Read problems statements in Mandarin and Russian. Translations in Vietnamese to be uploaded soon.
Nikitosh the painter has a 1-indexed array A of N elements. He wants to find the maximum value of expression
(A[l1] ⊕ A[l1 + 1] ⊕ ... ⊕ A[r1]) + (A[l2] ⊕ A[l2 + 1] ⊕ ... ⊕ A[r2]) where 1 ≤ l1 ≤ r1 < l2 ≤ r2 ≤ N.
Here, x ⊕ y means the bitwise XOR of x and y.
Because Nikitosh is a painter and not a mathematician, you need to help him in this task.
Input
The first line contains one integer N, denoting the number of elements in the array.
The second line contains N space-separated integers, denoting A1, A2, ... , AN.
Output
Output a single integer denoting the maximum possible value of the given expression.
Constraints
- 2 ≤ N ≤ 4*105
- 0 ≤ Ai ≤ 109
Subtasks
- Subtask 1 (40 points) : 2 ≤ N ≤ 104
- Subtask 2 (60 points) : Original constraints
Example
Input: 5 1 2 3 1 2 Output: 6
Explanation
Choose (l1, r1, l2, r2) = (1, 2, 3, 3) or (1, 2, 4, 5) or (3, 3, 4, 5).
很久之前做的一道Trie树搞异或的题啦。。。。
大致就是求一下前缀和后缀异或最大值然后更新答案就好了。
#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cmath> #include<cstring> #define ll int #define maxn 400005 using namespace std; long long ans=0; ll ci[35]; struct Trie{ ll ch[maxn*20][2]; ll tot,rot; void ins(ll x){ ll now=rot; for(int i=30;i>=0;i--){ ll c=(ci[i]&x)?1:0; if(!ch[now][c]) ch[now][c]=++tot; now=ch[now][c]; } } ll find_max(ll x){ ll now=rot,alr=0; for(int i=30;i>=0;i--){ ll c=(ci[i]&x)?1:0; if(ch[now][c^1]) alr+=ci[i],now=ch[now][c^1]; else now=ch[now][c]; } return alr; } }tr; ll a[maxn],w[maxn],n; ll lef[maxn],rig[maxn]; inline void init(){ ci[0]=1; for(int i=1;i<=30;i++) ci[i]=ci[i-1]<<1; tr.rot=tr.tot=1; } int main(){ init(); scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",a+i),w[i]=a[i]^w[i-1]; tr.ins(0); for(int i=1;i<=n;i++){ lef[i]=max(lef[i-1],tr.find_max(w[i])); tr.ins(w[i]); } memset(tr.ch,0,sizeof(tr.ch)); init(); for(int i=n;i;i--){ rig[i]=max(rig[i+1],tr.find_max(w[i])); tr.ins(w[i]); } for(int i=1;i<=n;i++) ans=max(ans,(long long)(lef[i]+rig[i])); printf("%lld ",ans); return 0; }