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internbootcamp/bootcamp/cmaximandmatrix/cmaximandmatrix.py

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+"""# 
+
+### 谜题描述
+Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):
+
+<image>
+
+Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.
+
+Expression (x xor y) means applying the operation of bitwise excluding \"OR\" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character \"^\", in Pascal — by \"xor\".
+
+Input
+
+A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).
+
+Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
+
+Output
+
+In a single line print a single integer — the answer to the problem. 
+
+Examples
+
+Input
+
+1 1
+
+
+Output
+
+1
+
+
+Input
+
+3 2
+
+
+Output
+
+1
+
+
+Input
+
+3 3
+
+
+Output
+
+0
+
+
+Input
+
+1000000000000 1048576
+
+
+Output
+
+118606527258
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+c = []
+
+def f(n, t):
+	x = 0
+	while 2 ** x <= n:
+		x += 1
+	x -= 1
+	ans = c[x][t + 1] if t + 1 <= x else 0
+	if n > 2 ** x and t > 0:
+		ans += f(n - 2 ** x, t - 1)
+		if t == 1:
+			ans += 1
+	return ans
+
+c.append([1])
+for i in range(1, 60):
+	q = [1]
+	for j in range(1, i):
+		q.append(c[-1][j] + c[-1][j - 1])
+	q.append(1)
+	c.append(q)
+
+n, t = map(int, raw_input().split())
+n += 1
+tt = 0
+while t % 2 == 0:
+	t /= 2
+	tt += 1
+if t > 1:
+	print 0
+else:
+	print f(n, tt)
+```
+
+
+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 re
+import random  # 添加缺失的random模块导入
+from bootcamp import Basebootcamp
+
+class Cmaximandmatrixbootcamp(Basebootcamp):
+    _c = None
+
+    def __init__(self, **params):
+        super().__init__(**params)
+
+    @classmethod
+    def _build_c(cls):
+        if cls._c is not None:
+            return
+        c = []
+        c.append([1])
+        for i in range(1, 60):
+            q = [1]
+            for j in range(1, i):
+                q.append(c[i-1][j] + c[i-1][j-1])
+            q.append(1)
+            c.append(q)
+        cls._c = c
+
+    @classmethod
+    def f(cls, n, t):
+        cls._build_c()
+        if n == 0 and t == 0:
+            return 1
+        if t < 0:
+            return 0
+        x = 0
+        while (2 ** x) <= n:
+            x += 1
+        x -= 1
+        max_k = x
+        if (t + 1) <= max_k:
+            ans = cls._c[x][t + 1]
+        else:
+            ans = 0
+        remaining = n - (2 ** x)
+        if remaining > 0 and t > 0:
+            ans += cls.f(remaining, t - 1)
+            if t == 1:
+                ans += 1
+        return ans
+
+    def case_generator(self):
+        max_n = 10**12
+        # 简化生成逻辑确保数值在合法范围
+        n = random.randint(1, max_n)
+        t_max = min(n + 1, 10**12)
+        t = random.randint(1, t_max)
+        return {"n": n, "t": t}
+
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        n = question_case["n"]
+        t = question_case["t"]
+        prompt = (
+            "Maxim loves to fill a matrix in a special way. Given two integers n and t, your task is to count how many integers m (1 ≤ m ≤ n) satisfy the following condition: The sum of values in the (m+1)th row (1-based) of the matrix equals t.\n\n"
+            "Matrix Construction Rules:\n"
+            "- The matrix is of size (m+1) × (m+1), where rows and columns are 0-based.\n"
+            "- Each cell at row i and column j contains the value (i XOR j).\n\n"
+            "Input Constraints:\n"
+            f"- 1 ≤ n, t ≤ 10^12\n"
+            f"- t ≤ n + 1\n\n"
+            "Your task is to compute the number of valid m values. Provide your answer as an integer enclosed within [answer] and [/answer] tags.\n\n"
+            "Example:\n"
+            "Input: 1 1\n"
+            "Output: [answer]1[/answer]\n\n"
+            f"Now, solve for n = {n} and t = {t}. Place your final answer within [answer] tags."
+        )
+        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):
+        n_original = identity["n"]
+        t_original = identity["t"]
+        n_code = n_original + 1
+        t = t_original
+        tt = 0
+        
+        # 处理t必须为2的幂次的条件
+        while t % 2 == 0 and t != 0:
+            t = t // 2
+            tt += 1
+        if t != 1:
+            correct_answer = 0
+        else:
+            correct_answer = cls.f(n_code, tt)
+        
+        try:
+            user_answer = int(solution)
+            return user_answer == correct_answer
+        except:
+            return False