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internbootcamp/bootcamp/cmagicformulas/cmagicformulas.py

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+"""# 
+
+### 谜题描述
+People in the Tomskaya region like magic formulas very much. You can see some of them below.
+
+Imagine you are given a sequence of positive integer numbers p1, p2, ..., pn. Lets write down some magic formulas:
+
+<image><image>
+
+Here, \"mod\" means the operation of taking the residue after dividing.
+
+The expression <image> means applying the bitwise xor (excluding \"OR\") operation to integers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by \"^\", in Pascal — by \"xor\".
+
+People in the Tomskaya region like magic formulas very much, but they don't like to calculate them! Therefore you are given the sequence p, calculate the value of Q.
+
+Input
+
+The first line of the input contains the only integer n (1 ≤ n ≤ 106). The next line contains n integers: p1, p2, ..., pn (0 ≤ pi ≤ 2·109).
+
+Output
+
+The only line of output should contain a single integer — the value of Q.
+
+Examples
+
+Input
+
+3
+1 2 3
+
+
+Output
+
+3
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+'''
+Created on Apr 25, 2014
+
+@author: szalivako
+'''
+
+def sumX(n):
+    if (n % 4 == 0):
+        return n
+    if (n % 4 == 1):
+        return 1
+    if (n % 4 == 2):
+        return n + 1
+    if (n % 4 == 3):
+        return 0
+    return 0
+
+n = int(raw_input())
+p = map(int, raw_input().split())
+
+Q = 0
+for i in range(n):
+    Q ^= p[i]
+if ((n - 1) % 2 == 1):
+    Q ^= 1
+
+for i in range(2, n + 1):
+    z = (n - i + 1) // i
+    if (z % 2 == 1):
+        Q ^= sumX(i - 1)
+    r = (n - i + 1) % i
+    if (r > 1):
+        Q ^= sumX(r - 1)
+    if ((n - i) % 2 == 1):
+        Q ^= i
+
+print Q
+```
+
+
+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 Cmagicformulasbootcamp(Basebootcamp):
+    def __init__(self, min_n=1, max_n=10, max_p=100):
+        self.min_n = min_n
+        self.max_n = max_n
+        self.max_p = max_p
+    
+    def case_generator(self):
+        n = random.randint(self.min_n, self.max_n)
+        p = [random.randint(0, self.max_p) for _ in range(n)]
+        return {'n': n, 'p': p}
+    
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        n = question_case['n']
+        p = question_case['p']
+        problem = (
+            "你被给定一个整数序列，需要计算魔法公式Q的值。规则如下：\n\n"
+            "1. 初始Q为所有数的异或结果。\n"
+            "2. 若n-1为奇数，Q异或1。\n"
+            "3. 对每个i从2到n：\n"
+            "   a) 计算z=(n-i+1)//i，若z为奇数，Q异或sumX(i-1)。\n"
+            "   b) 计算余数r=(n-i+1)%i，若r>1，Q异或sumX(r-1)。\n"
+            "   c) 若(n-i)为奇数，Q异或i。\n\n"
+            "sumX(k)的定义：\n"
+            "- k%4==0: sumX(k)=k\n"
+            "- k%4==1: sumX(k)=1\n"
+            "- k%4==2: sumX(k)=k+1\n"
+            "- k%4==3: sumX(k)=0\n\n"
+            "输入参数：\n"
+            f"n = {n}\n"
+            f"p = {p}\n\n"
+            "请计算Q的值，并将答案放在[answer]和[/answer]之间。"
+        )
+        return problem
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        last_answer = matches[-1].strip()
+        try:
+            return int(last_answer)
+        except ValueError:
+            return None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        n = identity['n']
+        p = identity['p']
+        
+        Q = 0
+        for num in p:
+            Q ^= num
+        
+        if (n - 1) % 2 == 1:
+            Q ^= 1
+        
+        for i in range(2, n + 1):
+            z = (n - i + 1) // i
+            if z % 2 == 1:
+                Q ^= cls.sumX(i - 1)
+            
+            r = (n - i + 1) % i
+            if r > 1:
+                Q ^= cls.sumX(r - 1)
+            
+            if (n - i) % 2 == 1:
+                Q ^= i
+        
+        return Q == solution
+    
+    @staticmethod
+    def sumX(k):
+        mod = k % 4
+        if mod == 0:
+            return k
+        elif mod == 1:
+            return 1
+        elif mod == 2:
+            return k + 1
+        else:  # mod == 3
+            return 0