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(ai ,aj) 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≤100000), indicating the length of the sequence.

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

It is guaranteed that the sum of n in all cases does not exceed 100000.

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

题意，求在a数组中找出s子集，要求子集中任意两个数异或之后比这两个数都小，求最大子集里面元素的个数

可以知道，2进制的0和1 才是1，其他是0，要求两个数异或要更小，则必须两个数二进制的最高位要都是1，所以，只需要知道在每个长度的二进制区间中，存在多少个数字，数字最多的那个就是答案

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#define sf scanf
#define pf printf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define scfff(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define prf(x) printf("%d\n",x)
#define mm(x,b) memset((x),(b),sizeof(x))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+7;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
using namespace std;
const double pi=acos(-1.0);
const int N=1e5+10;
int a[N];
int upp[32];//区间上限
int doww[32];//区间下限
void init()
{
	int x=2;
	upp[0]=1;
	doww[0]=1;
	rep(i,1,32)
	{
		doww[i]=x;
		upp[i-1]=x-1;
		x*=2;
	}
	upp[31]=x-1;
}
int num[32];//存每个区间的数量
int main()
{
	int re;scf(re);
	init();
	while(re--)
	{
		mm(num,0);
		int n;scf(n);
		rep(i,0,n)
		{
			scf(a[i]);
			rep(j,0,32)
			{
				if(a[i]>=doww[j]&&a[i]<=upp[j])
				{
					num[j]++;
					break;
				}
			}
		}
		int ans=0;
		rep(i,0,32)
		ans=max(ans,num[i]);
		prf(ans);
	}
	return 0;
}