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289 lines
9.3 KiB
Python
Executable file
289 lines
9.3 KiB
Python
Executable file
"""#
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### 谜题描述
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Patrick likes to play baseball, but sometimes he will spend so many hours hitting home runs that his mind starts to get foggy! Patrick is sure that his scores across n sessions follow the identity permutation (ie. in the first game he scores 1 point, in the second game he scores 2 points and so on). However, when he checks back to his record, he sees that all the numbers are mixed up!
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Define a special exchange as the following: choose any subarray of the scores and permute elements such that no element of subarray gets to the same position as it was before the exchange. For example, performing a special exchange on [1,2,3] can yield [3,1,2] but it cannot yield [3,2,1] since the 2 is in the same position.
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Given a permutation of n integers, please help Patrick find the minimum number of special exchanges needed to make the permutation sorted! It can be proved that under given constraints this number doesn't exceed 10^{18}.
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An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
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Input
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Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
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The first line of each test case contains integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation.
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The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ n) — the initial permutation.
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It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
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Output
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For each test case, output one integer: the minimum number of special exchanges needed to sort the permutation.
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Example
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Input
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2
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5
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1 2 3 4 5
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7
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3 2 4 5 1 6 7
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Output
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0
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2
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Note
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In the first permutation, it is already sorted so no exchanges are needed.
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It can be shown that you need at least 2 exchanges to sort the second permutation.
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[3, 2, 4, 5, 1, 6, 7]
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Perform special exchange on range (1, 5)
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[4, 1, 2, 3, 5, 6, 7]
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Perform special exchange on range (1, 4)
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[1, 2, 3, 4, 5, 6, 7]
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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 __future__ import division, print_function
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import os
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import sys
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from io import BytesIO, IOBase
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if sys.version_info[0] < 3:
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from __builtin__ import xrange as range
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from future_builtins import ascii, filter, hex, map, oct, zip
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# ------------------------------ TEMPLATE ABOVE THIS --------------------------------
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def main():
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t=int(input())
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for _ in range(t):
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n=int(input())
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A=list(map(int,input().split()))
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B=A[:]
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B.sort()
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cnt=0
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i=0
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while i<n:
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while i<n and A[i]==B[i]: i+=1
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flag=0
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while i<n and A[i]!=B[i]:
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i+=1
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flag=1
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if flag: cnt+=1
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if cnt<=1: print(cnt)
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else: print(2)
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return
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# ------------------------------ TEMPLATE BELOW THIS --------------------------------
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# region fastio
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BUFSIZE = 8192
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class FastIO(IOBase):
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newlines = 0
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def __init__(self, file):
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self._fd = file.fileno()
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self.buffer = BytesIO()
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self.writable = \"x\" in file.mode or \"r\" not in file.mode
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self.write = self.buffer.write if self.writable else None
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def read(self):
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while True:
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b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
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if not b:
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break
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ptr = self.buffer.tell()
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self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
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self.newlines = 0
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return self.buffer.read()
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def readline(self):
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while self.newlines == 0:
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b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
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self.newlines = b.count(b\"\n\") + (not b)
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ptr = self.buffer.tell()
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self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
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self.newlines -= 1
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return self.buffer.readline()
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def flush(self):
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if self.writable:
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os.write(self._fd, self.buffer.getvalue())
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self.buffer.truncate(0), self.buffer.seek(0)
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class IOWrapper(IOBase):
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def __init__(self, file):
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self.buffer = FastIO(file)
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self.flush = self.buffer.flush
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self.writable = self.buffer.writable
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self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))
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self.read = lambda: self.buffer.read().decode(\"ascii\")
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self.readline = lambda: self.buffer.readline().decode(\"ascii\")
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def print(*args, **kwargs):
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\"\"\"Prints the values to a stream, or to sys.stdout by default.\"\"\"
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sep, file = kwargs.pop(\"sep\", \" \"), kwargs.pop(\"file\", sys.stdout)
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at_start = True
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for x in args:
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if not at_start:
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file.write(sep)
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file.write(str(x))
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at_start = False
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file.write(kwargs.pop(\"end\", \"\n\"))
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if kwargs.pop(\"flush\", False):
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file.flush()
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if sys.version_info[0] < 3:
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sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
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else:
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sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
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def input(): return sys.stdin.readline().rstrip(\"\r\n\")
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# endregion
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if __name__ == \"__main__\":
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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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class Comkarandbaseballbootcamp(Basebootcamp):
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def __init__(self, min_n=5, max_n=10, answer_type=None):
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self.min_n = max(1, min_n)
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self.max_n = max(self.min_n, max_n)
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self.answer_type = answer_type # 先初始化属性
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if self.answer_type is not None and self.answer_type not in [0, 1, 2]:
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raise ValueError("answer_type must be 0, 1, 2, or None")
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def case_generator(self):
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# 确保answer_type有效性
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answer_type = self.answer_type if self.answer_type is not None else random.choice([0, 1, 2])
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# 处理最小尺寸约束
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min_n = self.min_n
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max_n = self.max_n
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if answer_type in {1, 2}:
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min_n = max(2, min_n)
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max_n = max(min_n, max_n)
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# 生成合法n值
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n = random.randint(min_n, max_n) if min_n <= max_n else min_n
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# 三种答案类型生成策略
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if answer_type == 0:
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arr = list(range(1, n+1))
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answer = 0
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elif answer_type == 1:
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# 生成单次交换的排列
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while True:
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arr = list(range(1, n+1))
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# 随机选择可交换区间
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start = random.randint(0, n-2)
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end = random.randint(start+1, n-1)
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sub = arr[start:end+1]
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# 生成错位排列（循环右移）
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derangement = sub[1:] + sub[:1]
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arr[start:end+1] = derangement
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if self._compute_answer(arr) == 1:
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answer = 1
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break
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else:
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# 生成需要两次交换的排列
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while True:
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arr = list(range(1, n+1))
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# 生成第一个错位区间
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start1 = random.randint(0, n-3)
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end1 = random.randint(start1+1, n-2)
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sub1 = arr[start1:end1+1]
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arr[start1:end1+1] = sub1[1:] + sub1[:1]
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# 生成第二个错位区间
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start2 = random.randint(end1+1, n-1)
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end2 = random.randint(start2+1, n-1) if start2 < n-1 else start2
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sub2 = arr[start2:end2+1]
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if len(sub2) >= 2:
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arr[start2:end2+1] = sub2[1:] + sub2[:1]
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if self._compute_answer(arr) == 2:
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answer = 2
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break
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return {
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'n': n,
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'arr': arr,
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'answer': answer
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}
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@staticmethod
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def _compute_answer(a):
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"""根据官方参考代码实现的答案验证逻辑"""
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B = sorted(a)
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cnt = 0
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i, n = 0, len(a)
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while i < n:
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while i < n and a[i] == B[i]:
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i += 1
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flag = 0
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while i < n and a[i] != B[i]:
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i += 1
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flag = 1
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if flag:
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cnt += 1
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return cnt if cnt <= 1 else 2
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@staticmethod
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def prompt_func(question_case):
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return f"""Patrick needs to sort his baseball score records using special exchanges. Given the permutation:
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n = {question_case['n']}
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{question_case['arr']}
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Calculate the minimum number of special exchanges required. Put your final answer within [answer] and [/answer] tags. For example: [answer]2[/answer]"""
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@staticmethod
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def extract_output(output):
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# 逆向搜索最后一个答案标签对
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end_pos = output.rfind('[/answer]')
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if end_pos == -1:
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return None
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start_pos = output.rfind('[answer]', 0, end_pos)
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if start_pos == -1:
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return None
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return output[start_pos+8:end_pos].strip()
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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return int(solution) == identity['answer']
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except (ValueError, KeyError):
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return False
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