Hossam and Combinatorics

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#算法

给定一个正整数序列，程序需找出所有满足绝对差值等于序列中最大绝对差值的数对(i,j)，并计算这类数对的数量。当最大值和最小值相等时，输出特定计算结果；否则，输出另一种计算结果。程序通过遍历序列找到最大值和最小值，然后计算符合条件的数对数目。

给定一个长度为n的序列a,要求计算一共有多少对(i,j)使得abs(a[i]-a[j])=序列a中最大的abs(a[x]-a[y]),其中i!=j,x!=y;

Hossam woke up bored, so he decided to create an interesting array with his friend Hazem.

Now, they have an array aa of nn positive integers, Hossam will choose a number ai and Hazem will choose a number aj.

Count the number of interesting pairs (ai,aj) that meet all the following conditions:

1≤i,j≤n;
i！=j;
The absolute difference ∣ai−aj∣ must be equal to the maximum absolute difference over all the pairs in the array. More formally, ∣ai−aj∣=max1≤p,q≤n∣ap−aq∣.

Input

The input consists of multiple test cases. The first line contains a single integer tt ( 1≤t≤100), which denotes the number of test cases. Description of the test cases follows.

The first line of each test case contains an integer nn (2≤n≤105).

The second line of each test case contains nn integers a1,a2,…,an (1 \le a_i \le 10^51≤ai≤105).

It is guaranteed that the sum of nn over all test cases does not exceed 10^5105.

Output

For each test case print an integer — the number of interesting pairs(ai,aj).

Sample 1

Input	Output
2 5 6 2 3 8 1 6 7 2 8 3 2 10	2 4

本题我们可以先求出最大最小值，然后求出最大最小值的数量下x，y。1.如果最大值等于最小值即输入的一组数中所有数相等，则输出等于（x-1）*y。2.如果最大值最小值不等，则输出等于x*y*2.（本题一定要注意数据范围）

int main()
{
	int t,n,max,min;
	long long j,k,x,y;
	scanf("%d",&t);
	while(t)
	{
		k=0;
		j=0;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		scanf("%d",&a[i]);
		max=a[0];
		min=a[0];
		for(int i=0;i<n;i++)
		{
			if(max<a[i])
			max=a[i];
			if(min>a[i])
			min=a[i];
		}
		for(int i=0;i<n;i++)
		{
		    if(a[i]==max)
			k++;
			if(a[i]==min)
			j++;	
		}
		x=(k-1)*j;
		y=k*j*2;
		if(max==min)
		printf("%lld\n",x);
		else
		printf("%lld\n",y);
		t--;
	}
	return 0;
}