CF 280B -——Maximum Xor Secondary（单调栈）

最新推荐文章于 2022-03-12 22:31:18 发布

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探讨了如何在给定的不重复整数序列中，高效寻找任意区间内的最大及次大元素的异或值，并进一步找出所有可能区间中该值的最大值。采用单调栈优化算法，将复杂度降至O(n)，实现快速求解。

Bike loves looking for the second maximum element in the sequence. The second maximum element in the sequence of distinct numbers x1, x2, ..., xk (k > 1) is such maximum element xj, that the following inequality holds: .

The lucky number of the sequence of distinct positive integers x1, x2, ..., xk (k > 1)is the number that is equal to the bitwise excluding OR of the maximum element of the sequence and the second maximum element of the sequence.

You've got a sequence of distinct positive integers s1, s2, ..., sn (n > 1). Let's denote sequence sl, sl + 1, ..., sr as s[l..r] (1 ≤ l < r ≤ n). Your task is to find the maximum number among all lucky numbers of sequences s[l..r].

Note that as all numbers in sequence s are distinct, all the given definitions make sence.

Input

The first line contains integer n (1 < n ≤ 105). The second line contains n distinct integers s1, s2, ..., sn (1 ≤ si ≤ 109).

Output

Print a single integer — the maximum lucky number among all lucky numbers of sequences s[l..r].

Examples

Input

5
5 2 1 4 3

Output

Input

5
9 8 3 5 7

Output

Note

For the first sample you can choose s[4..5] = {4, 3} and its lucky number is (4 xor 3) = 7. You can also choose s[1..2].

For the second sample you must choose s[2..5] = {8, 3, 5, 7}.

题意：给一个长度为 n 的序列，元素不重复，求任意区间中的最大元素与次大元素的异或值，然后再取这些值的最大值，注意区间长度任意。

题解：如果暴力的话是 O(n ^ 2) 的复杂度，显然不行，所以就需要用单调栈维护一下，O(n)解决，找到第一个比它大的数，看代码吧！！

#include <iostream>
#include <cstdio>
#include <stack>
using namespace std;
const int MAX = 1e5+100;
typedef long long ll;
ll a[MAX];
stack<ll> s;
int main(){
	int n;
	scanf("%d",&n);
	for (int i = 0; i < n;i++){
		scanf("%lld",&a[i]);
	}
	ll ans=0;
	for (ll i = n-1; i >= 0;i--){
		while(!s.empty()&&a[s.top()]<a[i]) s.pop();//单调栈找到第一个比它大的数 反向跑
		if(!s.empty()) ans=max(ans,a[s.top()]^a[i]);
		s.push(i);
	}
	for (ll i = 0; i < n;i++){
		while(!s.empty()&&a[s.top()]<a[i]) s.pop();//单调栈找到第一个比它大的数 正向跑
		if(!s.empty()) ans=max(ans,a[s.top()]^a[i]);
		s.push(i);
	}
// 必须两个方向跑，要不然会有遗漏
	printf("%lld\n",ans);
	return 0;
}