题意：

n≤1e6,ai≤2e6n\le1e6,a_i\le2e6n≤1e6,ai≤2e6

思路：

由于(aj&ak)(a_j \And a_k)(aj&ak)打的括号，所以应该放在一起考虑，现在我们可以枚举aia_iai，由于是或操作，所以我们肯定是从高位到低位贪心的选，如果当前iii右边有两个数他们这一位都是111，那么答案这一位一定是111。当然如果aia_iai的这一位也是111就不需要考虑这一位了，直接跳过就好。现在问题转换成了如何快速判断当前这位右边是否存在一个超集它这一位是111。
很容易想到用sosdpsosdpsosdp来预处理出所有超集，这样处理出来的是整个数组的，但是怎么判断当前位右边是否有至少两个呢？我们可以记一个超集的最右边的两个位置，当i≥i\gei≥当前第二大的位置的时候这一位就不能要。
实现起来就很简单辣。

// Problem: F. Bits And Pieces
// Contest: Codeforces - Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1208/problem/F
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=2000005,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int a[N];
PII pos[N];void add(int val,int id) {//if(id==-1) return;if(pos[val].X==-1) {pos[val].X=id;return;} else if(pos[val].Y==-1) {if(pos[val].X==id) return;pos[val].Y=id;if(id>pos[val].X) pos[val].Y=pos[val].X,pos[val].X=id;} else {if(id==pos[val].X||id==pos[val].Y) return;if(id>pos[val].X) pos[val].Y=pos[val].X,pos[val].X=id;else if(id>pos[val].Y) pos[val].Y=id; }
}int main()
{//  ios::sync_with_stdio(false);
//  cin.tie(0);scanf("%d",&n);memset(pos,-1,sizeof(pos));for(int i=1;i<=n;i++) scanf("%d",&a[i]),add(a[i],i);for(int i=0;i<21;i++) for(int j=0;j<N;j++) {if(j>>i&1) {add(j^(1<<i),pos[j].X);add(j^(1<<i),pos[j].Y);}}int ans=0;for(int i=1;i<=n-2;i++) {int now=0;for(int j=20;j>=0;j--) {if(a[i]>>j&1) continue;now+=1<<j;if(now>=N||pos[now].Y<=i||pos[now].Y==-1) now-=1<<j; }ans=max(ans,a[i]|now);}printf("%d\n",ans);return 0;
}
/**/

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2) F. Bits And Pieces sosdp预处理超集相关推荐

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)
Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2) 题号题目知识点 A XORinacci B Uniqueness ...
每日一套szuManthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)
构造矩阵:一开始随便写了一个矩阵发现为满足,每行每列的异或和为0又因为是4的倍数然后每个数加上16去构造剩下得部分 #include <iostream> #include <cs ...
Codefest 18 (rated, Div. 1 + Div. 2)-D-Valid BFS--思维--已知bfs序，求是否正确
Codefest 18 (rated, Div. 1 + Div. 2)-D-Valid BFSvj地址题意:给你一段bfs序列,让你判断是否正确思路:先给节点排序(很关键) 我们给每一个的点的邻 ...
Ozon Tech Challenge 2020 (Div.1 + Div.2, Rated) D. Kuroni and the Celebration 交互 + 思维
传送门文章目录题意: 思路: 题意: 给你一颗树,每次可以询问两个点的lcalcalca,询问次数不能超过⌊n2⌋\left \lfloor \frac{n}{2} \right \rfloor⌊ ...
Ozon Tech Challenge 2020 (Div.1 + Div.2, Rated, T-shirts + prizes!)
Ozon Tech Challenge 2020 (Div.1 + Div.2, Rated, T-shirts + prizes!) 题号题目知识点 A Kuroni and the Gifts ...
Deltix Round, Summer 2021 (open for everyone, rated, Div. 1 + Div. 2)
Deltix Round, Summer 2021 (open for everyone, rated, Div. 1 + Div. 2) 题号题目知识点 A A Variety of Opera ...
Deltix Round, Spring 2021 (open for everyone, rated, Div. 1 + Div. 2)
Deltix Round, Spring 2021 (open for everyone, rated, Div. 1 + Div. 2) 题号题目知识点 A Game of Life B Lor ...
Harbour.Space Scholarship Contest 2021-2022 (open for everyone, rated, Div. 1 + Div. 2)
Harbour.Space Scholarship Contest 2021-2022 (open for everyone, rated, Div. 1 + Div. 2) 题号题目知识点 A ...
Codeforces Round #504 (rated, Div. 1 + Div. 2, based on VK Cup 2018 Final)切题报告（A-B题）
这是我第二次参加CF比赛,也是第一次写博客,写得不好,望各位大佬海涵. A. Single Wildcard Pattern Matching 原题链接这道题,这是难到让人无法直视,太水了. 题目大 ...

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2) F. Bits And Pieces sosdp预处理超集

文章目录

题意：

思路：

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2) F. Bits And Pieces sosdp预处理超集相关推荐

最新文章

热门文章