CF1637D Yet Another Minimization Problem
710+

## 题意翻译

$$\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n(x_i+x_j)^2$$

– $t$ 组数据，$1\leqslant t\leqslant 40$。
– $1\leqslant n,\sum n\leqslant 100$。
– $1\leqslant a_i,b_i\leqslant 100$。

Translated by Eason_AC

## 题目描述

You are given two arrays $a$ and $b$ , both of length $n$ .

You can perform the following operation any number of times (possibly zero): select an index $i$ ( $1 \leq i \leq n$ ) and swap $a_i$ and $b_i$ .

Let’s define the cost of the array $a$ as $\sum_{i=1}^{n} \sum_{j=i + 1}^{n} (a_i + a_j)^2$ . Similarly, the cost of the array $b$ is $\sum_{i=1}^{n} \sum_{j=i + 1}^{n} (b_i + b_j)^2$ .

Your task is to minimize the total cost of two arrays.

## 输入输出格式

### 输入格式

Each test case consists of several test cases. The first line contains a single integer $t$ ( $1 \leq t \leq 40$ ) — the number of test cases. The following is a description of the input data sets.

The first line of each test case contains an integer $n$ ( $1 \leq n \leq 100$ ) — the length of both arrays.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $1 \leq a_i \leq 100$ ) — elements of the first array.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ( $1 \leq b_i \leq 100$ ) — elements of the second array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$ .

### 输出格式

For each test case, print the minimum possible total cost.

## 输入输出样例

### 输入样例 #1

3
1
3
6
4
3 6 6 6
2 7 4 1
4
6 7 2 4
2 5 3 5

### 输出样例 #1

0
987
914

## 说明

In the second test case, in one of the optimal answers after all operations $a = [2, 6, 4, 6]$ , $b = [3, 7, 6, 1]$ .

The cost of the array $a$ equals to $(2 + 6)^2 + (2 + 4)^2 + (2 + 6)^2 + (6 + 4)^2 + (6 + 6)^2 + (4 + 6)^2 = 508$ .

The cost of the array $b$ equals to $(3 + 7)^2 + (3 + 6)^2 + (3 + 1)^2 + (7 + 6)^2 + (7 + 1)^2 + (6 + 1)^2 = 479$ .

The total cost of two arrays equals to $508 + 479 = 987$ .

## 解题思路

$(a_i + a_j)^2 = (a_i)^2 + (a_j)^2 + 2 a_i a_j$，全部加起来，会发现每个$a_i$都需要加n-1次，因为他可以和其他n-1个数相乘。剩下的任意两项之和，用乘法分配律，可以发现，$a_i$需要乘上后面数字之和，或者$a_j$需要乘上前面所有数之和。