Closing The Gap

2000ms

262144K

Description:

There are $$$n$$$ block towers in a row, where tower $$$i$$$ has a height of $$$a_i$$$. You're part of a building crew, and you want to make the buildings look as nice as possible. In a single day, you can perform the following operation:

Choose two indices $$$i$$$ and $$$j$$$ ($$$1 \leq i, j \leq n$$$; $$$i \neq j$$$), and move a block from tower $$$i$$$ to tower $$$j$$$. This essentially decreases $$$a_i$$$ by $$$1$$$ and increases $$$a_j$$$ by $$$1$$$.

You think the ugliness of the buildings is the height difference between the tallest and shortest buildings. Formally, the ugliness is defined as $$$\max(a)-\min(a)$$$.

What's the minimum possible ugliness you can achieve, after any number of days?

Input:

The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Then $$$t$$$ cases follow.

The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 100$$$) — the number of buildings.

The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 10^7$$$) — the heights of the buildings.

Output:

For each test case, output a single integer — the minimum possible ugliness of the buildings.

Sample Input:

Sample Output:

0
0
1

Note:

In the first test case, the ugliness is already $$$0$$$.

In the second test case, you should do one operation, with $$$i = 1$$$ and $$$j = 3$$$. The new heights will now be $$$[2, 2, 2, 2]$$$, with an ugliness of $$$0$$$.

In the third test case, you may do three operations:

with $$$i = 3$$$ and $$$j = 1$$$. The new array will now be $$$[2, 2, 2, 1, 5]$$$,
with $$$i = 5$$$ and $$$j = 4$$$. The new array will now be $$$[2, 2, 2, 2, 4]$$$,
with $$$i = 5$$$ and $$$j = 3$$$. The new array will now be $$$[2, 2, 3, 2, 3]$$$.

The resulting ugliness is $$$1$$$. It can be proven that this is the minimum possible ugliness for this test.

Informação

Provedor Codeforces

Código CF1615A