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165 lines
4.7 KiB
Python
Executable file
165 lines
4.7 KiB
Python
Executable file
"""#
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### 谜题描述
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There is a new attraction in Singapore Zoo: The Infinite Zoo.
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The Infinite Zoo can be represented by a graph with an infinite number of vertices labeled 1,2,3,…. There is a directed edge from vertex u to vertex u+v if and only if u\&v=v, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). There are no other edges in the graph.
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Zookeeper has q queries. In the i-th query she will ask you if she can travel from vertex u_i to vertex v_i by going through directed edges.
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Input
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The first line contains an integer q (1 ≤ q ≤ 10^5) — the number of queries.
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The i-th of the next q lines will contain two integers u_i, v_i (1 ≤ u_i, v_i < 2^{30}) — a query made by Zookeeper.
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Output
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For the i-th of the q queries, output \"YES\" in a single line if Zookeeper can travel from vertex u_i to vertex v_i. Otherwise, output \"NO\".
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You can print your answer in any case. For example, if the answer is \"YES\", then the output \"Yes\" or \"yeS\" will also be considered as correct answer.
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Example
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Input
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5
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1 4
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3 6
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1 6
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6 2
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5 5
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Output
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YES
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YES
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NO
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NO
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YES
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Note
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The subgraph on vertices 1,2,3,4,5,6 is shown below.
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<image>
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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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for _ in range(input()):
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u,v = map(int,raw_input().split())
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if u>v:
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print \"NO\"
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continue
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x,y,f = 0,0,1
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for i in range(31):
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if u%2:
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x+=1
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if v%2:
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y+=1
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if y>x:
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f = 0
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break
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u/=2
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v/=2
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if f:
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print \"YES\"
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else:
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print \"NO\"
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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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class Dzookeeperandtheinfinitezoobootcamp(Basebootcamp):
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def __init__(self, u_min=1, u_max=(1 << 30)-1, v_min=1, v_max=(1 << 30)-1):
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self.u_min = u_min
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self.u_max = u_max
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self.v_min = v_min
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self.v_max = v_max
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def case_generator(self):
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# 生成YES案例（可达）的概率调整为40%
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if random.random() < 0.4:
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for _ in range(100): # 总尝试次数限制
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u = random.randint(self.u_min, self.u_max)
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mask_max = min(u, self.v_max - u)
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if mask_max < 0:
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continue
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for _ in range(100): # 单u尝试次数
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mask = random.randint(0, mask_max)
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v_prime = u & mask
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v = u + v_prime
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if self.v_min <= v <= self.v_max:
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return {'u': u, 'v': v}
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# 若无法生成有效YES案例，退回生成NO案例
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return self._generate_no_case()
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else:
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return self._generate_no_case()
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def _generate_no_case(self):
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""" 专门生成NO案例的方法 """
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# 首先生成u>v的情况（40%概率）
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if random.random() < 0.4:
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v = random.randint(self.v_min, self.u_max-1)
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u = random.randint(v+1, self.u_max)
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return {'u': u, 'v': v}
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# 生成u<=v但不可达的情况（最多尝试200次）
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for _ in range(200):
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u = random.randint(self.u_min, self.u_max)
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v = random.randint(u, self.v_max)
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if not self.is_reachable(u, v):
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return {'u': u, 'v': v}
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# 最终保障机制：生成u>v的简单案例
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v = random.randint(self.v_min, self.u_max-1)
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u = random.randint(v+1, self.u_max)
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return {'u': u, 'v': v}
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@staticmethod
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def prompt_func(question_case):
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u = question_case['u']
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v = question_case['v']
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return f"""Determine if path exists from {u} to {v} in Infinite Zoo.
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Rules:
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1. Edge u→(u+v') exists iff u & v' = v'
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2. Path follows edge directions
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Answer format: [answer]YES[/answer] or [answer]NO[/answer]"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.I)
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return matches[-1].strip().upper() if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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if solution not in {'YES', 'NO'}:
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return False
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return solution == ('YES' if cls.is_reachable(**identity) else 'NO')
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@staticmethod
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def is_reachable(u, v):
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if u > v:
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return False
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x = y = 0
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for _ in range(31):
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x += u & 1
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y += v & 1
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if y > x:
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return False
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u >>= 1
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v >>= 1
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return True
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