### PROBLEM LINK:

**Author:** Roman Rubanenko

**Tester:** Shiplu Hawlader

**Editorialist:** Praveen Dhinwa

### DIFFICULTY:

Medium

### PREREQUISITES:

Simple Math.

### PROBLEM:

Given N.

Let A = {1, 2, , N} and B = {N + 1, , 2 * N}.

Let C contains elements representing sum of pairs of A and B. Note that C is a multiset.

You will be given queries in which you are asked to find cnt of q in C.

### Explanation

Note that entries in C will be between N + 2 to 3N.

Given a q, how to find cnt of q in C?

Let us say that number selected from A is x and from B is y.

So x + y = q.

Rewrite as x = q - y.

We see that y should satisfy following inequalities.

N + 1 <= y <= 2N i.e. 1 <= q - y <= N.

You can see that count of valid y will be answer of our problem.

Can you get value of count of valid y from the above inequalities?

Yes, Rewrite both the inequalities as follows.

1 <= q - y <= N will become q - N <= y <= q - 1.

N + 1 <= y <= 2N

So we notice that y should be at least max(q - N, N + 1) and at most min(q - 1, 2N).

Now, can you find count of valid yâ€™s?

Yes, if y lies in the range [L, R], then answer will be R - L + 1.

Note that if L > R, then there is no solution.

**Pseudo Code**

```
long L = max(q - N, N + 1);
long R = min(q - 1, 2N);
long long ans = 0;
if (L > R)
ans = 0;
else
ans = R - L + 1;
```

**Few Small Points**

- Note that q can overflow from long data type because q can go upto 3 * 10^9. You should use long long data type.

**Complexity**:

For each query, we are just doing a constant number of operations. So for each test case, if there are queries, our time complexity

will be O(m) per test case.