CodeForces - 1615A Closing The Gap

Posted 2022-05-17 海岛Blog

A. Closing The Gap
time limit per test2 seconds
memory limit per test256 megabytes

There are n block towers in a row, where tower i has a height of ai. 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≤i,j≤n; i≠j), and move a block from tower i to tower j. This essentially decreases ai by 1 and increases aj 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≤t≤1000) — the number of test cases. Then t cases follow.

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

The second line of each test case contains n space separated integers a1,a2,…,an (1≤ai≤10⁷) — the heights of the buildings.

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

Example
input
3
3
10 10 10
4
3 2 1 2
5
1 2 3 1 5
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.

问题链接：CodeForces - 1615A Closing The Gap
问题简述：（略）
问题分析：（略）

AC的C语言程序如下：

/* CodeForces - 1615A Closing The Gap */

#include <stdio.h>

int main()

    int t, n, a;
    scanf("%d", &t);
    while (t--) 
        scanf("%d", &n);

        int sum = 0;
        for (int i = 1; i <= n; i++) 
            scanf("%d", &a);
            sum += a;
        

        puts(sum % n == 0 ? "0" : "1");
    

    return 0;