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question:The sum of digits. Input The first line contains T, the number of test cases. T lines follow each containing a single integer N. Output Output the sum of digits of each integer N for each test case in a new line. Constraints 1 ≤ T ≤ 10^5 1 ≤ N ≤ 10^7 SAMPLE INPUT 4 123 456 789 1001 SAMPLE OUTPUT 6 15 24 2
answer:def sum_of_digits(n): Returns the sum of the digits of the integer n. return sum(int(digit) for digit in str(n)) def process_test_cases(T, cases): Processes multiple test cases. T: number of test cases cases: a list of T integers Returns a list of integers representing the sum of digits of each input integer. results = [] for n in cases: results.append(sum_of_digits(n)) return results
question:Alex is organizing a coding competition and wants to create a system that rates the difficulty of the problems based on the number of participants who solved them. He needs your help to create a function that assigns a difficulty score to each problem. Every problem will have an initial difficulty score based on its tags and complexity, but that score might change depending on the number of participants who solved it. The new difficulty score is calculated by the following formula: [ text{New Difficulty} = text{Initial Difficulty} times (1 - frac{text{Number of Solvers}}{text{Total Participants}}) ] Your task is to write a program that processes the given input and outputs the new difficulty scores for all problems. Input - The first line contains two integers n and p (1 ≤ n ≤ 1000, 1 ≤ p ≤ 10000) — the number of problems and the total number of participants respectively. - The second line contains n integers ( d_1, d_2, ldots, d_n ) (1 ≤ ( d_i ) ≤ 1000) — the initial difficulty scores of the problems. - The third line contains n integers ( s_1, s_2, ldots, s_n ) (0 ≤ ( s_i ) ≤ p) — the number of participants who solved each problem. Output Print n floating-point numbers, the new difficulty scores of the problems, with an absolute or relative error not exceeding ( 10^{-6} ). Example Input 4 1000 500 600 800 900 200 250 400 450 Output 400.000000000 450.000000000 480.000000000 495.000000000 Input 3 500 700 300 1000 50 250 0 Output 630.000000000 150.000000000 1000.000000000
answer:def calculate_new_difficulties(n, p, initial_difficulties, solvers): new_difficulties = [] for i in range(n): new_difficulty = initial_difficulties[i] * (1 - solvers[i] / p) new_difficulties.append(new_difficulty) return new_difficulties # Example usage: # n = 4, p = 1000 # initial_difficulties = [500, 600, 800, 900] # solvers = [200, 250, 400, 450] # print(calculate_new_difficulties(n, p, initial_difficulties, solvers)) # Output: [400.0, 450.0, 480.0, 495.0]
question:Wildlife photographers often need to track animals for extended periods. Sometimes they leave cameras with limited battery in the wilderness and need to manage the power efficiently to maximize the number of photos taken. A camera has a daily power consumption array where each element represents the power consumed on that day. They know the battery's capacity beforehand and want to utilize this power such that the maximum number of photos is taken before the battery dies. Write a function that takes the power consumption array and the battery's capacity, and returns the maximum number of days the camera can function. Input The first line contains an integer n (1 ≤ n ≤ 100,000) — the number of days. The second line contains n integers a (1 ≤ a[i] ≤ 100) representing the power consumption for each day. The third line contains an integer k (1 ≤ k ≤ 10^9) representing the battery's capacity. Output In the only line, print the maximum number of days the camera can operate before the battery is exhausted. Examples Input: 5 10 20 30 40 50 70 Output: 3 Input: 6 5 5 5 5 5 5 10 Output: 2 Input: 4 1 2 3 4 10 Output: 4 Note In the first example, the battery capacity is 70, and the camera can operate for the first three days consuming a total of 60 units of power. In the second example, the battery can support the camera for only two days consuming a total of 10 units of power. In the third example, the battery capacity allows the camera to operate for all four days as the total consumption is 10 units.
answer:def max_days(power_consumption, battery_capacity): Returns the maximum number of days the camera can operate before the battery is exhausted. total_consumed = 0 days = 0 for consumption in power_consumption: if total_consumed + consumption <= battery_capacity: total_consumed += consumption days += 1 else: break return days
question:Farmer John is planning his farm. His farm is represented as a 2D grid of size n*m, and he wants to place some fences on the grid. Each cell of the grid can either have a fence or be empty. Farmer John wants to make sure that no two fences share a side. Determine the maximum number of fences Farmer John can place on the grid under this condition. A side is shared if two fences are adjacent vertically or horizontally. Input The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the dimensions of the grid. Output Print the maximum number of fences Farmer John can place on the grid. Examples Input 3 3 Output 5 Input 4 4 Output 8 Note In the first example, Farmer John can place fences in the positions: 1 0 1 0 1 0 1 0 1 In the second example, he can place fences in a checkerboard pattern: 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1
answer:def max_fences(n, m): Returns the maximum number of fences that can be placed on an n by m grid such that no two fences share a side. return (n * m + 1) // 2