answer:Your original OCR output contains several issues, including incorrect character recognition and incorrect reserved word identification. Here is the corrected Python code:```pythonimport redef user_password_check(password, username, previous_passwords): # Check length of password if len(password) < 12 or len(password) > 50: return 'Invalid' # Check if password contains any part of the username if username in password: return 'Invalid' # Check if password contains more than 3 consecutive identical characters if re.search(r'(.)1{3,}', password): return 'Invalid' # Check if password contains special character ('@', '#', '', '%', '&', '*') if not re.search(r'[@#%&*]', password): return 'Invalid' # Check if password contains more than 2 consecutive digits if re.search(r'd{3,}', password): return 'Invalid' # Check if password has more than 1 uppercase letter if len(re.findall(r'[A-Z]', password)) <= 1: return 'Invalid' # Check if password has more than 1 lowercase letter if len(re.findall(r'[a-z]', password)) <= 1: return 'Invalid' # Check if password has more than 2 digits if len(re.findall(r'd', password)) <= 2: return 'Invalid' if password in previous_passwords[-5:]: return 'Invalid' return 'Valid'```The above code defines a function to check the complexity of a new password. It verifies different conditions like length, special character, digit, and uppercase and lowercase letters usage. Also, it checks if the new password is among the last 5 used passwords.

answer:Sure, here's the Rust version of your Python function:```rustfn word_segments(s: &str, n: usize) -> Vec<String> { // return empty vector if the string is empty if s.is_empty() { return vec![] } // define vowels let vowels = "aeiouAEIOU"; // initialize an empty vector for the results let mut result = vec![]; // iterate over words in the string for word in s.split_whitespace() { // create a vector of characters from the word that are vowels let word_vowels: Vec<char> = word.chars().filter(|&c| vowels.contains(c)).collect(); // if vowel count in the word is n, add the word to the result if word_vowels.len() == n { result.push(word.to_string()); } else { // create segments of the word with length 'n' for i in (0..word.len()).step_by(n) { let segment = &word[i..i + n]; // check if the vowel count in the segment is 'n' let segment_vowels: Vec<char> = segment.chars().filter(|&c| vowels.contains(c)).collect(); if segment_vowels.len() == n { // if vowel count is 'n', add the segment to the result result.push(segment.to_string()); } } } } return result;}```Note: There's no precise analogue in Rust to Python's slicing syntax to get segments of a sequence. This is why we use the method step_by() over a range of indices, which doesn't exactly replicate Python's behaviour (it will panic if the end index is out of range). You may need to add in some more error checking in a real-world use context.

answer:Here is a Python code using the quadratic formula to solve the equation:```pythonimport math# This is a quadratic equation, has the form ax^2 + bx + c = 0a = 1 # coefficient of x^2b = 3 # coefficient of xc = -7 # constant term# computing the discriminantd = (b2) - (4*a*c)if d < 0: print("This equation has no real solution")elif d == 0: x = (-b) / (2*a) print("This equation has one solution: ", x)else: x1 = (-b - math.sqrt(d)) / (2*a) x2 = (-b + math.sqrt(d)) / (2*a) print("This equation has two solutions: ", x1, "and", x2)```When the script above is run, it will determine and display the solutions to the quadratic equation x² + 3x - 7 = 0.

answer:In PyTorch, you can use the torch.masked_select() function to perform the operation similar to logical indexing in numpy. However, this function will return a 1-D tensor from the original tensor, where the values in the mask parameter as True.Below is your solution you asked for while keeping the original shape:```pythonimport torch# The tensor you want to index/sliceB = torch.LongTensor([[999, 777, 114514], [9999, 7777, 1919810]])# The logical index you want to apply on the columns# Notice A_log is converted to bool tensor, because ByteTensor is deprecated.A_log = torch.tensor([1, 1, 0], dtype=torch.bool)# First adjust A_log to the same shape as B,# then apply it to B, finally transpose to keep the original shapeC = torch.masked_select(B.t(), A_log[:, None]).reshape(-1, B.shape[0]).t()print(C)```In this code, `A_log` is reshaped to be a 2D tensor with the same number of rows as `B` (by adding an extra dimension using `None`), then it's transposed and applied to `B`. The masked select picks out the corresponding elements in `B` that align with the True values in the adjusted `A_log`. The reshape command then reshapes the selected elements back into columns, and finally the transpose (`t()`) is used to return the tensor to its original orientation.Please note here, a new tensor is created, and your original tensor is unchanged. If you want to take advantage of memory saving/avoiding creating a new copy of tensor, you need to look into advanced indexing, but currently PyTorch does not offer a function like numpy's `np.ix_`, and you may need to consider re-organizing your data structure.

answer:Here is a Python solution using dynamic programming. The idea is to use a DP table to store the maximum number of nodes that can be monitored at each node, considering if a camera is placed at that node or not.This solution is a Depth-First Search-based DP solution where we traverse the tree and for each node, we calculate the maximum number of nodes that can be monitored in two scenarios: if we place a camera at the node and if we don't place a camera at the node.```pythonfrom collections import defaultdictclass Solution: def maxCountNodes(self, tree, costs, budget): dp = defaultdict(lambda: [-1, -1]) # DP Table edges = defaultdict(list) # adjacency list representation of the tree for u, v in tree[:-1]: # constructing the tree edges[u].append(v) edges[v].append(u) def dfs(node, parent): if dp[node][0] != -1: # if the result is already computed return dp[node] # Calculating the maximum nodes assuming we installed a camera at this node withCamera = costs[node] for child in edges[node]: if child == parent: continue withCamera += dfs(child, node)[1] # Considering the case where we do not place a camera at the node withoutCamera = 0 for child in edges[node]: if child == parent: continue withoutCamera += max(dfs(child, node)) # Check if we have enough budget to place a camera at the node if withCamera <= budget: dp[node][0] = max(dp[node][0], withCamera, withoutCamera) else: dp[node][0] = withoutCamera dp[node][1] = withoutCamera # maximum nodes without camera on node return dp[node] maxCount = max(dfs(0, -1)) # start dfs traversal from root node return maxCount```This solution runs in `O(n)` time as we are running DFS on the tree, which visits each node exactly once, where `n` is the number of nodes in the tree. The space complexity is also `O(n)` due to storing the DP table.However, note that this solution doesn't guarantee the maximum number of node monitoring with the minimal cost. A more optimal solution might be required for that.

answer:Assuming you are in an SQL environment, here is a sample query:Below is an assumption of the databases:`Employee` table:| EmployeeID | Name ||------------|--------|| 1 | John || 2 | Emma || 3 | Robert |`Salary` table:| EmployeeID | Salary | Date ||------------|--------|------------|| 1 | 5000 | 2022-05-01 || 2 | 6000 | 2022-04-01 || 3 | 7000 | 2022-03-01 |`Department` table:| DepartmentID | Department ||--------------|------------|| 1 | HR || 2 | IT || 3 | Finance |`Employee-Department` mapping table:| EmployeeID | DepartmentID ||------------|--------------|| 1 | 1 || 2 | 2 || 3 | 3 |You can join these tables and use grouping to find out the maximum salary, the corresponding month and the department.```sqlSELECT S.Date as Month, MAX(Salary) as Maximum_Salary, D.Department FROM Salary S JOIN Employee-Department ED ON S.EmployeeID = ED.EmployeeIDJOIN Department D ON ED.DepartmentID = D.DepartmentIDGROUP BY S.Date, D.Department;```If you want to know which employee belonged to the department and earned the highest salary, you can use window functions:```sqlSELECT E.Name, S.Date AS Month, S.Salary AS Maximum_Salary, D.Department FROM ( SELECT *, RANK() OVER(PARTITION BY Date ORDER BY Salary DESC) as r FROM Salary ) S JOIN Employee-Department ED ON S.EmployeeID = ED.EmployeeIDJOIN Department D ON ED.DepartmentID = D.DepartmentIDJOIN Employee E ON S.EmployeeID = E.EmployeeIDWHERE S.r = 1;```These queries bring the whole range of dates where the maximum salary was earned, the department as well as the employee who earned it.