/*
思路同：CODE[VS] 1048 石子归并
dp[i][k] := i到k的珠子，所能获得的最大能量
dp[i][k] = max(dp[i][k]， dp[i][j] + dp[j+1][i+k] + arr[i]*arr[j+1]*arr[i+k+1])
注意：
此题为循环“石子归并”，空间开为两倍（将循环转换为线性解决），输出结果max(dp[i][i+(N-1)])
*/
1 #include <iostream>
2 #include <cstdlib>
3 #include <cstdio>
4 #include <algorithm>
5 #include <locale>
6 #include <cmath>
7 #include <vector>
8 using namespace std;
9 const int INF = 0x3f3f3f3f;
10 const int MaxN = 220;
11
12 int N, arr[MaxN];
13 int dp[MaxN][MaxN] = {0};
14
15 void Solve() {
16 int len = (N << 1);
17 for (int k = 1; k < len; ++k) {
18 for (int i = 0; (i + k) < len; ++i) {
19 for (int j = i; j < (i + k); ++j) {
20 dp[i][i + k] = max(dp[i][i + k], dp[i][j] + dp[j + 1][i + k] + arr[i] * arr[j + 1] * arr[i + k + 1]);
21 }
22 }
23 }
24 int ans = 0;
25 for (int i = 0; i < N; ++i) {
26 ans = max(ans, dp[i][i + N - 1]);
27 }
28 cout << ans << endl;
29 }
30
31 int main() {
32
33 cin >> N;
34 for (int i = 0; i < N; ++i) {
35 cin >> arr[i];
36 arr[i + N] = arr[i];
37 }
38
39 Solve();
40
41
42
43 #ifdef HOME
44 cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
45 #endif
46 return 0;
47 }