#include <bits/stdc++.h> // By Lucky Ox
#define int long long
#define endl "\n"
#define pii pair<int, int>
#define PI atan(1.0) * 4
using namespace std;
using i128 = __int128;
typedef unsigned long long ull;
const long long mod = 1000000000000002493, INF = 0x3f3f3f3f3f3f3f3f;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int x, int y) { return x / gcd(x, y) * y; }
int q_pow(int a, int k, int p) { int res = 1; while (k) { if (k & 1) res = res * a % p; k >>= 1; a = a * a % p; } return res; }
int len_10(int x) { int len = 0; while(x) { x /= 10; len ++ ; } return len; }
int P(int x, int p){ return (x % p + p) % p; }
int to_int(string s) { int val = 0; for(int i = 0; i < (int)s.size(); i ++ ){val *= 10; val += s[i] - '0';}return val;}//注意:s是空串也会返回0
void print(i128 x) {if(x > 9) print(x / 10); putchar(x % 10 + '0'); }//i128输出
//用__lg()来求一个数二进制下的位数 返回的len 表示这个数是[0, 1, ...., len] 比如10 __lg(10) = 3, 1010 [3, 2, 1, 0]
//__builtin_popcountll(int x) 求二进制下x中1的数量 __buitlin_ctzll(int x) 求二进制下末尾0的个数
//(n & (1 << i))的值可能会是1, 2, 4...... (n >> i) & 1的值一定是1

const int N = 1e2 + 10;
int n;
int a[N];
int dp[N * 2][N * 2];

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i ++ ) {
        cin >> a[i];
        a[i + n] = a[i];
    }

    for (int len = 3; len <= n + 1; len ++ )
        for (int l = 1; l + len - 1 <= n * 2; l ++ ) {
            int r = l + len - 1;
            for (int k = l + 1; k < r; k ++ )
                dp[l][r] = max(dp[l][r], dp[l][k] + dp[k][r] + a[l] * a[k] * a[r]);
        }

    int res = 0;
    for (int i = 1; i <= n; i ++ ) res = max(res, dp[i][i + n]);
    cout << res << endl;
}

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);
    cout << fixed << setprecision(10);
    int T = 1;
    //cin >> T;
    while (T -- ) solve();
    return 0;
}

代码如上 这样子过不了的