BZOJ4260 Codechef Rebxor

Main topic:

Given a length ofNThe sequence, ask1≤l1≤r1<l2≤r2≤n Makes(⊙R1I=L1AI) + (⊕r2 i =l 2a i) Maximum, output this maximum value.

Ideas:

Use Trie to find the prefix xor sum and suffix xor, and then find the prefix xor and the suffix XOR and the largest, before and after the sum, to find the maximum value. It is also possible to use the trie to find a different or more persistent.

Code:

Trie

1#include <cstdio>2#include <cstring>3#include <iostream>4 #defineM 4000095 using namespacestd;6 intcnt,ans,child[m<<5][2],a[m],lmax[m],rmax[m];7 8 voidBuildintx)9 {Ten     inti,w,now=0; One      for(i=1<< -; i;i>>=1) A     { -w= (x&i)?1:0; -         if(!child[now][w]) child[now][w]=++CNT; thenow=Child[now][w]; -     } - } -  + intCalintx) - { +     inti,w,now=0, ans=0; A      for(i=1<< -; i;i>>=1) at     { -w= (x&i)?0:1; -         if(Child[now][w]) ans+=i,now=Child[now][w]; -         Elsenow=child[now][!W]; -     } -     returnans; in } -  to intMain () + { -     intN,i,now; scanf"%d",&n); the      for(i=1; i<=n;i++) scanf ("%d",&a[i]); *      for(Build (now=0), i=1; i<=n;i++) $     {Panax NotoginsengBuild (now^=a[i]); -Lmax[i]=max (Cal (now), lmax[i-1]); the     } +memset (Child,0,sizeof(child)); A      for(Build (now=cnt=0), i=n;i;i--) the     { +Build (now^=a[i]); -Rmax[i]=max (Cal (now), rmax[i+1]); $     } $      for(i=0; i<=n;i++) Ans=max (ans,lmax[i]+rmax[i+1]); -printf"%d\n", ans); -     return 0; the}

Durable Trie:

1#include <cstdio>2#include <iostream>3 #defineM 4000094 using namespacestd;5 6 intsum[m<<5],lc[m<<5],rc[m<<5],a[m],root[m<<5],cnt;7 8 intRead ()9 {Ten     intx=0; One     CharCh=GetChar (); A      while(ch<'0'|| Ch>'9') ch=GetChar (); -      while(ch>='0'&& ch<='9') x= (x<<3) + (x<<1) +ch- -, ch=GetChar (); -     returnx; the } -  - voidBuildint&cur,int_cur,intXintk) - { +sum[cur=++cnt]=sum[_cur]+1;if(k<0)return; -     if(x& (1<<k)) Lc[cur]=lc[_cur],build (rc[cur],rc[_cur],x,k-1); +     ElseRc[cur]=rc[_cur],build (lc[cur],lc[_cur],x,k-1); A } at  - intAskintXintYintval) - { -     inti,ans=0; -      for(i= -; i>=0; i--) -         if(val& (1<<i)) in             if(Sum[lc[y]]-sum[lc[x]]) ans|=1<<i,x=lc[x],y=Lc[y]; -             Elsex=rc[x],y=Rc[y]; to         Else if(Sum[rc[y]]-sum[rc[x]]) ans|=1<<i,x=rc[x],y=Rc[y]; +              ElseX=LC[X], y=Lc[y]; -     returnans; the } *  $ intMain ()Panax Notoginseng { -     intN=read (), i,mx=0, ans=0; the      for(i=1; i<=n;i++) A[i]=read () ^a[i-1]; +      for(i=1; i<=n;i++) Build (root[i],root[i-1],a[i], -); A      for(i=n-1; i;i--) the     { +mx=Max (Mx,ask (root[i],root[n],a[i)); -Ans=max (Ans,mx+ask (root[1],root[i],a[i])); $     } $printf"%d", ans); -     return 0; -}

BZOJ4260 Codechef Rebxor