Here is a C++ solution for the problem using dynamic programming: ```cpp #include using namespace std; int dp[25][25][5][5]; int solve(int l, int r, int a, int b, vector& arr) { if (l r) return 0; if (dp[l][r][a][b] != -1) return dp[l][r][a][b]; int res = 0; for (int i = 0; i 2) break; for (int j = 0; j 2 || i + j 2) break; res = max(res, solve(l + 1, r - j, a + i, max(0, j - 1), arr) + i * arr[l] + j * arr[r]); } } return dp[l][r][a][b] = res; } int main() { int n; cin n; vector arr(n); for (int i = 0; i arr[i]; memset(dp, -1, sizeof(dp)); int ans = 0; for (int i = 0; i

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Here is a C++ solution for the problem using dynamic programming: ```cpp #include using namespace std; int dp[25][25][5][5]; int solve(int l, int r, int a, int b, vector& arr) { if (l > r) return 0; if (dp[l][r][a][b] != -1) return dp[l][r][a][b]; int res = 0; for (int i = 0; i < 3; ++i) { if (a + i > 2) break; for (int j = 0; j < 3; ++j) { if (b + j > 2 || i + j > 2) break; res = max(res, solve(l + 1, r - j, a + i, max(0, j - 1), arr) + i * arr[l] + j * arr[r]); } } return dp[l][r][a][b] = res; } int main() { int n; cin >> n; vector arr(n); for (int i = 0; i < n; ++i) cin >> arr[i]; memset(dp, -1, sizeof(dp)); int ans = 0; for (int i = 0; i < n; ++i) { rotate(arr.begin(), arr.begin() + 1, arr.end()); ans = max(ans, solve(0, n - 1, 0, 0, arr)); } cout << ans << endl; return 0; } ``` This code works by iterating over all possible states and calculating the maximum amount of money that can be robbed for each state. The states are defined by the left and right boundaries of the vaults that are being considered, and the number of vaults that have been robbed on the left and right sides. The dynamic programming array `dp` is used to store the maximum amount of money that can be robbed for each state, and the `solve` function is used to calculate this value recursively. The `rotate` function is used to consider all possible starting points for the robbery, and the maximum amount of money that can be robbed is updated for each starting point.

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Here is a C++ solution for the problem using dynamic programming: ```cpp #include using namespace std; int dp[25][25][5][5]; int solve(int l, int r, int a, int b, vector& arr) { if (l > r) return 0; if (dp[l][r][a][b] != -1) return dp[l][r][a][b]; int res = 0; for (int i = 0; i < 3; ++i) { if (a + i > 2) break; for (int j = 0; j < 3; ++j) { if (b + j > 2 || i + j > 2) break; res = max(res, solve(l + 1, r - j, a + i, max(0, j - 1), arr) + i * arr[l] + j * arr[r]); } } return dp[l][r][a][b] = res; } int main() { int n; cin >> n; vector arr(n); for (int i = 0; i < n; ++i) cin >> arr[i]; memset(dp, -1, sizeof(dp)); int ans = 0; for (int i = 0; i < n; ++i) { rotate(arr.begin(), arr.begin() + 1, arr.end()); ans = max(ans, solve(0, n - 1, 0, 0, arr)); } cout << ans << endl; return 0; } ``` This code works by iterating over all possible states and calculating the maximum amount of money that can be robbed for each state. The states are defined by the left and right boundaries of the vaults that are being considered, and the number of vaults that have been robbed on the left and right sides. The dynamic programming array `dp` is used to store the maximum amount of money that can be robbed for each state, and the `solve` function is used to calculate this value recursively. The `rotate` function is used to consider all possible starting points for the robbery, and the maximum amount of money that can be robbed is updated for each starting point.

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