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. jQuery中append()、prepend()与after()、before()的区别

    转载 未曾见海  https://www.cnblogs.com/afei-qwerty/p/6682963.html 在jQuery中,添加元素有append(),prepend() 和 after ...

  2. StringJoiner

    示例一 public class StringJoinerTest1 { public static void main(String[] args) { StringJoiner joiner = ...

  3. C#处理List<object>重复数据的问题

    private class ListDistinct : IEqualityComparer<object> { public bool Equals(object x, object y ...

  4. PAT——1018. 锤子剪刀布

    大家应该都会玩“锤子剪刀布”的游戏:两人同时给出手势,胜负规则如图所示: 现给出两人的交锋记录,请统计双方的胜.平.负次数,并且给出双方分别出什么手势的胜算最大. 输入格式: 输入第1行给出正整数N( ...

  5. Context initialization failed org.springframework.beans.factory.BeanCreationException

    严重: Context initialization failed org.springframework.beans.factory.BeanCreationException: Error cre ...

  6. eclipse安装tomcat时只有locahost,不显示server name

    Eclipseh中无法安装Tomcat,报错信息如下 Cannot create a server using the selected type   原因:以前安装的tomcat目录改变 解决方法: ...

  7. ABAP开发顾问必备:SAP ABAP开发技术总结[转载]

    转载自SAP师太技术博客,原文链接:http://www.cnblogs.com/jiangzhengjun/p/4260224.html 在原文上增加了链接,此文及此文的链接版权都归SAP师太所有. ...

  8. 05.odoo12开源框架学习

    博客为日常工作学习积累总结: 1.odoo12学习 参考博客:https://alanhou.org/centos-odoo-12/ CentOS 7快速安装配置 Odoo 12 添加新用户必做,不然 ...

  9. LAMP+Varnish的实现

    基于Keepalived+Varnish+Nginx实现的高可用LAMP架构 注意:各节点的时间需要同步(ntpdate ntp1.aliyun.com),关闭firewalld(systemctl ...

  10. python爬虫学习笔记(2)-----代理模式

    一.UserAgent UserAgent 中文意思是用户代理,简称UA,它是一个特殊字符串头,使得服务器能够识别用户 设置UA的两种方式: 1.heads from urllib import re ...