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question:Problem Statement You are given a set of N points on a 2D plane. Each point is represented by its coordinates (x, y) . Your task is to find a non-degenerate triangle with the maximum possible area that can be formed using any three of these points. If there are multiple triangles with the same maximum area, you can return any of them. If no such triangle exists, return a special value indicating that no non-degenerate triangle can be formed. A triangle is non-degenerate if its area is greater than zero. Constraint * 3 leq N leq 50 * -1000 leq x_i, y_i leq 1000 Input The input format is as follows: N x_1 y_1 x_2 y_2 ... x_N y_N Output Output the maximum possible area of a non-degenerate triangle formed by any three points, rounded to 2 decimal places, or print "No triangle" if no non-degenerate triangle can be formed. Example Input 4 0 0 0 1 1 0 1 1 Output 0.50 Input 5 0 0 1 0 2 0 3 0 4 0 Output No triangle
answer:import itertools import math def max_triangle_area(points): def area(p1, p2, p3): return 0.5 * abs(p1[0] * (p2[1] - p3[1]) + p2[0] * (p3[1] - p1[1]) + p3[0] * (p1[1] - p2[1])) max_area = 0 for p1, p2, p3 in itertools.combinations(points, 3): max_area = max(max_area, area(p1, p2, p3)) return max_area def process_input(N, coordinates): points = [tuple(map(int, coordinates[i:i+2])) for i in range(0, 2*N, 2)] max_area = max_triangle_area(points) if max_area > 0: return round(max_area, 2) else: return "No triangle"
question:The city of Byteland has a grid of size m by n, and its mayor enjoys organizing events by building new vertical roads exclusively on particular rows to facilitate better traffic movement. Each new vertical road spans the entire grid and aligns with one of the columns. However, the mayor wants to ensure that no two new vertical roads are built on consecutive columns to avoid traffic congestion in adjacent paths. Design a program that, given the grid's dimensions and a list of the indices of rows where the roads will be built, determines the maximum number of vertical roads that can be constructed. # Input The first line of input consists of two integers, m and n (1 ≤ m, n ≤ 10000) — the dimensions of the grid (number of rows and columns). The second line consists of an integer k (0 ≤ k ≤ m) — the number of rows where the roads will be constructed. Each of the next k lines contains an integer r (0 ≤ r < m) — the index of a row on which a new vertical road can be built. # Output Print a single integer representing the maximum number of vertical roads that can be built without any two roads being on consecutive columns. # Example Input ``` 5 5 3 1 3 4 ``` Output ``` 2 ``` Note In this example, the grid size is 5x5 and vertical roads can be built on rows 1, 3, and 4. Thus, the maximum number of vertical roads that can be built on non-consecutive columns is 2. For example, they can be built on columns 0 and 2 or columns 1 and 3, ensuring no two consecutive columns have a vertical road.
answer:def max_vertical_roads(m, n, k, rows): Determine the maximum number of vertical roads that can be constructed on a grid of dimensions m x n with the given rows where roads can be built. No two roads can be on consecutive columns. # If there are no available rows to build vertical roads, return 0 if k == 0: return 0 # Calculate how many non-consecutive columns can be used max_roads = (n + 1) // 2 return max_roads
question:For a given list of integers, perform the following operations to manage the list. The operations include adding an integer, removing an integer, finding the maximum integer, and finding the median of the list. Note that adding and removing operations need to maintain the list in sorted order. * add(x): Add the integer x to the list. * remove(x): Remove one occurrence of the integer x from the list. If the integer x is not present in the list, ignore the operation. * max(): Print the maximum integer from the list. * median(): Print the median of the integers in the list. If the list has an odd number of elements, the median is the middle element. If the list has an even number of elements, the median is the average of the two middle elements. Constraints * 1 leq q leq 200,000 * 1 leq x leq 1,000,000,000 Input The input is given in the following format. q query_1 query_2 : query_q Each query query_i is given by 0 x or 1 x or 2 or 3 where the first digits 0, 1, 2, and 3 represent add, remove, max, and median operations respectively. Output For each max operation, print the corresponding maximum integer. For each median operation, print the corresponding median value. Example Input ``` 11 0 7 0 3 0 5 0 9 2 3 1 5 2 3 1 10 3 ``` Output ``` 9 5 9 7 ```
answer:import bisect class IntegerListManager: def __init__(self): self.integers = [] def add(self, x): bisect.insort(self.integers, x) def remove(self, x): if x in self.integers: self.integers.remove(x) def max(self): if self.integers: return self.integers[-1] return None def median(self): if not self.integers: return None n = len(self.integers) if n % 2 == 1: return self.integers[n // 2] else: return (self.integers[(n // 2) - 1] + self.integers[n // 2]) / 2
question:You are given a set of integers, each representing the times (in seconds) at which tasks will arrive. Each task, when it arrives, will add a load of `1` to the system. Once a task is processed, it will remain in the system for `t` seconds before being removed from the system. Your goal is to determine the maximum load on the system at any point in time, given the arrival times of the tasks and the duration `t`. # Input - The first line contains a single integer `t` (1 ≤ t ≤ 10^4) — the number of test cases. - The first line of each test case contains two integers `n` and `d` (1 ≤ n ≤ 10^5 and 1 ≤ d ≤ 10^9) — where `n` is the number of tasks and `d` is the duration a task will remain in the system. - The second line of each test case contains `n` integers `a_1, a_2, ..., a_n` (1 ≤ a_i ≤ 10^9) — the arrival times of the tasks. It is guaranteed that the sum of `n` over all test cases does not exceed 10^5. # Output For each test case, output a single integer: the maximum load on the system at any point in time. # Example Input ``` 2 5 3 1 2 3 4 10 6 2 1 1 1 1 1 1 ``` Output ``` 4 6 ``` Explanation In the first test case, tasks arrive at times 1, 2, 3, 4, and 10. Each task will be processed and stay in the system for 3 seconds. The maximum system load occurs at time 4, where tasks that arrived at times 1, 2, 3, and 4 are all still in the system. In the second test case, all tasks arrive at the same time and stay in the system for 2 seconds, leading to a maximum load of 6.
answer:def max_system_load(t, test_cases): Determines the maximum load on the system at any point in time for each test case. Arguments: t -- Number of test cases test_cases -- List containing tuples (n, d, arrivals) for each test case Returns: List of integers where each integer is the maximum load for the corresponding test case. results = [] for case in test_cases: n, d, arrivals = case arrivals.sort() max_load = 0 current_load = 0 events = [] for arrival in arrivals: events.append((arrival, 'arrival')) events.append((arrival + d, 'departure')) events.sort() for time, event_type in events: if event_type == 'arrival': current_load += 1 else: current_load -= 1 max_load = max(max_load, current_load) results.append(max_load) return results