【BZOJ3261】最大异或和（可持久化Trie）

题意：

思路：可持久化Trie板子题，支持序列插入和询问

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 typedef unsigned int uint;

 typedef unsigned long long ull;

 typedef long double ld;

 typedef pair<int,int> PII;

 typedef pair<ll,ll> Pll;

 typedef vector<int> VI;

 typedef vector<PII> VII;

 typedef pair<ll,ll>P;

 #define N  12000010

 #define M  6000010

 #define INF 1e9

 #define fi first

 #define se second

 #define MP make_pair

 #define pb push_back

 #define pi acos(-1)

 #define mem(a,b) memset(a,b,sizeof(a))

 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)

 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)

 #define lowbit(x) x&(-x)

 #define Rand (rand()*(1<<16)+rand())

 #define id(x) ((x)<=B?(x):m-n/(x)+1)

 #define ls p<<1

 #define rs p<<1|1

 #define fors(i) for(auto i:e[x]) if(i!=p)

 const int MOD=1e8+,inv2=(MOD+)/;

       int p=1e4+;

       double eps=1e-;

       int dx[]={-,,,};

       int dy[]={,,-,};

 char ch[];

 int t[N][],s[N],root[N],b[N],cnt;

 int read()

 {

    int v=,f=;

    char c=getchar();

    while(c<||<c) {if(c=='-') f=-; c=getchar();}

    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();

    return v*f;

 }

 ll readll()

 {

    ll v=,f=;

    char c=getchar();

    while(c<||<c) {if(c=='-') f=-; c=getchar();}

    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();

    return v*f;

 }

 void ins(int p1,int &p2,int z,int d)

 {

     int p=(z>>d)&;

     p2=++cnt;

     s[p2]=s[p1]+;

     if(d<) return;

     t[p2][p^]=t[p1][p^];

     ins(t[p1][p],t[p2][p],z,d-);

 }

 int query(int p1,int p2,int z,int d)

 {

     if(d<) return ;

     int p=(z>>d)&;

     if(s[t[p2][p^]]-s[t[p1][p^]]) return (<<d)+query(t[p1][p^],t[p2][p^],z,d-);

      else return query(t[p1][p],t[p2][p],z,d-);

 }

 int main()

 {

     //freopen("1.in","r",stdin);

     //freopen("1.out","w",stdout);

     cnt=;

     ins(,root[],,);

     int n=read(),m=read();

     n++;

     rep(i,,n)

     {

         int x=read();

         b[i]=b[i-]^x;

         ins(root[i-],root[i],b[i],);

     }

     while(m--)

     {

         scanf("%s",ch);

         if(ch[]=='A')

         {

             int x=read();

             n++;

             b[n]=b[n-]^x;

             ins(root[n-],root[n],b[n],);

         }

          else

          {

              int l=read(),r=read(),x=read();

              int ans=query(root[l-],root[r],b[n]^x,);

              printf("%d\n",ans);

          }

     }

     return ;

 }