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  • Bits And Pieces

    F. Bits And Pieces
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You are given an array aa of nn integers.

    You need to find the maximum value of ai|(aj&ak)ai|(aj&ak) over all triplets (i,j,k)(i,j,k) such that i<j<ki<j<k.

    Here && denotes the bitwise AND operation, and || denotes the bitwise OR operation.

    Input

    The first line of input contains the integer nn (3n1063≤n≤106), the size of the array aa.

    Next line contains nn space separated integers a1a1, a2a2, ..., anan (0ai21060≤ai≤2⋅106), representing the elements of the array aa.

    Output

    Output a single integer, the maximum value of the expression given in the statement.

    Examples
    input
    Copy
    3
    2 4 6
    
    output
    Copy
    6
    
    input
    Copy
    4
    2 8 4 7
    
    output
    Copy
    12
    
    Note

    In the first example, the only possible triplet is (1,2,3)(1,2,3). Hence, the answer is 2|(4&6)=62|(4&6)=6.

    In the second example, there are 44 possible triplets:

    1. (1,2,3)(1,2,3), value of which is 2|(8&4)=22|(8&4)=2.
    2. (1,2,4)(1,2,4), value of which is 2|(8&7)=22|(8&7)=2.
    3. (1,3,4)(1,3,4), value of which is 2|(4&7)=62|(4&7)=6.
    4. (2,3,4)(2,3,4), value of which is 8|(4&7)=128|(4&7)=12.

    The maximum value hence is 1212.

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    #include <bits/stdc++.h>
    
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    const int maxn = (1 << 24);
    const int M = 23;
    int n, cnt[maxn];
    
    inline void update(ll cur, int pos = 0) {
        if (cnt[cur] >= 2)return;
        cnt[cur]++;
        for (register int i = pos; i <= M; ++i) {
            if (cur & (1 << i)) {
                update(cur ^ (1 << i), i);
            }
        }
    }
    
    int main() {
        //freopen("1.txt", "r", stdin);
        scanf("%d", &n);
        vector<ll> e(n);
        for (auto &i : e) {
            scanf("%lld", &i);
        }
        //for(register int i=0;i<3;++i)printf("%lld
    ",e[i]);
        update(e[n - 1]);
        update(e[n - 2]);
        ll res = 0;
    
        for (register int i = n - 3; i >= 0; --i) {
            ll head = e[i];
            int nx = 0;
            for (register int j = M; j >= 0; --j) {
                if (!(head & (1 << j)) && cnt[nx | (1 << j)] >= 2) {
                    nx |= (1 << j);
                }
                //printf("debug res = %lld
    ",res);
            }
            res = max(res, head | nx);
    
            update(e[i]);
        }
        printf("%lld
    ", res);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/czy-power/p/11411187.html
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