
change(rt[n],rt[n-1],30,S);
}else{
scanf("%d%d%d",&l,&r,&x);
l++;
r++;
printf("%d\n",ask(rt[l-2],rt[r-1],30,S^x));
}
}
return 0;
}
/*
5 5
2 6 4 3 6
A 1
Q 3 5 4
A 4
Q 5 7 0
Q 3 6 6
*/