嗯……同样搞个前缀异或和，然后将x与sum异或一下，就是在[l-1,r-1]中找x^sum的最大异或值了。同样可持久化Trie搞搞即可（模板还是没背全啊……sad
1 /************************************************************** 2 Problem: 3261 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:3888 ms 7 Memory:239552 kb 8 ****************************************************************/ 9 10 //BZOJ 3261 11 #include<cstdio> 12 #include<cstring> 13 #include<cstdlib> 14 #include<iostream> 15 #include<algorithm> 16 #define rep(i,n) for(int i=0;i<n;++i) 17 #define F(i,j,n) for(int i=j;i<=n;++i) 18 #define D(i,j,n) for(int i=j;i>=n;--i) 19 #define pb push_back 20 using namespace std; 21 typedef long long LL; 22 inline int getint(){ 23 int r=1,v=0; char ch=getchar(); 24 for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-1; 25 for(; isdigit(ch);ch=getchar()) v=v*10-'0'+ch; 26 return r*v; 27 } 28 const int N=20000010; 29 /*******************template********************/ 30 31 int n,m,t[N][2],tot,rt[1000010],id[N]; 32 inline void Ins(int pre,int x,int k){ 33 int now=rt[k]=++tot; id[tot]=k; 34 D(i,30,0){ 35 int j=(x>>i)&1; 36 t[now][j^1]=t[pre][j^1]; 37 t[now][j]=++tot; id[tot]=k; now=tot; 38 pre=t[pre][j]; 39 } 40 } 41 inline int ask(int l,int r,int x){ 42 int ans=0,now=rt[r]; 43 D(i,30,0){ 44 int j=((x>>i)&1)^1; 45 if (id[t[now][j]]>=l) ans|=1<<i; else j^=1; 46 now=t[now][j]; 47 } 48 return ans; 49 } 50 51 int main(){ 52 #ifndef ONLINE_JUDGE 53 freopen("3261.in","r",stdin); 54 freopen("3261.out","w",stdout); 55 #endif 56 n=getint(); m=getint(); 57 id[0]=-1; 58 Ins(rt[0],0,0); 59 int sum=0; 60 F(i,1,n){ 61 sum^=getint(); 62 Ins(rt[i-1],sum,i); 63 } 64 char cmd[5]; int x,l,r; 65 F(i,1,m){ 66 scanf("%s",cmd); 67 if (cmd[0]=='A'){ 68 sum^=getint(); 69 Ins(rt[n],sum,n+1); n++; 70 }else{ 71 l=getint(); r=getint(); x=getint(); 72 printf("%d\n",ask(l-1,r-1,sum^x)); 73 } 74 } 75 return 0; 76 }