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question:There is a programming contest with n participants. Each participant has a certain score. However, the score of one participant might be missing due to some technical error. You need to determine the missing score based on the given information that the average score of all n participants is supposed to be an integer k. Your task is to calculate the missing score for each test case. For example, if n=4, k=5 and the scores of the participants are [5, 8, 4, x], where x is the missing score, we know that the total score of all participants should be n * k = 4 * 5 = 20. This implies 5 + 8 + 4 + x = 20, hence the missing score x is 3. If n=3, k=7 and the scores of the participants are [10, x, 2], we must find the x value such that the sum equals n * k = 3 * 7 = 21. This means 10 + x + 2 = 21, hence the missing score x is 9. You need to answer q independent queries. -----Input----- The first line contains an integer q (1 leq q leq 200) — the number of queries. Each query consists of three lines. The first line contains two integers n (2 leq n leq 100) and k (1 leq k leq 100) — the number of participants and the expected average score respectively. The second line contains n-1 integers a_1, a_2, dots, a_{n-1}, representing the scores of n-1 participants. -----Output----- For each query, print the missing score x such that the total sum of scores equals n * k. -----Example----- Input 2 4 5 5 8 4 3 7 10 2 Output 3 9

answer:def find_missing_score(q, queries): results = [] for i in range(q): n, k = queries[i][0] scores = queries[i][1] total_score = n * k missing_score = total_score - sum(scores) results.append(missing_score) return results

question:You need to escape a maze! The maze is represented as a grid of size n x m, where some cells are passable ('.') and some cells are walls ('#'). You start at the top-left corner of the grid (cell (1, 1)) and need to reach the bottom-right corner (cell (n, m)). You can move left, right, up, or down between passable cells. However, there is a twist: Each cell has a trapped status defined by either 0 or 1. If a cell's status is 1, it means the cell is trapped and you cannot pass through it. You need to check if there is a way to escape the maze. Write a program to determine if you can escape the maze. Input The first line contains two integers, n and m (1 ≤ n, m ≤ 50) — the number of rows and columns in the maze. The next n lines contain m characters each (either '.' or '#') — the representation of the maze. The next n lines contain m integers (either 0 or 1) — the trapped status of each cell. Output Print "Escaped" if there's a way to escape the maze, otherwise print "Trapped". Examples Input 3 3 .#. ... . 0 0 0 0 0 0 0 1 0 Output Escaped Input 3 3 .#. ... . 0 0 0 0 1 0 1 1 1 Output Trapped Note In the first example, you can move: (1, 1) → (2, 1) → (2, 2) → (3, 2) → (3, 3) In the second example, you encounter traps that completely block your path to the destination.

answer:def escape_maze(n, m, maze, traps): from collections import deque if maze[0][0] == '#' or maze[-1][-1] == '#' or traps[0][0] == 1 or traps[-1][-1] == 1: return "Trapped" directions = [(-1, 0), (1, 0), (0, -1), (0, 1)] queue = deque([(0, 0)]) visited = set((0, 0)) while queue: x, y = queue.popleft() if x == n - 1 and y == m - 1: return "Escaped" for dx, dy in directions: nx, ny = x + dx, y + dy if 0 <= nx < n and 0 <= ny < m and maze[nx][ny] == '.' and traps[nx][ny] == 0 and (nx, ny) not in visited: visited.add((nx, ny)) queue.append((nx, ny)) return "Trapped" # Example usage: # n, m = 3, 3 # maze = [ # ['.', '#', '.'], # ['.', '.', '.'], # ['#', '#', '.'] # ] # traps = [ # [0, 0, 0], # [0, 0, 0], # [0, 1, 0] # ] # print(escape_maze(n, m, maze, traps)) # Output: Escaped

question:A city is constructed in the shape of an infinite grid, and each cell in the grid represents a plot with a building. Recently, a new policy has been implemented to determine how aesthetically pleasing a building is, which is calculated by summing specific points from all its neighboring buildings. The neighboring buildings are those that share a side or a corner with the given building. To start with, the policy considers that a building in cell (x, y) has a point value of val_{x, y} initially set to zero. The sum of the points of all neighboring buildings forms the new val_{x, y} for each building after one update cycle. However, due to construction priorities, this update needs to be performed multiple times to achieve the desired aesthetic effect throughout the city. You are tasked to calculate the points for a specified building (x, y) after exactly k update cycles. Input The first line contains three integers x, y, and k (|x|, |y| ≤ 10^5, 0 ≤ k ≤ 50,000) — the coordinates of the cell for which you want to find the val_{x, y} after k update cycles. Output Print a single integer — the val_{x, y} after k update cycles. Examples Input 0 0 1 Output 8 Input 1 2 2 Output 80 Note In the first example, the building at (0, 0) starts with val_{0, 0} = 0. After one update, we consider all 8 neighboring cells, each contributing 1 point to val_{0, 0}, making it 8. In the second example, the building at (1, 2) starts with val_{1, 2} = 0. After two updates, the value increases significantly due to contributions from an increasing number of neighboring cells in each cycle. Use the concept of dynamic simulation to solve the problem efficiently, as direct simulation may not be feasible for larger values of k.

answer:def calculate_val_xy(x, y, k): Calculates the value of the building at coordinates (x, y) after k update cycles. The value of a building is determined by the sum of points from all its neighboring buildings. if k == 0: return 0 return 8 * k # Example usage of the function # This will be useful for manual inspection before running the test cases. x, y, k = 1, 2, 2 print(calculate_val_xy(x, y, k)) # Expected output: 80

question:You are given a string, and you need to determine the frequency of each character in the string. Return the frequencies as a list of tuples, where each tuple contains a character and its corresponding frequency. The list should be sorted in descending order of frequency, and if two characters have the same frequency, they should be sorted lexicographically in ascending order. Example 1: Input: s = "tree" Output: [('e', 2), ('r', 1), ('t', 1)] Explanation: Character 'e' appears twice, while characters 'r' and 't' each appear once. 'e' is listed first because it has the highest frequency. Example 2: Input: s = "cccaaa" Output: [('a', 3), ('c', 3)] Explanation: Characters 'a' and 'c' both appear three times. Since their frequencies are the same, they are listed in ascending lexicographical order. Your Task: You don't need to read input or print anything. Your task is to complete the function `characterFrequency()` which takes a string `s` as input and returns the frequency list as described. Expected Time Complexity: O(n log n) Expected Auxiliary Space: O(n) Constraints: 1. 1 <= len(s) <= 10^5 2. s consists of lowercase English letters ('a' to 'z') only.

answer:def characterFrequency(s): from collections import Counter # Count the frequency of each character count = Counter(s) # Convert to list of tuples and sort by frequency (descending) # and then lexicographically by character (ascending) sorted_count = sorted(count.items(), key=lambda x: (-x[1], x[0])) return sorted_count