设a[i]为前缀和，则i~j的异或和为a[j]^a[i]，对于2个只需把另一个当成后缀就可以了
求max(a[j]^a[i])的话就用tire维护就可以了
1 #include<bits/stdc++.h> 2 #define inc(i,l,r) for(int i=l;i<=r;i++) 3 #define dec(i,l,r) for(int i=l;i>=r;i--) 4 #define link(x) for(edge *j=h[x];j;j=j->next) 5 #define mem(a) memset(a,0,sizeof(a)) 6 #define inf 1e9 7 #define ll long long 8 #define succ(x) (1<<x) 9 #define lowbit(x) (x&(-x)) 10 #define NM 400000+5 11 using namespace std; 12 int read(){ 13 int x=0,f=1;char ch=getchar(); 14 while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} 15 while(isdigit(ch))x=x*10+ch-'0',ch=getchar(); 16 return x*f; 17 } 18 const int m=31; 19 int n,a[NM],ans,_t,l[NM],r[NM]; 20 struct node{ 21 node *c[2]; 22 }N[(m+1)*NM],*o=N,*root; 23 void ins(int x){ 24 node *r=root; 25 dec(i,m,0){ 26 int t=(1<<i)&x?1:0; 27 if(!r->c[t])r->c[t]=++o; 28 r=r->c[t]; 29 } 30 } 31 int find(int x){ 32 int s=0;node *r=root; 33 dec(i,m,0){ 34 int t=(1<<i)&x?0:1; 35 if(r->c[t]){ 36 s+=1<<i; 37 r=r->c[t]; 38 }else r=r->c[t^1]; 39 } 40 return s; 41 } 42 int main(){ 43 freopen("data.in","r",stdin); 44 n=read(); 45 inc(i,1,n)a[i]=read(); 46 root=++o;ins(_t); 47 inc(j,1,n){ 48 // node *t=o; 49 _t^=a[j]; 50 l[j]=max(l[j-1],find(_t)); 51 ins(_t); 52 // printf("%d ",o-N); 53 } 54 mem(N);root=o=N;_t=0;ins(_t); 55 dec(i,n,1){ 56 _t^=a[i]; 57 r[i]=max(r[i-1],find(_t)); 58 ins(_t); 59 } 60 //inc(i,1,n)printf("%d ",l[i]);printf("\n"); 61 //inc(i,1,n)printf("%d ",r[i]);printf("\n"); 62 inc(i,1,n-1)ans=max(ans,l[i]+r[i+1]); 63 printf("%d\n",ans); 64 return 0; 65 }
