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internbootcamp/bootcamp/dmaxmedian/dmaxmedian.py

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+"""# 
+
+### 谜题描述
+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.
+
+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.
+
+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.
+
+Input
+
+The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5).
+
+The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n).
+
+Output
+
+Output one integer m — the maximum median you can get.
+
+Examples
+
+Input
+
+
+5 3
+1 2 3 2 1
+
+
+Output
+
+
+2
+
+Input
+
+
+4 2
+1 2 3 4
+
+
+Output
+
+
+3
+
+Note
+
+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.
+
+In the second example median([3..4]) = 3.
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+from collections import Counter, defaultdict, deque
+import bisect
+from sys import stdin, stdout
+from itertools import repeat
+import math
+
+
+def inp(force_list=False):
+    re = map(int, raw_input().split())
+    if len(re) == 1 and not force_list:
+        return re[0]
+    return re
+
+def inst():
+    return raw_input().strip()
+
+def gcd(x, y):
+   while(y):
+       x, y = y, x % y
+   return x
+
+
+mod = int(1e9)+7
+
+def quickm(a, b):
+    base = a
+    re = 1
+    while b:
+        if b&1:
+            re *= base
+            re %= mod
+        b >>= 1
+        base *= base
+        base %= mod
+    return re
+
+def inv(num):
+    return quickm(num, mod-2)
+
+
+
+def my_main():
+    kase = 1 #inp()
+    pans = []
+    for _ in range(kase):
+        n, k = inp()
+        da = inp(True)
+        l, r = min(da), max(da)+1
+        while l < r-1:
+            mid = (l+r)/2
+            def ck(mid):
+                ps = [0]
+                for j in [(-1 if i<mid else 1) for i in da]:
+                    ps.append(j+ps[-1])
+                ok = 0
+                mps = [-100000] * (n+1)
+                mps[-1] = ps[-1]
+                for i in range(n-1, -1, -1):
+                    mps[i] = max(mps[i+1], ps[i])
+                for i in range(n-k+1):
+                    if mps[i+k] - ps[i] > 0:
+                        ok = 1
+                        break
+                return ok
+            # print l, r
+            if ck(mid):
+                l, r = mid, r
+            else:
+                l, r = l, mid
+        print l
+
+
+    # print '\n'.join(pans)
+
+my_main()
+```
+
+
+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。
+"""
+
+from bootcamp import Basebootcamp
+from bootcamp import Basebootcamp
+import random
+import re
+
+def calculate_max_median(n, k, array):
+    """优化后的中位数计算函数"""
+    left, right = min(array), max(array)
+    answer = left  # 初始化
+    
+    while left <= right:
+        mid = (left + right) // 2
+        prefix = [0]*(n+1)
+        min_prefix = float('inf')
+        
+        # 计算前缀和
+        for i in range(n):
+            prefix[i+1] = prefix[i] + (1 if array[i] >= mid else -1)
+        
+        # 寻找有效窗口
+        valid = False
+        for i in range(k, n+1):
+            if prefix[i] - min_prefix > 0:
+                valid = True
+                break
+            min_prefix = min(min_prefix, prefix[i - k + 1])
+        
+        if valid:
+            answer = mid
+            left = mid + 1
+        else:
+            right = mid - 1
+    
+    return answer
+
+class Dmaxmedianbootcamp(Basebootcamp):  # 修正类名
+    def __init__(self, **params):
+        super().__init__(**params)
+        default_params = {
+            'min_n': 5,
+            'max_n': 20,
+            'max_val': 20,
+            'ensure_solvable': True  # 保证生成有解的案例
+        }
+        self.params = {**default_params, **params}
+    
+    def case_generator(self):
+        """生成有效案例的优化版本"""
+        n = random.randint(self.params['min_n'], self.params['max_n'])
+        k = random.randint(1, n)
+        
+        # 生成有解数组的逻辑
+        while True:
+            arr = [random.randint(1, self.params['max_val']) for _ in range(n)]
+            if len(set(arr)) >= 2:  # 确保至少有两个不同值
+                break
+        
+        return {
+            'n': n,
+            'k': k,
+            'array': arr.copy(),
+            'answer': calculate_max_median(n, k, arr)
+        }
+    
+    @staticmethod
+    def prompt_func(case):
+        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。
+
+输入：
+{case['n']} {case['k']}
+{' '.join(map(str, case['array']))}
+
+规则：
+1. 中位数定义：排序后第⌊(长度+1)/2⌋个元素
+2. 子数组必须连续且长度≥k
+3. 输出最大可能的中位数
+
+请将最终答案放在[answer]标签内，如：[answer]42[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output)
+        try:
+            return int(matches[-1]) if matches else None
+        except:
+            return None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        return solution == identity['answer']