Exam 2 Maximum Xor with Prefix and Suffix - Works Application 16

Maximum Xor with Prefix and Suffix

Description

We have N numbers as an array, you need to find a prefix array and a suffix arrar,which we can get the maximum xor value with all elements in them. Notice that for prefix[0,1] and suffix [r,n-1],do not intersect (l 小于 r), and they can be empty.

Limits

-Memory limit per test : 256 megabytes

-Time limit per test : The faster the better

Compile & Environment

C++

g++ Main.cc –o Main –fno-asm –Wall –lm –static –std=c++0x –DONLINE_JUDGE

Java

J2SE8

Maximum stack size is 50m

Input

The first line is one Number N( 1<=N<=100000)

The second line contains N numbers ai (0 <= ai <=1e12) separated by space,

Which represents the array.

Output

Just output the maximum xor result.

Sample Test

Input

3

1 2 3

Out

3

Code

#include<iostream>
#include <algorithm>

long long pre[100005];
long long suf[100005];
long long arr[100005];
long long res[100005];
using namespace std;

long long cal(long long zu[100005], int m, int n){
long long res = 0;
for (int i = m; i <= n; i++){
res = max(res, zu[i]);
}
return res;
}

int main(){
int N;
cin >> N;
int i = 1;
for (int i = 1; i <= N; i++){
cin >> arr[i];
}
pre[0] = pre[N + 1] = suf[0] = suf[N + 1] = 0;
for (int i = 1; i<N; i++){
pre[i] = pre[i - 1] ^ arr[i];
}
for (int i = N; i >= 1; i--){
suf[i] = suf[i + 1] ^ arr[i];
}
long long ans = 0;
for (int i = 1; i <= N; i++){
ans = max(ans, max(cal(pre, 0, i), cal(suf, i + 1, N)));

}
cout << ans << endl;
return 0;
}

ca3d251f-0e7b-4258-8e68-8f7c9caa676+7