K XOR Clique

BaoBao has a sequence a​1​​,a​2​​,...,a​n​​. He would like to find a subset S of {1,2,...,n} such that ∀i,j∈S, a​i​​⊕a​j​​<min(a​i​​,a​j​​) and ∣S∣ is maximum, where ⊕ means bitwise exclusive or.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤10​5​​), indicating the length of the sequence.

The second line contains n integers: a​1​​,a​2​​,...,a​n​​ (1≤a​i​​≤10​9​​), indicating the sequence.

It is guaranteed that the sum of n in all cases does not exceed 10​5​​.

Output

For each test case, output an integer denoting the maximum size of S.

Sample Input
3
3
1 2 3
3
1 1 1
5
1 2323 534 534 5
Sample Output
2
3
2
给出n个数字,要求输出一个最长集合的长度,在这个集合中任意两个数两两异或后结果比原来小
相当于集合中每个数的二进制形式长度相等
 #include<bits/stdc++.h>
using namespace std;
int s[];
int bit_width(unsigned int n)
{
unsigned int i = ; do {
++i;
} while ((n >> i)); return i;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,a;
scanf("%d",&n);
memset(s,,sizeof(s));
int len1,len2,num=,maxx=;
for(int i=;i<n;i++)
{
scanf("%d",&a);
len1=bit_width(a);
s[len1]++; }
for(int i=;i<;i++)
{
maxx=max(maxx,s[i]);
}
printf("%d\n",maxx);
}
return ;
}

The 2018 ACM-ICPC Asia Qingdao Regional Contest K XOR Clique的更多相关文章

  1. Gym - 101981K The 2018 ICPC Asia Nanjing Regional Contest K.Kangaroo Puzzle 暴力或随机

    题面 题意:给你1个20*20的格子图,有的是障碍有的是怪,你可以每次指定上下左右的方向,然后所有怪都会向那个方向走, 如果2个怪撞上了,就融合在一起,让你给不超过5w步,让所有怪都融合 题解:我们可 ...

  2. The 2018 ACM-ICPC Asia Qingdao Regional Contest(部分题解)

    摘要: 本文是The 2018 ACM-ICPC Asia Qingdao Regional Contest(青岛现场赛)的部分解题报告,给出了出题率较高的几道题的题解,希望熟悉区域赛的题型,进而对其 ...

  3. The 2018 ACM-ICPC Asia Qingdao Regional Contest

    The 2018 ACM-ICPC Asia Qingdao Regional Contest 青岛总体来说只会3题 C #include<bits/stdc++.h> using nam ...

  4. ACM ICPC Central Europe Regional Contest 2013 Jagiellonian University Kraków

    ACM ICPC Central Europe Regional Contest 2013 Jagiellonian University Kraków Problem A: Rubik’s Rect ...

  5. 2019-2020 ICPC, Asia Jakarta Regional Contest (Online Mirror, ICPC Rules, Teams Preferred)

    2019-2020 ICPC, Asia Jakarta Regional Contest (Online Mirror, ICPC Rules, Teams Preferred) easy: ACE ...

  6. The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online J - Press the Button(思维)

    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4056 题意 有一个按钮.一个灯.一个计时器和一个计数器,每按一次按钮,计时 ...

  7. The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online -C:Halting Problem(模拟)

    C Halting Problem In computability theory, the halting problem is the problem of determining, from a ...

  8. The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online Solution

    A    Live Love 水. #include<bits/stdc++.h> using namespace std; typedef long long ll; ; const i ...

  9. 2018-2019, ICPC, Asia Yokohama Regional Contest 2018 K

    传送门:https://codeforces.com/gym/102082/attachments 题解: 代码: /** * ┏┓ ┏┓ * ┏┛┗━━━━━━━┛┗━━━┓ * ┃ ┃ * ┃ ━ ...

随机推荐

  1. Sublime Text 3 for Mac 3176 序号版

    —– BEGIN LICENSE —–sgbteamSingle User LicenseEA7E-11532598891CBB9 F1513E4F 1A3405C1 A865D53F115F202E ...

  2. Anaconda安装与常用命令及方法(深度学习入门1)

    Anaconda是一个软件发行版,它附带了 conda.Python 和 150 多个科学包及其依赖项. 安装Anaconda Anaconda分为Linux.Windows.Mac等版本,去 htt ...

  3. CPP-基础:临界区

    VC windows api 多线程---临界区 临界区(Critical Section)是一段独占对某些共享资源访问的代码,在任意时刻只允许一个线程对共享资源进行访问.如果有多个线程试图同时访问临 ...

  4. 如何遍历Map操作总结

    Map<Integer, String> map = new HashMap<Integer, String>(); map.put(1, "123"); ...

  5. Knowledge Point 20180305 详解精度问题

    1.1 精度与基本数据类型运算的深度解析 我们在探讨Java基本数据类型时多次提到过精度的问题,那么计算机中的精度究竟是什么样的,为什么我们有时候的计算和我们预期的不同呢?下面我们通过精度来了解: 1 ...

  6. Spark Streaming编程示例

    近期也有开始研究使用spark streaming来实现流式处理.本文以流式计算word count为例,简单描述如何进行spark streaming编程. 1. 依赖的jar包 参考<分别用 ...

  7. 20181030NOIP模拟赛T3

    2017种树 2017共有N棵树从0到N-1标号.现要把这些树种在一条直线上,第i棵树的种植位置X[i]如下确定: X[0] = X[0] MOD L: X[i] = (X[i-1]*A+B) MOD ...

  8. chromium之lazy_instance

    先看看介绍 // The LazyInstance<Type, Traits> class manages a single instance of Type, // which will ...

  9. linux下pip错误 ImportError: No module named 'pip_internal'

    wget https://bootstrap.pypa.io/get-pip.py --no-check-certificate sudo python get-pip.py

  10. CentOS中的 yum upgrade 和 yum update 的区别

    通过 man yum 的帮助信息了解 yum update 和 yum upgrade: update If run without any packages, update will update ...