mirror of
https://github.com/InternLM/InternBootcamp.git
synced 2026-04-22 16:49:04 +00:00
166 lines
4.5 KiB
Python
Executable file
166 lines
4.5 KiB
Python
Executable file
"""#
|
||
|
||
### 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
|