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internbootcamp/bootcamp/dzookeeperandtheinfinitezoo/dzookeeperandtheinfinitezoo.py

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+"""# 
+
+### 谜题描述
+There is a new attraction in Singapore Zoo: The Infinite Zoo.
+
+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.
+
+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.
+
+Input
+
+The first line contains an integer q (1 ≤ q ≤ 10^5) — the number of queries.
+
+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.
+
+Output
+
+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\".
+
+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.
+
+Example
+
+Input
+
+
+5
+1 4
+3 6
+1 6
+6 2
+5 5
+
+
+Output
+
+
+YES
+YES
+NO
+NO
+YES
+
+Note
+
+The subgraph on vertices 1,2,3,4,5,6 is shown below.
+
+<image>
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+for _ in range(input()):
+    u,v = map(int,raw_input().split())
+    
+    if u>v:
+        print \"NO\"
+        continue
+    
+    x,y,f = 0,0,1
+    
+    for i in range(31):
+        if u%2:
+            x+=1
+        if v%2:
+            y+=1
+        if y>x:
+            f = 0
+            break
+        u/=2
+        v/=2
+        
+    if f:
+        print \"YES\"
+    else:
+        print \"NO\"
+```
+
+
+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
+
+class Dzookeeperandtheinfinitezoobootcamp(Basebootcamp):
+    def __init__(self, u_min=1, u_max=(1 << 30)-1, v_min=1, v_max=(1 << 30)-1):
+        self.u_min = u_min
+        self.u_max = u_max
+        self.v_min = v_min
+        self.v_max = v_max
+
+    def case_generator(self):
+        # 生成YES案例（可达）的概率调整为40%
+        if random.random() < 0.4:
+            for _ in range(100):  # 总尝试次数限制
+                u = random.randint(self.u_min, self.u_max)
+                mask_max = min(u, self.v_max - u)
+                if mask_max < 0:
+                    continue
+                for _ in range(100):  # 单u尝试次数
+                    mask = random.randint(0, mask_max)
+                    v_prime = u & mask
+                    v = u + v_prime
+                    if self.v_min <= v <= self.v_max:
+                        return {'u': u, 'v': v}
+            # 若无法生成有效YES案例，退回生成NO案例
+            return self._generate_no_case()
+        else:
+            return self._generate_no_case()
+    
+    def _generate_no_case(self):
+        """ 专门生成NO案例的方法 """
+        # 首先生成u>v的情况（40%概率）
+        if random.random() < 0.4:
+            v = random.randint(self.v_min, self.u_max-1)
+            u = random.randint(v+1, self.u_max)
+            return {'u': u, 'v': v}
+        # 生成u<=v但不可达的情况（最多尝试200次）
+        for _ in range(200):
+            u = random.randint(self.u_min, self.u_max)
+            v = random.randint(u, self.v_max)
+            if not self.is_reachable(u, v):
+                return {'u': u, 'v': v}
+        # 最终保障机制：生成u>v的简单案例
+        v = random.randint(self.v_min, self.u_max-1)
+        u = random.randint(v+1, self.u_max)
+        return {'u': u, 'v': v}
+
+    @staticmethod
+    def prompt_func(question_case):
+        u = question_case['u']
+        v = question_case['v']
+        return f"""Determine if path exists from {u} to {v} in Infinite Zoo.
+Rules:
+1. Edge u→(u+v') exists iff u & v' = v'
+2. Path follows edge directions
+
+Answer format: [answer]YES[/answer] or [answer]NO[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.I)
+        return matches[-1].strip().upper() if matches else None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        if solution not in {'YES', 'NO'}:
+            return False
+        return solution == ('YES' if cls.is_reachable(**identity) else 'NO')
+
+    @staticmethod
+    def is_reachable(u, v):
+        if u > v:
+            return False
+        x = y = 0
+        for _ in range(31):
+            x += u & 1
+            y += v & 1
+            if y > x:
+                return False
+            u >>= 1
+            v >>= 1
+        return True