【CodeForces】 Codeforces Round #672 (Div. 2) B.Rock and Lever （思维&位运算）

B. Rock and Lever

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
“You must lift the dam. With a lever. I will give it to you.
You must block the canal. With a rock. I will not give the rock to you.”

Danik urgently needs rock and lever! Obviously, the easiest way to get these things is to ask Hermit Lizard for them.

Hermit Lizard agreed to give Danik the lever. But to get a stone, Danik needs to solve the following task.

You are given a positive integer n, and an array a of positive integers. The task is to calculate the number of such pairs (i,j) that i<j and ai & aj≥ai⊕aj, where & denotes the bitwise AND operation, and ⊕ denotes the bitwise XOR operation.

Danik has solved this task. But can you solve it?

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1≤t≤10) denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains one positive integer n (1≤n≤105) — length of the array.

The second line contains n positive integers ai (1≤ai≤109) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 105.

Output

For every test case print one non-negative integer — the answer to the problem.

Example

input

5
5
1 4 3 7 10
3
1 1 1
4
6 2 5 3
2
2 4
1
1

output

1
3
2
0
0

Note
In the first test case there is only one pair: (4,7): for it 4 & 7=4, and 4⊕7=3.

In the second test case all pairs are good.

In the third test case there are two pairs: (6,5) and (2,3).

In the fourth test case there are no good pairs.

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题意：

在 一些序列中 有多少组 (i,j) 满足 i<j and ai & aj≥ai⊕aj
，也就是二者与运算 大于等于 二者异或运算

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思路：

主要是要找出何时 两个数的 与运算 才会 大于等于 异或运算

发现：
① 两个数的二进制位数相同时，首位与必定是1，而首位异或必定是0，所以此时满足条件

② 两个数的二进制位数不同时，首位与必定是0，而首位异或必定是1 ，此时不满足条件

所以只需要找出 二进制位数相同的数 就满足条件

记录二进制位数相同的个数，假如个数为4，那么能组合的不重复的组数就有 （4*3）/2 ，即 n *（n-1）/ 2

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代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=5e4+50;
ll t,n,x;
 
int main()
{	   
	cin>>t;
	while(t--)
	{
		map<ll,ll> mp; //记录相同位数的个数
		ll ans=0;
		cin>>n;
		for(int i=0;i<n;i++)
		{
			cin>>x;
			ll y=x;
			
			ll sum=0;
			while(y>0) //获得二进制位数
			{
				sum++;
				y>>=1;
			}
			mp[sum]++;
		}
		
		for(auto x:mp)
		{
			ans+=x.second*(x.second-1)/2; //累加组数
		}
		cout<<ans<<endl;
	}
	return 0;
}