【CodeForces】 Codeforces Round #672 (Div. 2) B.Rock and Lever （思維&位運算）

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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題意：

在 一些序列中 有多少組 (i,j) 滿足 i<j and ai & aj≥ai⊕aj
，也就是二者與運算 大於等於 二者異或運算

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思路：

主要是要找出何時 兩個數的 與運算 才會 大於等於 異或運算

發現：
① 兩個數的二進位制位數相同時，首位與必定是1，而首位異或必定是0，所以此時滿足條件

② 兩個數的二進位制位數不同時，首位與必定是0，而首位異或必定是1 ，此時不滿足條件

所以只需要找出 二進位制位數相同的數 就滿足條件

記錄二進位制位數相同的個數，假如個數為4，那麼能組合的不重複的組數就有 （4*3）/2 ，即 n *（n-1）/ 2

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程式碼：

#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; //記錄相同位數的個數
		ll ans=0;
		cin>>n;
		for(int i=0;i<n;i++)
		{
			cin>>x;
			ll y=x;
			
			ll sum=0;
			while(y>0) //獲得二進位制位數
			{
				sum++;
				y>>=1;
			}
			mp[sum]++;
		}
		
		for(auto x:mp)
		{
			ans+=x.second*(x.second-1)/2; //累加組數
		}
		cout<<ans<<endl;
	}
	return 0;
}