init-commit

internbootcamp/bootcamp/bshifting/bshifting.py

@@ -0,0 +1,166 @@
+"""# 
+
+### 谜题描述
+John Doe has found the beautiful permutation formula.
+
+Let's take permutation p = p1, p2, ..., pn. Let's define transformation f of this permutation: 
+
+<image>
+
+where k (k > 1) is an integer, the transformation parameter, r is such maximum integer that rk ≤ n. If rk = n, then elements prk + 1, prk + 2 and so on are omitted. In other words, the described transformation of permutation p cyclically shifts to the left each consecutive block of length k and the last block with the length equal to the remainder after dividing n by k. 
+
+John Doe thinks that permutation f(f( ... f(p = [1, 2, ..., n], 2) ... , n - 1), n) is beautiful. Unfortunately, he cannot quickly find the beautiful permutation he's interested in. That's why he asked you to help him.
+
+Your task is to find a beautiful permutation for the given n. For clarifications, see the notes to the third sample.
+
+Input
+
+A single line contains integer n (2 ≤ n ≤ 106).
+
+Output
+
+Print n distinct space-separated integers from 1 to n — a beautiful permutation of size n.
+
+Examples
+
+Input
+
+2
+
+
+Output
+
+2 1 
+
+
+Input
+
+3
+
+
+Output
+
+1 3 2 
+
+
+Input
+
+4
+
+
+Output
+
+4 2 3 1 
+
+Note
+
+A note to the third test sample: 
+
+  * f([1, 2, 3, 4], 2) = [2, 1, 4, 3]
+  * f([2, 1, 4, 3], 3) = [1, 4, 2, 3]
+  * f([1, 4, 2, 3], 4) = [4, 2, 3, 1]
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+#include <bits/stdc++.h>
+using namespace std;
+const int di[] = {-1, 0, 1, 0};
+const int dj[] = {0, 1, 0, -1};
+const long long MOD = 1e9 + 7;
+const long long INF = 1e9;
+const double EPS = 1e-5;
+int main() {
+  ios::sync_with_stdio(false), cin.tie(0);
+  int n;
+  cin >> n;
+  deque<int> num(n);
+  for (int i = 0; i < n; i++) num[i] = i + 1;
+  for (int shft = 2; shft <= n; shft++) {
+    int last = num[0];
+    for (int i = shft; i < n; i += shft) swap(num[i], last);
+    num.pop_front();
+    num.push_back(last);
+  }
+  for (int i = 0; i < n; i++) cout << num[i] << \" \n\"[i == n - 1];
+  cin.ignore(), cin.get();
+}
+```
+
+
+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
+from collections import deque
+from bootcamp import Basebootcamp
+
+class Bshiftingbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        """
+        参数有效性检查：确保2 ≤ min_n ≤ max_n ≤ 1e6
+        """
+        self.min_n = max(2, params.get('min_n', 2))
+        self.max_n = min(10**6, params.get('max_n', 100))
+        super().__init__(**params)
+
+    def case_generator(self):
+        """
+        严格保证生成的n在[2, 1e6]范围内
+        """
+        n = random.randint(self.min_n, self.max_n)
+        return {'n': n}
+
+    @staticmethod
+    def prompt_func(question_case):
+        n = question_case['n']
+        return f"""Find the beautiful permutation for n={n} by applying sequential transformations for k=2 to n.
+Each transformation divides the permutation into k-length blocks and left-rotates each block.
+Output format: space-separated numbers within [answer] tags.
+
+Example for n=4:
+[answer]
+4 2 3 1
+[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        # 增强提取逻辑：允许任意分隔符和空格
+        matches = re.findall(r'\[answer](.*?)\[/answer]', output, re.DOTALL | re.IGNORECASE)
+        if not matches:
+            return None
+            
+        # 提取最后一个answer块并规范化格式
+        raw_answer = matches[-1].strip()
+        processed = re.sub(r'[^\d]', ' ', raw_answer)  # 非数字转空格
+        processed = re.sub(r'\s+', ' ', processed).strip()  # 合并空格
+        return processed
+
+    @classmethod
+    def _generate_correct_answer(cls, n):
+        """严格遵循C++参考实现逻辑"""
+        num = deque(range(1, n+1))
+        for k in range(2, n+1):
+            if len(num) == 0:
+                break
+            last = num[0]
+            i = k
+            while i < n:  # 注意这里是原始n值，不是当前deque长度
+                if i < len(num):
+                    num[i], last = last, num[i]
+                i += k
+            num.popleft()
+            num.append(last)
+        return list(num)
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        n = identity['n']
+        try:
+            # 严格验证数字集合和顺序
+            user_ans = list(map(int, solution.split()))
+            correct = cls._generate_correct_answer(n)
+            return user_ans == correct
+        except:
+            return False