CF1615A.Closing The Gap

普及/提高-

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题目描述

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?

输入格式

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.

输出格式

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

输入输出样例

输入#1
```
3
3
10 10 10
4
3 2 1 2
5
1 2 3 1 5
```
输出#1
```
0
0
1
```

说明/提示

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.

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