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}.

題意：
給你一個含有n個數(shù)的數(shù)組，讓你找一個連續(xù)的區(qū)間，這個區(qū)間中的最大值異或上次大值得到的數(shù)值最大。
思路：

我們可以知道，一個數(shù)a[i] ，最多可以當(dāng)兩個區(qū)間的有效次大值

即a[i] 左邊右邊第一個比a[i] 大的數(shù)，與a[i] 構(gòu)成的2個區(qū)間。

那么我們可以維護一個單調(diào)遞減的單調(diào)棧，棧中每一個數(shù) a[i] 左邊的數(shù)就是數(shù)組中左邊第一個比a[i]大的數(shù)，

把a[i] 從棧中彈出的數(shù)，就是 在數(shù)組中 a[i] 右邊第一個比a[i] 大的數(shù)。

這樣我們就可以把所有有效的區(qū)間中最大值和次大值都得出，更新答案即可。

細(xì)節(jié)見代碼：

#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';}} }

轉(zhuǎn)載于:https://www.cnblogs.com/qieqiemin/p/11434777.html

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