The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there are n boys enrolling in. At the beginning, the n boys stand in a row and go to the stage one by one. However, the director suddenly knows that very boy has a value of diaosi D, if the boy is k-th one go to the stage, the unhappiness of him will be (k-1)*D, because he has to wait for (k-1) people. Luckily, there is a dark room in the Small hall, so the director can put the boy into the dark room temporarily and let the boys behind his go to stage before him. For the dark room is very narrow, the boy who first get into dark room has to leave last. The director wants to change the order of boys by the dark room, so the summary of unhappiness will be least. Can you help him?

Input　　

The first line contains a single integer T, the number of test cases.For each case, the first line is n (0 < n <= 100)
　　The next n line are n integer D1-Dn means the value of diaosi of boys (0 <= Di <= 100)
Output

For each test case, output the least summary of unhappiness .

Sample Input

2
　　
5
1
2
3
4
5

5
5
4
3
2
2

Sample Output

Case #1: 20
Case #2: 24

題意：
   諸如You Are One的電視節目非常受歡迎。為了滿足仍然單身的男孩的需要，TJUT自行舉辦了表演。表演在小型大廳舉行，因此吸引了很多男孩
和女孩。現在有n個男孩報名參加。起初，n個男孩排成一列，一個個地進入舞臺。但是，導演突然知道，這個男孩的價值是diaosi D，如果男孩第k
個進入舞臺，他的不幸將是（k-1）* D，因為他必須等待（ k-1）個人。幸運的是，在小禮堂中有一間暗室，因此導演可以將男孩暫時放到暗室中，
並讓身後的男孩在他之前登上舞臺。由於暗室非常狹窄，因此首先進入暗室的男孩必須最後離開。
 

導演想在黑暗的房間裡改變男孩的順序，所以對不幸的總結最少。你能幫他嗎？

輸入：
   第一行包含一個整數T，即測試用例的數量。對於每種情況，第一行都是n（0 <n <= 100）
   接下來的n行是n個整數D1-Dn表示男孩的diaosi值（0 <= Di <= 100）

輸出：

對於每個測試用例，請輸出最少的不滿意摘要。

知識點： 區間DP 暗室裡的順序相當於一個棧

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 const int inf = 10000000;
 6 int n,val[105],sum[105],dp[105][105]; 
 7 int main()
 8 {
 9 
10     int t,i,j,k,l,cas = 1;
11     scanf("%d",&t);
12     while(t--)
13     {
14         scanf("%d",&n);
15         memset(sum,0,sizeof(sum));
16         for(i = 1; i<=n; i++)
17         {
18             scanf("%d",&val[i]);
19             sum[i] = sum[i-1]+val[i];
20         }
21         memset(dp,0,sizeof(dp));
22         for(i = 1; i<=n; i++)
23         {
24             for(j = i+1; j<=n; j++)
25                 dp[i][j] = inf;
26         }
27         for(l = 1; l<n; l++)
28         {
29             for(i = 1; i<=n-l; i++)
30             {
31                 j = i+l;
32                 for(k = 1; k<=j-i+1; k++)
33                     dp[i][j] = min(dp[i][j],dp[i+1][i+k-1]+dp[i+k][j]+k*(sum[j]-sum[i+k-1])+val[i]*(k-1));
34             }
35         }
36         printf("Case #%d: %d\n",cas++,dp[1][n]);
37     }
38     return 0;
39 
40 }