Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem C
Chopsticks
Input: Standard Input
Output: Standard Output

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 L_i on the next line in non-decreasing order indicating the lengths of the chopsticks.(1<=L_i<=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
__________________________________________________________________________________________

參考 http://www.cppblog.com/rakerichard/archive/2010/02/19/108081.html

網路上蒐的另外一個用 for 迴圈跑的,
http://www.cppblog.com/syhd142/articles/118232.html
我感到疑惑, 對於 1 個人的化, 那個做法是錯誤的

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define oo 100000000
int DP[1010][5001];
int L[5001], bad[5001];
int cmp(const void *i, const void *j) {
    return *(int *)j - *(int *)i;
}
int min(int x, int y) {
    return x < y ? x : y;
}
int findDP(int k, int n) {
    if(DP[k][n] != -1)
        return DP[k][n];
    if(n < 3*k) {
        DP[k][n] = oo;
    } else if(k == 0) {
        DP[k][n] = 0;
    } else {
        DP[k][n] = min(findDP(k, n-1), findDP(k-1, n-2)+bad[n]);
    }
    return DP[k][n];
}
int main() {
    int t, n, k, i, j;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &k, &n);
        for(i = 1; i <= n; i++)
            scanf("%d", &L[i]);
        k += 8;
        qsort(L+1, n, sizeof(int), cmp);
        for(i = 2; i <= n; i++)
            bad[i] = (L[i]-L[i-1])*(L[i]-L[i-1]);
        for(i = 0; i <= k; i++)
            for(j = 0; j <= n; j++)
                DP[i][j] = -1;
        printf("%d\n", findDP(k, n));
    }
    return 0;
}