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internbootcamp/bootcamp/cmemoryanddeevolution/cmemoryanddeevolution.py

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+"""# 
+
+### 谜题描述
+Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length x, and he wishes to perform operations to obtain an equilateral triangle of side length y.
+
+In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
+
+What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y?
+
+Input
+
+The first and only line contains two integers x and y (3 ≤ y < x ≤ 100 000) — the starting and ending equilateral triangle side lengths respectively.
+
+Output
+
+Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y if he starts with the equilateral triangle of side length x.
+
+Examples
+
+Input
+
+6 3
+
+
+Output
+
+4
+
+
+Input
+
+8 5
+
+
+Output
+
+3
+
+
+Input
+
+22 4
+
+
+Output
+
+6
+
+Note
+
+In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do <image>.
+
+In the second sample test, Memory can do <image>.
+
+In the third sample test, Memory can do: <image>
+
+<image>.
+
+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
+
+y, x = map(int, raw_input().split())
+current = [ x, x, x ]
+goal = [ y, y, y ]
+steps = 0
+while current != goal:
+    best_index, best_value = -1, -1
+    for i, a in enumerate(current):
+        b, c = current[(i + 1) % 3], current[(i + 2) % 3]
+        x = min(y, b + c - 1)
+        if x > a and x > best_value:
+            best_index, best_value = i, x
+    current[best_index] = best_value
+    steps += 1
+print(steps)
+```
+
+
+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
+import re
+from bootcamp import Basebootcamp
+
+class Cmemoryanddeevolutionbootcamp(Basebootcamp):
+    def __init__(self, min_y=3, max_x=100000):
+        self.min_y = min_y
+        self.max_x = max_x
+
+    def case_generator(self):
+        y = random.randint(self.min_y, self.max_x - 1)
+        x = random.randint(y + 1, self.max_x)
+        return {'x': x, 'y': y}
+
+    @staticmethod
+    def prompt_func(question_case):
+        x = question_case['x']
+        y = question_case['y']
+        prompt = f"""Memory starts with an equilateral triangle with side length {x} and wants to transform it into one with side length {y}. Each second, he can modify one side's length to any positive integer, provided the new triangle remains non-degenerate (sum of any two sides exceeds the third). All side lengths must be integers. What is the minimum number of seconds required?
+
+Output the answer as an integer enclosed within [answer] and [/answer]. Example: [answer]4[/answer]"""
+        return prompt
+
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        last_match = matches[-1].strip()
+        try:
+            return int(last_match)
+        except ValueError:
+            return None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        # 交换x和y以适配参考代码逻辑
+        original_x = identity['x']
+        original_y = identity['y']
+        x = original_y  # 参考代码中的输入xy是颠倒的
+        y = original_x
+        
+        current = [x, x, x]
+        goal = [y, y, y]
+        steps = 0
+        
+        while current != goal:
+            best_index = -1
+            best_value = -1
+            for i in range(3):
+                a = current[i]
+                b = current[(i+1) % 3]
+                c = current[(i+2) % 3]
+                new_val = min(y, b + c - 1)
+                if new_val > a and new_val > best_value:
+                    best_index = i
+                    best_value = new_val
+            
+            if best_index == -1:  # 无解
+                return False
+            
+            current[best_index] = best_value
+            steps += 1
+            
+            # 安全阀防止无限循环
+            if steps > 1000:
+                return False
+        
+        return solution == steps