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internbootcamp/bootcamp/dtopsecrettask/dtopsecrettask.py

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+"""# 
+
+### 谜题描述
+A top-secret military base under the command of Colonel Zuev is expecting an inspection from the Ministry of Defence. According to the charter, each top-secret military base must include a top-secret troop that should... well, we cannot tell you exactly what it should do, it is a top secret troop at the end. The problem is that Zuev's base is missing this top-secret troop for some reasons.
+
+The colonel decided to deal with the problem immediately and ordered to line up in a single line all n soldiers of the base entrusted to him. Zuev knows that the loquacity of the i-th soldier from the left is equal to qi. Zuev wants to form the top-secret troop using k leftmost soldiers in the line, thus he wants their total loquacity to be as small as possible (as the troop should remain top-secret). To achieve this, he is going to choose a pair of consecutive soldiers and swap them. He intends to do so no more than s times. Note that any soldier can be a participant of such swaps for any number of times. The problem turned out to be unusual, and colonel Zuev asked you to help.
+
+Determine, what is the minimum total loquacity of the first k soldiers in the line, that can be achieved by performing no more than s swaps of two consecutive soldiers.
+
+Input
+
+The first line of the input contains three positive integers n, k, s (1 ≤ k ≤ n ≤ 150, 1 ≤ s ≤ 109) — the number of soldiers in the line, the size of the top-secret troop to be formed and the maximum possible number of swap operations of the consecutive pair of soldiers, respectively.
+
+The second line of the input contains n integer qi (1 ≤ qi ≤ 1 000 000) — the values of loquacity of soldiers in order they follow in line from left to right.
+
+Output
+
+Print a single integer — the minimum possible total loquacity of the top-secret troop.
+
+Examples
+
+Input
+
+3 2 2
+2 4 1
+
+
+Output
+
+3
+
+
+Input
+
+5 4 2
+10 1 6 2 5
+
+
+Output
+
+18
+
+
+Input
+
+5 2 3
+3 1 4 2 5
+
+
+Output
+
+3
+
+Note
+
+In the first sample Colonel has to swap second and third soldiers, he doesn't really need the remaining swap. The resulting soldiers order is: (2, 1, 4). Minimum possible summary loquacity of the secret troop is 3. In the second sample Colonel will perform swaps in the following order:
+
+  1. (10, 1, 6 — 2, 5) 
+  2. (10, 1, 2, 6 — 5) 
+
+
+
+The resulting soldiers order is (10, 1, 2, 5, 6). 
+
+Minimum possible summary loquacity is equal to 18.
+
+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;
+int n, k, s, ar[1 << 20];
+int dp[3][200][200 * 200];
+int main() {
+  ios_base::sync_with_stdio(0);
+  cin >> n >> k >> s;
+  for (int i = 1; i <= n; i++) cin >> ar[i];
+  if (s > n * n / 2 + 10) s = n * n / 2 + 10;
+  for (int done = 0; done <= s; done++)
+    for (int pref = 0; pref <= k; pref++) dp[0][pref][done] = 1e9;
+  dp[0][0][0] = 0;
+  for (int i = 1; i <= n; i++) {
+    for (int j = 0; j <= s; j++)
+      for (int q = 0; q <= k; q++) dp[i % 2][q][j] = 1e9;
+    for (int pref = 0; pref < i && pref <= k; pref++)
+      for (int done = 0; done <= s; done++) {
+        dp[i % 2][pref][done] =
+            min(dp[i % 2][pref][done], dp[1 - i % 2][pref][done]);
+        int ndone = done + i - pref - 1;
+        if (ndone <= s)
+          dp[i % 2][pref + 1][ndone] = min(dp[i % 2][pref + 1][ndone],
+                                           dp[1 - i % 2][pref][done] + ar[i]);
+      }
+  }
+  int ans = 1e9;
+  for (int i = 0; i <= s; i++) ans = min(ans, dp[n % 2][k][i]);
+  cout << ans << endl;
+  return 0;
+}
+```
+
+
+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
+from bootcamp import Basebootcamp
+
+def calculate_min_loquacity(n, k, s, q):
+    adjusted_s = min(s, (n*n)//2 + 10)  # 严格模拟参考代码的调整逻辑
+    INF = float('inf')
+    
+    # 初始化DP数组，使用滚动数组优化
+    dp = [[[INF] * (adjusted_s + 1) for _ in range(k+1)] for __ in range(2)]
+    dp[0][0][0] = 0  # 初始状态
+
+    for i in range(1, n+1):
+        current = i % 2
+        prev = 1 - current
+        
+        # 重置当前层
+        for j in range(k+1):
+            for t in range(adjusted_s + 1):
+                dp[current][j][t] = INF
+        
+        # 状态转移
+        for pref in range(0, min(i-1, k)+1):
+            for done in range(adjusted_s + 1):
+                if dp[prev][pref][done] == INF:
+                    continue
+                
+                # 情况1：不选当前士兵
+                if dp[current][pref][done] > dp[prev][pref][done]:
+                    dp[current][pref][done] = dp[prev][pref][done]
+                
+                # 情况2：选当前士兵
+                new_pref = pref + 1
+                if new_pref > k:
+                    continue
+                
+                swaps_needed = i - new_pref  # 与参考代码完全一致的计算方式
+                new_done = done + swaps_needed
+                
+                if new_done <= adjusted_s:
+                    new_value = dp[prev][pref][done] + q[i-1]
+                    if new_value < dp[current][new_pref][new_done]:
+                        dp[current][new_pref][new_done] = new_value
+        
+    # 寻找最终答案
+    final_layer = n % 2
+    return min(dp[final_layer][k][:adjusted_s+1])
+
+class Dtopsecrettaskbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        self.params = params
+        # 参数校验确保合法性
+        self.params.setdefault('min_n', 3)
+        self.params.setdefault('max_n', 150)
+        self.params.setdefault('min_k', 1)
+        self.params.setdefault('max_k', 150)
+        self.params.setdefault('min_s', 1)
+        self.params.setdefault('max_s', 10**9)
+        self.params.setdefault('q_min', 1)
+        self.params.setdefault('q_max', 10**6)
+        
+        # 确保参数间约束关系
+        self.params['min_n'] = max(self.params['min_n'], 1)
+        self.params['max_n'] = min(self.params['max_n'], 150)
+        self.params['min_k'] = max(self.params['min_k'], 1)
+        self.params['max_k'] = min(self.params['max_k'], self.params['max_n'])
+
+    def case_generator(self):
+        # 确保k <= n的约束
+        n = random.randint(self.params['min_n'], self.params['max_n'])
+        k = random.randint(
+            self.params['min_k'], 
+            min(n, self.params['max_k'])
+        )
+        s = random.randint(self.params['min_s'], self.params['max_s'])
+        
+        # 生成loquacity值时保证至少k个非极大值
+        q = [
+            random.randint(self.params['q_min'], self.params['q_max'])
+            for _ in range(n)
+        ]
+        # 插入k个较小值以确保有解
+        for _ in range(k):
+            q[random.randint(0, n-1)] = random.randint(
+                self.params['q_min'], 
+                self.params['q_max']//100
+            )
+        
+        return {
+            'n': n,
+            'k': k,
+            's': s,
+            'q': q,
+            'expected': calculate_min_loquacity(n, k, s, q)
+        }
+
+    @staticmethod
+    def prompt_func(question_case):
+        return f"""作为军事参谋，你需要解决秘密部队优化问题。请根据以下输入数据计算最小loquacity总和：
+
+输入格式：
+第一行：n k s
+第二行：q1 q2 ... qn
+
+当前输入：
+{question_case['n']} {question_case['k']} {question_case['s']}
+{' '.join(map(str, question_case['q']))}
+
+要求：
+1. 最多进行s次相邻交换
+2. 最终前k个士兵的loquacity总和最小
+3. 答案需用[answer]标签包裹，如：[answer]123[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        import re
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        try:
+            return int(matches[-1].strip())
+        except (ValueError, TypeError):
+            return None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        return solution == identity['expected']