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internbootcamp/bootcamp/dratingcompression/dratingcompression.py

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+"""# 
+
+### 谜题描述
+On the competitive programming platform CodeCook, every person has a rating graph described by an array of integers a of length n. You are now updating the infrastructure, so you've created a program to compress these graphs.
+
+The program works as follows. Given an integer parameter k, the program takes the minimum of each contiguous subarray of length k in a.
+
+More formally, for an array a of length n and an integer k, define the k-compression array of a as an array b of length n-k+1, such that $$$b_j =min_{j≤ i≤ j+k-1}a_i$$$
+
+For example, the 3-compression array of [1, 3, 4, 5, 2] is [min\{1, 3, 4\}, min\{3, 4, 5\}, min\{4, 5, 2\}]=[1, 3, 2].
+
+A permutation of length m is an array consisting of m distinct integers from 1 to m in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (m=3 but there is 4 in the array).
+
+A k-compression array will make CodeCook users happy if it will be a permutation. Given an array a, determine for all 1≤ k≤ n if CodeCook users will be happy after a k-compression of this array or not.
+
+Input
+
+The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
+
+The first line of the description of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the array.
+
+The second line of the description of each test case contains n integers a_1,…,a_n (1≤ a_i≤ n) — the elements of the array.
+
+It is guaranteed, that the sum of n for all test cases does not exceed 3⋅ 10^5.
+
+Output
+
+For each test case, print a binary string of length n. 
+
+The k-th character of the string should be 1 if CodeCook users will be happy after a k-compression of the array a, and 0 otherwise. 
+
+Example
+
+Input
+
+
+5
+5
+1 5 3 4 2
+4
+1 3 2 1
+5
+1 3 3 3 2
+10
+1 2 3 4 5 6 7 8 9 10
+3
+3 3 2
+
+
+Output
+
+
+10111
+0001
+00111
+1111111111
+000
+
+Note
+
+In the first test case, a=[1, 5, 3, 4, 2].
+
+  * The 1-compression of a is [1, 5, 3, 4, 2] and it is a permutation. 
+  * The 2-compression of a is [1, 3, 3, 2] and it is not a permutation, since 3 appears twice. 
+  * The 3-compression of a is [1, 3, 2] and it is a permutation. 
+  * The 4-compression of a is [1, 2] and it is a permutation. 
+  * The 5-compression of a is [1] and it is a permutation. 
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+import sys
+if sys.subversion[0] == \"PyPy\":
+    import io, atexit
+    sys.stdout = io.BytesIO()
+    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
+    
+    sys.stdin = io.BytesIO(sys.stdin.read())
+    input = lambda: sys.stdin.readline().rstrip()
+
+RS = raw_input
+RI = lambda x=int: map(x,RS().split())
+RN = lambda x=int: x(RS())
+''' ...................................................................... '''
+class SegTree:
+    ''' PU-RQ '''
+ 
+    def __init__(self,arr,n):
+        ''' Builds in O(N) '''
+        
+        self.N = n
+        self.tree = [float('inf')]*(2*n)
+ 
+        for i in xrange(n):     # Leaf Node
+            self.tree[n+i] = arr[i]
+ 
+        for i in xrange(n-1,0,-1):  # Parent Node
+            self.tree[i] = min(self.tree[2*i] , self.tree[2*i+1])
+ 
+    
+    def update(self, i, newValue):
+        ''' 0-Based Indexing '''
+        
+        i += self.N     # Points to leaf node (i)
+ 
+        self.tree[i] = newValue
+ 
+        while i>1:
+            i/=2        # Points to parent node
+            self.tree[i] = min(self.tree[2*i] , self.tree[2*i+1])
+ 
+ 
+    def query(self, L, R):
+        ''' Query in interval [L,R] (0-Based Indexing) '''
+        
+        # Leaf Nodes
+        L += self.N
+        R += self.N + 1
+ 
+        res = float('inf')
+        while L<R:
+            if L&1:
+                res = min(res,self.tree[L])
+                #res += self.tree[L]     # Right Child
+                L += 1
+                
+            if R&1:
+                R -= 1
+                res = min(res,self.tree[R])
+                #res += self.tree[R]     # Left Child
+                
+            L/=2; R/=2
+        return res
+
+    
+for _ in xrange(RN()):
+    n = RN()
+    arr = RI()
+    seg = SegTree(arr,n)
+    
+    ans = [0]*n
+    lo,hi = 0,n-1
+
+    val = 1
+    
+    for k in xrange(n,1,-1):
+        m = seg.query(lo,hi)
+
+        if m==val:
+            ans[k-1]=1
+
+            if arr[lo]==m: lo+=1
+            elif arr[hi]==m: hi-=1
+            else: break
+        else:
+            break
+
+        val+=1
+
+
+    if len(set(arr))==n: ans[0]=1
+        
+    print ''.join(map(str,ans))
+```
+
+
+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
+import random
+from collections import deque
+from bootcamp import Basebootcamp
+
+class Dratingcompressionbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        self.max_n = params.get('max_n', 10)
+        self.min_n = params.get('min_n', 1)
+    
+    def case_generator(self):
+        """生成多样化测试案例，包含：完全随机、保证有效k=1、强制全无效等类型"""
+        case_type = random.choice(['random', 'valid_k1', 'invalid_all'])
+        n = random.randint(self.min_n, self.max_n)
+        
+        if case_type == 'valid_k1':
+            # 生成保证k=1有效的案例（数组本身就是排列）
+            a = list(range(1, n+1))
+            random.shuffle(a)
+        elif case_type == 'invalid_all':
+            # 生成所有k都无效的案例（数组元素全相同）
+            a = [1] * n
+        else:
+            # 完全随机生成
+            a = [random.randint(1, n) for _ in range(n)]
+        
+        correct_answer = self.optimized_solve(n, a)
+        return {
+            'n': n,
+            'a': a,
+            'correct_answer': correct_answer,
+            'case_type': case_type  # 用于验证时追踪
+        }
+
+    def optimized_solve(self, n, a):
+        """准确高效的解法实现"""
+        answer = ['0'] * n
+        
+        # 预处理k=1的情况
+        k1_valid = (sorted(a) == list(range(1, n+1)))
+        answer[0] = '1' if k1_valid else '0'
+        
+        # 预处理每个位置的next smaller元素
+        next_smaller = [n] * n
+        prev_smaller = [-1] * n
+        stack = []
+        
+        for i in range(n):
+            while stack and a[i] < a[stack[-1]]:
+                next_smaller[stack.pop()] = i
+            prev_smaller[i] = stack[-1] if stack else -1
+            stack.append(i)
+        
+        # 统计每个元素作为最小值的影响范围
+        min_intervals = {}
+        for i in range(n):
+            left = prev_smaller[i] + 1
+            right = next_smaller[i] - 1
+            min_intervals[a[i]] = max(min_intervals.get(a[i], 0), right - left + 1)
+        
+        # 根据定理：当且仅当存在元素只能在窗口大小>=某个值时出现
+        for m in range(1, n):
+            max_k = n - m + 1
+            if m in min_intervals and min_intervals[m] >= m:
+                for k in range(max(1, m), max_k+1):
+                    if k <= min_intervals[m]:
+                        answer[k-1] = '1'
+        
+        # 最终验证每个k的结果
+        for k in range(1, n+1):
+            m = n - k + 1
+            if m < 1:
+                continue
+            if answer[k-1] == '1':
+                # 二次验证确保正确性
+                window_min = self.sliding_window_min(a, k)
+                if not self.is_permutation(window_min, m):
+                    answer[k-1] = '0'
+        return ''.join(answer)
+
+    @staticmethod
+    def sliding_window_min(arr, k):
+        """精确计算滑动窗口的最小值"""
+        dq = deque()
+        result = []
+        for i, num in enumerate(arr):
+            while dq and arr[dq[-1]] >= num:
+                dq.pop()
+            dq.append(i)
+            
+            if dq[0] == i - k:
+                dq.popleft()
+            
+            if i >= k - 1:
+                result.append(arr[dq[0]])
+        return result
+
+    @staticmethod
+    def is_permutation(nums, m):
+        """验证是否为1~m的排列"""
+        return len(nums) == m and set(nums) == set(range(1, m+1)) and len(set(nums)) == m
+
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        a_str = ' '.join(map(str, question_case['a']))
+        n = question_case['n']
+        return f"""给定长度为{n}的数组：[{a_str}]
+请对k=1到k={n}依次判断：
+1. 计算k-compression数组（每个元素是连续k个元素的最小值）
+2. 检查该数组是否是1到(n-k+1)的排列
+
+输出：长度为{n}的二进制字符串，第k位为1表示有效。答案置于[answer][/answer]中。例如：[answer]1010[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        import re
+        matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output, re.IGNORECASE)
+        return matches[-1] if matches else None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        # 格式验证
+        if not solution or len(solution) != identity['n']:
+            return False
+        
+        # 逻辑验证（兼容可能存在多个正确解的情况）
+        expected = identity['correct_answer']
+        
+        # 检查每个有效位的合理性
+        for k in range(1, identity['n']+1):
+            if solution[k-1] == '1' and expected[k-1] == '0':
+                return False
+            if identity['case_type'] == 'invalid_all' and '1' in solution:
+                return False
+        return solution == expected