Maximum Xor Secondary CodeForces - 281D （单调栈）

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
7
Input
5
9 8 3 5 7
Output
15
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：
给你一个含有n个数的数组，让你找一个连续的区间，这个区间中的最大值异或上次大值得到的数值最大。
思路：

我们可以知道，一个数a[i] ，最多可以当两个区间的有效次大值

即a[i] 左边右边第一个比a[i] 大的数，与a[i] 构成的2个区间。

那么我们可以维护一个单调递减的单调栈，栈中每一个数 a[i] 左边的数就是数组中左边第一个比a[i]大的数，

把a[i] 从栈中弹出的数，就是 在数组中 a[i] 右边第一个比a[i] 大的数。

这样我们就可以把所有有效的区间中最大值和次大值都得出，更新答案即可。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
ll a[maxn];
stack<ll> st;
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    gbtb;
    cin >> n;
    repd(i, 1, n) {
        cin >> a[i];
    }
    ll ans = 0ll;
    repd(i, 1, n) {
        if (st.empty()) {
            st.push(a[i]);
        }else
        {
            while(st.size()&&st.top()<a[i])
            {
                ans=max(ans,(st.top()^a[i]));
                st.pop();
            }
            if(st.size())
                ans=max(ans,(st.top()^a[i]));
            st.push(a[i]);
        }
    }
    ll x;
    if(st.size())
    {
        x=st.top();
        st.pop();
    }
    while(st.size())
    {
        ans=max(ans,(x^st.top()));
        x=st.top();
        st.pop();
    }
    cout<<ans<<endl;
    return 0;
}

inline void getInt(int *p)
{
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    } else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}