题目描述

Highway 201 is the most busy street in Rockport. Traffic cars cause a lot of hindrances to races, especially when there are a lot of them. The track which passes through this highway can be divided into $n$ sub-tracks. You are given an array aa where $a_i$ represents the number of traffic cars in the $i$ -th sub-track. You define the inconvenience of the track as $\sum\limits_{i=1}^{n} \sum\limits_{j=i+1}^{n} \lvert a_i-a_j\rvert$ , where $|x|$ is the absolute value of $x$ .

You can perform the following operation any (possibly zero) number of times: choose a traffic car and move it from its current sub-track to any other sub-track.

Find the minimum inconvenience you can achieve.

输入格式

The first line of input contains a single integer $t$ $( 1\leq t\leq 10\,000)$ — the number of test cases.

The first line of each test case contains a single integer $n$ $( 1\leq n\leq 2\cdot 10^5)$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ $( 0\leq a_i\leq 10^9)$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot 10^5$ .

输出格式

For each test case, print a single line containing a single integer: the minimum inconvenience you can achieve by applying the given operation any (possibly zero) number of times.

输入输出样例

输入 #1

3
3
1 2 3
4
0 1 1 0
10
8 3 6 11 5 2 1 7 10 4

输出 #1

0
4
21

说明/提示

For the first test case, you can move a car from the $3$ -rd sub-track to the $1$ -st sub-track to obtain $0$ inconvenience.

For the second test case, moving any car won't decrease the inconvenience of the track.

分析

要求数列和不变,对数列进行任意操作,求出数列中任意两个数差值的绝对值的和的最小值。我们自然可以想到尽可能地让数值均分,均分过后的数列一定只会有 $1$ 种或 $2$ 种数值,作差即可得到答案。

对于作差,我们也可以进行推导优化。假设均分过后的数列有 $2$ 种数值,那么较大数值的个数为 $\sum_{i=1}^{n} a[i]\;\bmod\; n $,因为只存在 $2$ 种数值,所以较大数值与较小数值作差的绝对值一定为 $1$,那么我们的答案就是较大数值的个数乘较小数值的个数即 $[n\;\cdot\;(\sum_{i=1}^{n} a[i])\;\bmod\; n]-[(\sum_{i=1}^{n} a[i])\;\bmod\; n]^2$

代码

#include<bits/stdc++.h>
#define int long long
using namespace std;
int t,n,a[200001];
int sum,op;
signed main(){
    scanf("%lld",&t);
    while(t--){
        scanf("%lld",&n);
        sum=0;
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
            sum+=a[i];
        }
        op=sum%n;
        printf("%lld\n",(n-op)*op);
    }
}
最后修改:2021 年 07 月 19 日 07 : 11 PM
如果觉得我的文章对你有用,请随意赞赏