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question:A group of scientists are developing a powerful algorithm that can generate a personalized workout plan for users based on their fitness levels and goals. The algorithm determines the sequence of exercises, but sometimes it produces duplicate consecutive workouts which is not ideal. They want to simplify the workout plan by compressing consecutive duplicate workouts into a single workout followed by the count of repetitions. For example, if the workout sequence is "pushup pushup situp situp situp squat squat", it should be compressed to "pushup2 situp3 squat2". Write a program that takes multiple workout sequences and compresses each one as described. INPUT: The first line of input is T, the number of workout sequences. Each of the following T lines contains a sequence of workouts in a space-separated string format. OUTPUT: For each workout sequence, print the compressed workout sequence on a new line. CONSTRAINTS: 1 ≤ T ≤ 50 1 ≤ Length of each workout sequence ≤ 1000 Each workout sequence contains only lowercase alphabetic characters separated by spaces representing different workouts. SAMPLE INPUT: 3 pushup pushup situp situp situp squat squat jump jump jump situp pushup run run pushup pushup pushup run run run SAMPLE OUTPUT: pushup2 situp3 squat2 jump3 situp1 pushup1 run2 pushup3 run3 Explanation: In the first sequence, "pushup" appears twice consecutively, "situp" appears three times consecutively, and "squat" appears twice consecutively. In the second sequence, "jump" appears three times, "situp" appears once, and "pushup" appears once. In the third sequence, "run" appears twice, "pushup" appears three times consecutively, and then "run" appears three times.

answer:def compress_workout_sequences(T, sequences): results = [] for sequence in sequences: workouts = sequence.split() compressed = [] current_workout = workouts[0] count = 1 for i in range(1, len(workouts)): if workouts[i] == current_workout: count += 1 else: compressed.append(f"{current_workout}{count}") current_workout = workouts[i] count = 1 compressed.append(f"{current_workout}{count}") results.append(" ".join(compressed)) return results

question:A data center contains R racks numbered from 1 to R, and each rack has C servers numbered from 1 to C. The servers are organized in a grid where each rack corresponds to a row and each server in a rack corresponds to a column. When a server is down, it needs to be replaced and for this task, a technician must visit the particular rack and column of the grid. You’re given multiple sets of operations performed in this data center, where each operation can be one of the following: 1. "REPLACE r c" - which means the server in rack r and column c has been replaced. 2. "QUERY r c" - queries the status of the server in rack r and column c, whether it has been replaced or not. Your task is to process these operations and determine the outcome for each "QUERY" operation. **INPUT FORMAT:** ``` Line 1: Number of test cases T Next each test case contains: Line 1: Two integers, R (number of racks) and C (number of columns) Followed by multiple lines representing operations, ending with "END" ``` Each operation line contains either: - "REPLACE r c" - "QUERY r c" **CONSTRAINTS:** - 1 ≤ T ≤ 10 - 1 ≤ R, C ≤ 1000 - 1 ≤ r ≤ R - 1 ≤ c ≤ C - The maximum number of operations per test case is 100000 **OUTPUT FORMAT:** For each test case output results for each "QUERY" operation in separate lines, printing "YES" if the server has been replaced and "NO" otherwise. **SAMPLE INPUT:** ``` 2 3 3 REPLACE 1 1 QUERY 1 1 QUERY 2 2 REPLACE 3 3 QUERY 3 3 END 2 2 QUERY 1 1 REPLACE 1 1 QUERY 1 1 END ``` **SAMPLE OUTPUT:** ``` YES NO YES NO YES ``` **Explanation:** Test case 1: - The server at (1, 1) is replaced. QUERY operation at (1, 1) returns "YES". - QUERY operation at (2, 2) returns "NO" because it hasn't been replaced. - The server at (3, 3) is replaced. QUERY operation at (3, 3) returns "YES". Test case 2: - QUERY operation at (1, 1) returns "NO" because it hasn't been replaced yet. - The server at (1, 1) is replaced. QUERY operation at (1, 1) returns "YES".

answer:def data_center_operations(test_cases): results = [] for case in test_cases: R, C, operations = case replaced = set() for op in operations: if op.startswith("REPLACE"): _, r, c = op.split() replaced.add((int(r), int(c))) elif op.startswith("QUERY"): _, r, c = op.split() if (int(r), int(c)) in replaced: results.append("YES") else: results.append("NO") return results

question:Raina is on a hike through a beautiful forest, and she wants to reach a picnic spot as quickly as possible. There are n trails in the forest, and each trail has a specific length and an elevation change. Raina can walk different speeds depending on whether she is going uphill or downhill. Given the length and elevation change of each trail, compute the minimum time Raina will need to reach her picnic spot if she can start from any point at the beginning of any trail. Input The first line of the input contains two integers uh and dh (1 ≤ uh, dh ≤ 100), where uh is Raina’s speed going uphill (in units per hour), and dh is her speed going downhill (in units per hour). The second line contains a single integer n (1 ≤ n ≤ 1000), the number of trails. Each of the following n lines contains two integers li and ei (1 ≤ li ≤ 100, -100 ≤ ei ≤ 100), where li is the length of the i-th trail, and ei is the elevation change of the i-th trail. Output Print a single real value—the minimum time Raina needs to reach her picnic spot. Your answer will be considered correct if its absolute or relative error does not exceed 10^-6. Examples Input 4 6 3 5 -10 6 6 7 2 Output 1.3333333333333333 Input 2 3 2 8 20 10 -30 Output 3.3333333333333333 Note In the first sample, for trail 1, Raina will move downhill (elevation change is negative), so the time will be 5 / 6. For trail 2, Raina will move uphill, so the time will be 6 / 4. For trail 3, Raina will move uphill too, so the time will be 7 / 4. The minimum of these times is 5 / 6 which equals approximately 0.8333333333 hours. In the second sample, for trail 1, time will be 8 / 2 = 4 (uphill). For the second trail, Raina will move downhill, so the time will be 10 / 3 which equals approximately 3.3333333333 hours.

answer:def minimum_hike_time(uh, dh, trails): min_time = float('inf') for length, elevation in trails: if elevation > 0: time = length / uh else: time = length / dh min_time = min(min_time, time) return min_time

question:In a small town, there is a single main street with n houses lined up one after another. Each house is painted with one of the k possible colors. The town has received complaints from some residents who believe that having two adjacent houses painted the same color is unappealing. Therefore, the mayor has decided to repaint some of the houses in order to ensure that no two adjacent houses share the same color. Repainting a house of any color takes the same effort, regardless of the initial color of the house. Given the initial colors of the houses, determine the minimum number of houses that need to be repainted to achieve the goal. Input The first line contains two space-separated integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100) — the number of houses and the number of possible colors. The second line contains n space-separated integers, each integer c_i (1 ≤ c_i ≤ k) representing the initial color of each house. Output Print a single integer — the minimum number of houses that need to be repainted. Examples Input 5 3 1 1 2 3 3 Output 2 Input 4 2 1 2 2 1 Output 1 Note In the first example, repainting the second house to any color other than 1, and repainting the fifth house to any color other than 3 will result in no two adjacent houses having the same color. In the second example, repainting the third house to the first color will achieve the goal.

answer:def min_repaints(n, k, colors): Given n houses and k possible colors, calculate the minimum number of repaints needed such that no two adjacent houses have the same color. :param n: int - number of houses :param k: int - number of possible colors :param colors: list of int - initial colors of the houses :return: int - minimum number of repaints needed repaints = 0 for i in range(1, n): if colors[i] == colors[i - 1]: repaints += 1 return repaints

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