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224 lines
5.3 KiB
Python
Executable file
224 lines
5.3 KiB
Python
Executable file
"""#
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### 谜题描述
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You are a given an array a of length n. Find a subarray a[l..r] with length at least k with the largest median.
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A median in an array of length n is an element which occupies position number ⌊ (n + 1)/(2) ⌋ after we sort the elements in non-decreasing order. For example: median([1, 2, 3, 4]) = 2, median([3, 2, 1]) = 2, median([2, 1, 2, 1]) = 1.
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Subarray a[l..r] is a contiguous part of the array a, i. e. the array a_l,a_{l+1},…,a_r for some 1 ≤ l ≤ r ≤ n, its length is r - l + 1.
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Input
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The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5).
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The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n).
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Output
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Output one integer m — the maximum median you can get.
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Examples
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Input
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5 3
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1 2 3 2 1
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Output
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2
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Input
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4 2
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1 2 3 4
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Output
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3
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Note
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In the first example all the possible subarrays are [1..3], [1..4], [1..5], [2..4], [2..5] and [3..5] and the median for all of them is 2, so the maximum possible median is 2 too.
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In the second example median([3..4]) = 3.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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from collections import Counter, defaultdict, deque
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import bisect
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from sys import stdin, stdout
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from itertools import repeat
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import math
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def inp(force_list=False):
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re = map(int, raw_input().split())
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if len(re) == 1 and not force_list:
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return re[0]
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return re
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def inst():
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return raw_input().strip()
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def gcd(x, y):
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while(y):
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x, y = y, x % y
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return x
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mod = int(1e9)+7
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def quickm(a, b):
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base = a
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re = 1
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while b:
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if b&1:
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re *= base
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re %= mod
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b >>= 1
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base *= base
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base %= mod
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return re
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def inv(num):
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return quickm(num, mod-2)
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def my_main():
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kase = 1 #inp()
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pans = []
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for _ in range(kase):
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n, k = inp()
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da = inp(True)
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l, r = min(da), max(da)+1
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while l < r-1:
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mid = (l+r)/2
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def ck(mid):
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ps = [0]
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for j in [(-1 if i<mid else 1) for i in da]:
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ps.append(j+ps[-1])
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ok = 0
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mps = [-100000] * (n+1)
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mps[-1] = ps[-1]
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for i in range(n-1, -1, -1):
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mps[i] = max(mps[i+1], ps[i])
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for i in range(n-k+1):
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if mps[i+k] - ps[i] > 0:
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ok = 1
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break
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return ok
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# print l, r
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if ck(mid):
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l, r = mid, r
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else:
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l, r = l, mid
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print l
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# print '\n'.join(pans)
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my_main()
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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from bootcamp import Basebootcamp
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import random
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import re
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def calculate_max_median(n, k, array):
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"""优化后的中位数计算函数"""
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left, right = min(array), max(array)
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answer = left # 初始化
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while left <= right:
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mid = (left + right) // 2
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prefix = [0]*(n+1)
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min_prefix = float('inf')
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# 计算前缀和
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for i in range(n):
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prefix[i+1] = prefix[i] + (1 if array[i] >= mid else -1)
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# 寻找有效窗口
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valid = False
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for i in range(k, n+1):
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if prefix[i] - min_prefix > 0:
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valid = True
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break
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min_prefix = min(min_prefix, prefix[i - k + 1])
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if valid:
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answer = mid
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left = mid + 1
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else:
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right = mid - 1
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return answer
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class Dmaxmedianbootcamp(Basebootcamp): # 修正类名
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def __init__(self, **params):
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super().__init__(**params)
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default_params = {
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'min_n': 5,
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'max_n': 20,
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'max_val': 20,
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'ensure_solvable': True # 保证生成有解的案例
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}
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self.params = {**default_params, **params}
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def case_generator(self):
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"""生成有效案例的优化版本"""
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n = random.randint(self.params['min_n'], self.params['max_n'])
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k = random.randint(1, n)
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# 生成有解数组的逻辑
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while True:
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arr = [random.randint(1, self.params['max_val']) for _ in range(n)]
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if len(set(arr)) >= 2: # 确保至少有两个不同值
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break
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return {
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'n': n,
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'k': k,
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'array': arr.copy(),
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'answer': calculate_max_median(n, k, arr)
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}
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@staticmethod
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def prompt_func(case):
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return f"""给定长度为n的数组，请找出长度≥k的连续子数组的最大中位数。
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输入：
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{case['n']} {case['k']}
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{' '.join(map(str, case['array']))}
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规则：
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1. 中位数定义：排序后第⌊(长度+1)/2⌋个元素
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2. 子数组必须连续且长度≥k
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3. 输出最大可能的中位数
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请将最终答案放在[answer]标签内，如：[answer]42[/answer]"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output)
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try:
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return int(matches[-1]) if matches else None
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except:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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return solution == identity['answer']
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