Time Limit: 10 Seconds Memory Limit: 32768 KB
In China, people use a pair of chopsticks to get food on the table, but Mr. L is a bit different. He uses a set of three chopsticks -- one pair, plus an EXTRA long chopstick to get some big food by piercing it through the food. As you may guess, the length of the two shorter chopsticks should be as close as possible, but the length of the extra one is not important, as long as it's the longest. To make things clearer, for the set of chopsticks with lengths A,B,C(A<=B<=C), (A-B)^2 is called the 'badness' of the set.
It's December 2nd, Mr.L's birthday! He invited K people to join his birthday party, and would like to introduce his way of using chopsticks. So, he should prepare K+8 sets of chopsticks(for himself, his wife, his little son, little daughter, his mother, father, mother-in-law, father-in-law, and K other guests). But Mr.L suddenly discovered that his chopsticks are of quite different lengths! He should find a way of composing the K+8 sets, so that the total badness of all the sets is minimized.
Input
The first line in the input contains a single integer T, indicating the number of test cases(1<=T<=20). Each test case begins with two integers K, N(0<=K<=1000, 3K+24<=N<=5000), the number of guests and the number of chopsticks. There are N positive integers Li on the next line in non-decreasing order indicating the lengths of the chopsticks.(1<=Li<=32000).
Output
For each test case in the input, print a line containing the minimal total badness of all the sets.
Sample Input
1
1 40
1 8 10 16 19 22 27 33 36 40 47 52 56 61 63 71 72 75 81 81 84 88 96 98 103 110 113 118 124 128 129 134 134 139 148 157 157 160 162 164
Sample Output
23
Note
For the sample input, a possible collection of the 9 sets is:
8,10,16; 19,22,27; 61,63,75; 71,72,88; 81,81,84; 96,98,103; 128,129,148; 134,134,139; 157,157,160
題目大意：有一個人發明了一種新的用筷子的方法，每次用三根筷子，筷子長度不一（A<=B<=C），規定三根筷子的“不合適度”用(A-B)^2來表示。現在有n根長度不一的筷子，問選出(m+8)副筷子的最小不合適度之和是多少。輸入時按照筷子長度非降序輸入。
分析：經典動態規劃問題。不難得出，若第i根筷子是一組筷子中最短的，最優方案中該組筷子裏第二短的一定是第i+1根筷子。於是能得出以下的動態規劃：
定義f[i,j]表示後i根筷子組成j組筷子時的最小不合適度，顯然初值爲f[i,0]=0,f[n-2,1]=(a[n-2]-a[n-1])^2。
f[i,j]=min(f[i+1,j],f[i+2,j-1]+(a[i]-a[i+1])^2)。注意n-i+1要大於等於3*j。還有就是要倒推。
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