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internbootcamp/bootcamp/cvladikandfractions/cvladikandfractions.py

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+"""# 
+
+### 谜题描述
+Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.
+
+Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
+
+If there is no such answer, print -1.
+
+Input
+
+The single line contains single integer n (1 ≤ n ≤ 104).
+
+Output
+
+If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
+
+If there are multiple answers, print any of them.
+
+Examples
+
+Input
+
+3
+
+
+Output
+
+2 7 42
+
+
+Input
+
+7
+
+
+Output
+
+7 8 56
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+n = int(raw_input())
+z = 0
+if(n==1):
+    print -1
+else:
+    for x in xrange(2,1000000001):
+        for y in xrange(x+1,1000000001):
+            if((1/float(x)+1/float(y))>=2/float(n)):
+                break
+            z = float(n*x*y)/float(2*x*y-n*y-n*x)
+            if z==float(int(z)) and z!=x and z!=y:
+                print x, y, int(z)
+                break
+            else:
+                if(z<y):
+                    z = 0
+                    break
+                z = 0
+        if(z>0):
+            break
+```
+
+
+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 Cvladikandfractionsbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        super().__init__(**params)  # 显式调用基类初始化
+        self.max_n = params.get('max_n', 10000)  # 默认上限为10^4
+    
+    def case_generator(self):
+        # 生成n的范围为1到max_n，包含所有可能的问题实例
+        # 当n=1时答案应为-1，其他n≥2根据题目逻辑应有解
+        n = random.randint(1, self.max_n)
+        return {"n": n}
+    
+    @staticmethod
+    def prompt_func(question_case):
+        n = question_case["n"]
+        problem_text = (
+            f"Vladik and Chloe are determining who is better at math. Vladik claims that for any positive integer n, "
+            f"the fraction 2/n can be expressed as the sum of three distinct positive unit fractions. Help Vladik prove this "
+            f"by finding three distinct positive integers x, y, z such that 1/x + 1/y + 1/z = 2/{n}. The numbers x, y, z must "
+            f"not exceed 1e9. If it's impossible, output -1.\n\n"
+            "Provide your answer as three space-separated integers or -1 enclosed within [answer] and [/answer] tags. "
+            "Example: [answer]2 7 42[/answer] or [answer]-1[/answer] if no solution exists."
+        )
+        return problem_text
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        last_answer = matches[-1].strip()
+        if last_answer == '-1':
+            return -1
+        parts = last_answer.split()
+        if len(parts) != 3:
+            return None
+        try:
+            x, y, z = map(int, parts)
+            return (x, y, z)
+        except ValueError:
+            return None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        n = identity["n"]
+        if n == 1:
+            return solution == -1  # n=1时只有-1正确
+        else:
+            # 其他n必须返回有效三元组
+            if solution == -1:
+                return False  # n≥2时不接受-1
+            if not isinstance(solution, tuple) or len(solution) != 3:
+                return False
+            x, y, z = solution
+            if x <= 0 or y <= 0 or z <= 0:
+                return False
+            if x > 1e9 or y > 1e9 or z > 1e9:
+                return False
+            if x == y or x == z or y == z:
+                return False
+            # 关键数学验证：n(xy + yz + zx) == 2xyz
+            left = n * (x * y + y * z + z * x)
+            right = 2 * x * y * z
+            return left == right