【dp】you are the one

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4283

题解： 当最优解下， a1在j的位置排出， 则a2 ——aj-1 和 aj——an为两个独立事件，

　　状态转移方程：

 dp[i][i + j] = min(dp[i][i + j], dp[i + 1][i + k] + k *arr[i] + dp[i + k + 1][i + j] +( k + 1) *sum[i + k + 1][i + j]) 做法和取石子类似。

 /***Good Luck***/
 #define _CRT_SECURE_NO_WARNINGS
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <stack>
 #include <map>
 #include <queue>
 #include <vector>
 #include <set>
 #include <functional>
 #include <cmath>
 #include <numeric>

 #define Zero(a) memset(a, 0, sizeof(a))
 #define Neg(a)  memset(a, -1, sizeof(a))
 #define All(a) a.begin(), a.end()
 #define PB push_back
 #define inf 0x3f3f3f3f
 #define inf2 0x7fffffffffffffff
 #define ll long long
 using namespace std;
 //#pragma comment(linker, "/STACK:102400000,102400000")
 void get_val(int &a) {
     int value = , s = ;
     char c;
     while ((c = getchar()) == ' ' || c == '\n');
     if (c == '-') s = -s; else value = c - ;
     while ((c = getchar()) >= '' && c <= '')
         value = value *  + c - ;
     a = s * value;
 }
 const int maxn = ;
 int arr[maxn];
 int n;
 int dp[maxn][maxn];
 int sum[maxn][maxn];

 int main() {
     //freopen("data.out", "w", stdout);
     //freopen("data.in", "r", stdin);
     //cin.sync_with_stdio(false);
     int T;
     cin >> T;
     for (int ii = ; ii <= T; ++ii) {
         cin >> n;
         for (int i = ; i <= n; ++i) scanf("%d", arr + i);
         Zero(dp);
         Zero(sum);
         for (int i = ; i <= n; ++i) {
             int t = ;
             for (int j = ; j + i <= n; ++j) {
                 t += arr[i + j];
                 sum[i][i + j] = t;
             }
         }
         for (int j = ; j <= n - ; ++j) {
             for (int i = ; i + j <= n; ++i) {
                 dp[i][i + j] = ( << ) - ;
                 for (int k = ; k <= j; ++k) {
                     dp[i][i + j] = min(dp[i][i + j], dp[i + ][i + k] + k *arr[i] + dp[i + k + ][i + j] +( k + ) *sum[i + k + ][i + j]);
                 }
             }
         }
         printf("Case #%d: ", ii);
         cout << dp[][n] << endl;
     }
     return ;
 }