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  • Maximum Xor Secondary CodeForces

    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), '', 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 == '
    ');
        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';
            }
        }
    }
    
    
    
    
    
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  • 原文地址:https://www.cnblogs.com/qieqiemin/p/11434777.html
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