C. Lunar New Year and Number Division
greedy; implementation; math; sortings
Lunar New Year is approaching, and Bob is struggling with his homework – a number division problem.There are \(n\) positive integers \(a_1, a_2, \ldots, a_n\) on Bob's homework paper, where \(n\) is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least \(2\) numbers. Suppose the numbers are divided into \(m\) groups, and the sum of the numbers in the \(j\)-th group is \(s_j\). Bob's aim is to minimize the sum of the square of \(s_j\), that is $$$\(\sum_{j = 1}^{m} s_j^2.\)$$$Bob is puzzled by this hard problem. Could you please help him solve it?
The first line contains an even integer \(n\) (\(2 \leq n \leq 3 \cdot 10^5\)), denoting that there are \(n\) integers on Bob's homework paper.The second line contains \(n\) integers \(a_1, a_2, \ldots, a_n\) (\(1 \leq a_i \leq 10^4\)), describing the numbers you need to deal with.
A single line containing one integer, denoting the minimum of the sum of the square of \(s_j\), which is $$$\(\sum_{i = j}^{m} s_j^2,\)\( where \)m$$$ is the number of groups.
In the first sample, one of the optimal solutions is to divide those \(4\) numbers into \(2\) groups \(\{2, 8\}, \{5, 3\}\). Thus the answer is \((2 + 8)^2 + (5 + 3)^2 = 164\).In the second sample, one of the optimal solutions is to divide those \(6\) numbers into \(3\) groups \(\{1, 2\}, \{1, 2\}, \{1, 2\}\). Thus the answer is \((1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27\).