Akash and Equal Mean Problem Code: CHFMIST | CodeChef Solution | PYTH 3.6
Akash goes to class everyday. His classes are held on the 9th floor so he uses a lift. One day N people get on the lift. However, since the lift can carry at most N−2 people, 2 of them need to get off.
Given an array A of the weights of the N people, Akash wonders how many ways of removing 2 people exist such that the mean of weights of the people in the lift remains constant. In other words, the mean of weights of the N−2 people remaining in the lift must be the same as the mean of weights of all the N people who were initially on the lift. Help Akash find how many such ways exist.
Input Format
The first line contains a single integer T - the number of test cases. Then the test cases follow.
The first line of each test case contains an integer N - the number of people who get on the lift initially.
The second line of each test case contains N space-separated integers A1,A2,…,AN denoting the weights of the N people on the lift.
Output Format
For each test case, print a single integer denoting the number of ways of removing 2 people such that the mean reamins the same.
Constraints
1≤T≤1000
3≤N≤105
1≤Ai≤109
Sum of N over all test cases does not exceed 2⋅105
Sample Input 1
3
4
1 3 5 7
4
1 7 7 7
6
1 1 1 1 1 1
Sample Output 1
2
0
15
Explanation
Test case 1: The mean of weights of all N students is 1+3+5+74=164=4.
Consider removing the people with weights (1,7). The mean of weights of the N−2 people remaining in the lift is 3+52=82=4, which is same as the mean of all N people.
Consider removing the people with weights (3,5). The mean of weights of the N−2 people remaining in the lift is 1+72=82=4, which is same as the mean of all N people.
There is no other way to remove 2 people such that the mean reamins the same. Hence the answer is 2.
Test case 2: The mean of weights of all N students is 1+7+7+74=5.5. It can be seen that there is no way to remove 2 people such that the mean reamins the same. Hence the answer is 0.
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