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303 lines
9.4 KiB
Python
Executable file
303 lines
9.4 KiB
Python
Executable file
"""#
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### 谜题描述
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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.
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The program works as follows. Given an integer parameter k, the program takes the minimum of each contiguous subarray of length k in a.
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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 $$$b_j =min_{j≤ i≤ j+k-1}a_i$$$
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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].
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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).
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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.
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Input
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The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
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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.
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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.
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It is guaranteed, that the sum of n for all test cases does not exceed 3⋅ 10^5.
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Output
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For each test case, print a binary string of length n.
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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.
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Example
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Input
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5
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5
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1 5 3 4 2
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4
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1 3 2 1
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5
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1 3 3 3 2
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10
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1 2 3 4 5 6 7 8 9 10
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3
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3 3 2
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Output
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10111
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0001
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00111
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1111111111
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000
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Note
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In the first test case, a=[1, 5, 3, 4, 2].
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* The 1-compression of a is [1, 5, 3, 4, 2] and it is a permutation.
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* The 2-compression of a is [1, 3, 3, 2] and it is not a permutation, since 3 appears twice.
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* The 3-compression of a is [1, 3, 2] and it is a permutation.
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* The 4-compression of a is [1, 2] and it is a permutation.
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* The 5-compression of a is [1] and it is a permutation.
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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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import sys
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if sys.subversion[0] == \"PyPy\":
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import io, atexit
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sys.stdout = io.BytesIO()
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atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
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sys.stdin = io.BytesIO(sys.stdin.read())
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input = lambda: sys.stdin.readline().rstrip()
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RS = raw_input
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RI = lambda x=int: map(x,RS().split())
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RN = lambda x=int: x(RS())
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''' ...................................................................... '''
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class SegTree:
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''' PU-RQ '''
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def __init__(self,arr,n):
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''' Builds in O(N) '''
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self.N = n
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self.tree = [float('inf')]*(2*n)
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for i in xrange(n): # Leaf Node
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self.tree[n+i] = arr[i]
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for i in xrange(n-1,0,-1): # Parent Node
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self.tree[i] = min(self.tree[2*i] , self.tree[2*i+1])
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def update(self, i, newValue):
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''' 0-Based Indexing '''
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i += self.N # Points to leaf node (i)
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self.tree[i] = newValue
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while i>1:
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i/=2 # Points to parent node
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self.tree[i] = min(self.tree[2*i] , self.tree[2*i+1])
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def query(self, L, R):
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''' Query in interval [L,R] (0-Based Indexing) '''
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# Leaf Nodes
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L += self.N
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R += self.N + 1
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res = float('inf')
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while L<R:
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if L&1:
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res = min(res,self.tree[L])
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#res += self.tree[L] # Right Child
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L += 1
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if R&1:
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R -= 1
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res = min(res,self.tree[R])
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#res += self.tree[R] # Left Child
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L/=2; R/=2
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return res
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for _ in xrange(RN()):
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n = RN()
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arr = RI()
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seg = SegTree(arr,n)
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ans = [0]*n
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lo,hi = 0,n-1
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val = 1
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for k in xrange(n,1,-1):
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m = seg.query(lo,hi)
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if m==val:
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ans[k-1]=1
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if arr[lo]==m: lo+=1
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elif arr[hi]==m: hi-=1
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else: break
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else:
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break
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val+=1
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if len(set(arr))==n: ans[0]=1
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print ''.join(map(str,ans))
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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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import random
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from collections import deque
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from bootcamp import Basebootcamp
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class Dratingcompressionbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.max_n = params.get('max_n', 10)
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self.min_n = params.get('min_n', 1)
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def case_generator(self):
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"""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"""
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case_type = random.choice(['random', 'valid_k1', 'invalid_all'])
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n = random.randint(self.min_n, self.max_n)
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if case_type == 'valid_k1':
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# 生成保证k=1有效的案例（数组本身就是排列）
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a = list(range(1, n+1))
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random.shuffle(a)
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elif case_type == 'invalid_all':
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# 生成所有k都无效的案例（数组元素全相同）
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a = [1] * n
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else:
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# 完全随机生成
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a = [random.randint(1, n) for _ in range(n)]
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correct_answer = self.optimized_solve(n, a)
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return {
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'n': n,
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'a': a,
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'correct_answer': correct_answer,
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'case_type': case_type # 用于验证时追踪
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}
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def optimized_solve(self, n, a):
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"""准确高效的解法实现"""
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answer = ['0'] * n
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# 预处理k=1的情况
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k1_valid = (sorted(a) == list(range(1, n+1)))
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answer[0] = '1' if k1_valid else '0'
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# 预处理每个位置的next smaller元素
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next_smaller = [n] * n
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prev_smaller = [-1] * n
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stack = []
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for i in range(n):
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while stack and a[i] < a[stack[-1]]:
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next_smaller[stack.pop()] = i
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prev_smaller[i] = stack[-1] if stack else -1
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stack.append(i)
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# 统计每个元素作为最小值的影响范围
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min_intervals = {}
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for i in range(n):
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left = prev_smaller[i] + 1
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right = next_smaller[i] - 1
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min_intervals[a[i]] = max(min_intervals.get(a[i], 0), right - left + 1)
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# 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
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for m in range(1, n):
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max_k = n - m + 1
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if m in min_intervals and min_intervals[m] >= m:
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for k in range(max(1, m), max_k+1):
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if k <= min_intervals[m]:
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answer[k-1] = '1'
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# 最终验证每个k的结果
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for k in range(1, n+1):
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m = n - k + 1
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if m < 1:
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continue
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if answer[k-1] == '1':
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# 二次验证确保正确性
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window_min = self.sliding_window_min(a, k)
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if not self.is_permutation(window_min, m):
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answer[k-1] = '0'
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return ''.join(answer)
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@staticmethod
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def sliding_window_min(arr, k):
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"""精确计算滑动窗口的最小值"""
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dq = deque()
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result = []
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for i, num in enumerate(arr):
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while dq and arr[dq[-1]] >= num:
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dq.pop()
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dq.append(i)
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if dq[0] == i - k:
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dq.popleft()
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if i >= k - 1:
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result.append(arr[dq[0]])
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return result
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@staticmethod
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def is_permutation(nums, m):
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"""验证是否为1~m的排列"""
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return len(nums) == m and set(nums) == set(range(1, m+1)) and len(set(nums)) == m
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@staticmethod
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def prompt_func(question_case) -> str:
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a_str = ' '.join(map(str, question_case['a']))
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n = question_case['n']
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return f"""给定长度为{n}的数组：[{a_str}]
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请对k=1到k={n}依次判断：
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1. 计算k-compression数组（每个元素是连续k个元素的最小值）
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2. 检查该数组是否是1到(n-k+1)的排列
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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]"""
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@staticmethod
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def extract_output(output):
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import re
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matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output, re.IGNORECASE)
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return matches[-1] if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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# 格式验证
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if not solution or len(solution) != identity['n']:
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return False
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# 逻辑验证（兼容可能存在多个正确解的情况）
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expected = identity['correct_answer']
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# 检查每个有效位的合理性
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for k in range(1, identity['n']+1):
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if solution[k-1] == '1' and expected[k-1] == '0':
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return False
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if identity['case_type'] == 'invalid_all' and '1' in solution:
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return False
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return solution == expected
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