init-commit

internbootcamp/bootcamp/echoosingballs/echoosingballs.py

@@ -0,0 +1,243 @@
+"""# 
+
+### 谜题描述
+There are n balls. They are arranged in a row. Each ball has a color (for convenience an integer) and an integer value. The color of the i-th ball is ci and the value of the i-th ball is vi.
+
+Squirrel Liss chooses some balls and makes a new sequence without changing the relative order of the balls. She wants to maximize the value of this sequence.
+
+The value of the sequence is defined as the sum of following values for each ball (where a and b are given constants):
+
+  * If the ball is not in the beginning of the sequence and the color of the ball is same as previous ball's color, add (the value of the ball)  ×  a. 
+  * Otherwise, add (the value of the ball)  ×  b. 
+
+
+
+You are given q queries. Each query contains two integers ai and bi. For each query find the maximal value of the sequence she can make when a = ai and b = bi.
+
+Note that the new sequence can be empty, and the value of an empty sequence is defined as zero.
+
+Input
+
+The first line contains two integers n and q (1 ≤ n ≤ 105; 1 ≤ q ≤ 500). The second line contains n integers: v1, v2, ..., vn (|vi| ≤ 105). The third line contains n integers: c1, c2, ..., cn (1 ≤ ci ≤ n).
+
+The following q lines contain the values of the constants a and b for queries. The i-th of these lines contains two integers ai and bi (|ai|, |bi| ≤ 105).
+
+In each line integers are separated by single spaces.
+
+Output
+
+For each query, output a line containing an integer — the answer to the query. The i-th line contains the answer to the i-th query in the input order.
+
+Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
+
+Examples
+
+Input
+
+6 3
+1 -2 3 4 0 -1
+1 2 1 2 1 1
+5 1
+-2 1
+1 0
+
+
+Output
+
+20
+9
+4
+
+
+Input
+
+4 1
+-3 6 -1 2
+1 2 3 1
+1 -1
+
+
+Output
+
+5
+
+Note
+
+In the first example, to achieve the maximal value:
+
+  * In the first query, you should select 1st, 3rd, and 4th ball. 
+  * In the second query, you should select 3rd, 4th, 5th and 6th ball. 
+  * In the third query, you should select 2nd and 4th ball. 
+
+
+
+Note that there may be other ways to achieve the maximal value.
+
+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, v[100000], c[100000];
+int q, a, b;
+long long f[100000 + 1];
+bool m[100000 + 1];
+int x, y;
+long long query(long long a, long long b) {
+  memset(f, 0, sizeof(f));
+  memset(m, 0, sizeof(m));
+  f[0] = x = y = 0;
+  long long res = 0;
+  for (int i = 0; i < n; ++i) {
+    long long other = c[i] == x ? f[y] : f[x];
+    if (m[c[i]]) {
+      long long old = f[c[i]];
+      f[c[i]] = max(old, max(old + v[i] * a, other + v[i] * b));
+    } else {
+      f[c[i]] = other + v[i] * b;
+      m[c[i]] = 1;
+    }
+    res = max(res, f[c[i]]);
+    if (c[i] == x || c[i] == y) {
+      if (f[x] < f[y]) {
+        int tmp = x;
+        x = y;
+        y = tmp;
+      }
+    } else {
+      if (f[c[i]] > f[x]) {
+        y = x;
+        x = c[i];
+      } else if (f[c[i]] > f[y]) {
+        y = c[i];
+      }
+    }
+  }
+  return res;
+}
+int main(int argc, char *argv[]) {
+  cin >> n >> q;
+  for (int i = 0; i < n; ++i) {
+    cin >> v[i];
+  }
+  for (int i = 0; i < n; ++i) {
+    cin >> c[i];
+  }
+  for (int i = 0; i < q; ++i) {
+    cin >> a >> b;
+    cout << query(a, b) << 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
+
+class Echoosingballsbootcamp(Basebootcamp):
+    def __init__(self, n_min=3, n_max=6, v_min=-5, v_max=5, c_max=3, q_min=1, q_max=3):
+        self.n_min = n_min
+        self.n_max = n_max
+        self.v_min = v_min
+        self.v_max = v_max
+        self.c_max = c_max
+        self.q_min = q_min
+        self.q_max = q_max
+
+    def case_generator(self):
+        n = random.randint(self.n_min, self.n_max)
+        v = [random.randint(self.v_min, self.v_max) for _ in range(n)]
+        c = [random.randint(1, self.c_max) for _ in range(n)]
+        q = random.randint(self.q_min, self.q_max)
+        queries = [(random.randint(-5, 5), random.randint(-5, 5)) for _ in range(q)]
+        expected_outputs = [self._compute_query(v, c, a, b) for a, b in queries]
+        return {
+            'n': n,
+            'q': q,
+            'v': v,
+            'c': c,
+            'queries': queries,
+            'expected_outputs': expected_outputs
+        }
+
+    @staticmethod
+    def _compute_query(v_list, c_list, a, b):
+        f = {}
+        x = None  # Largest color
+        y = None  # Second largest color
+        res = 0
+        
+        for vi, ci in zip(v_list, c_list):
+            other = 0
+            if x is not None:
+                if ci == x:
+                    other = f.get(y, 0) if y is not None else 0
+                else:
+                    other = f.get(x, 0)
+
+            current = f.get(ci, -float('inf'))
+            new_val = other + vi * b
+            if current != -float('inf'):
+                new_val = max(new_val, current + vi * a)
+            
+            f[ci] = max(current, new_val) if current != -float('inf') else new_val
+            res = max(res, f[ci])
+
+            # Update color rankings
+            colors = sorted(f.items(), key=lambda x: -x[1])
+            x = colors[0][0] if colors else None
+            y = colors[1][0] if len(colors) > 1 else None
+
+        return max(res, 0)
+
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        n = question_case['n']
+        q = question_case['q']
+        v_str = ' '.join(map(str, question_case['v']))
+        c_str = ' '.join(map(str, question_case['c']))
+        queries = '\n'.join([f"{a} {b}" for a, b in question_case['queries']])
+        return f"""你是一个算法竞赛选手，需要解决以下问题：
+
+给定{n}个球按序排列，每个球有颜色c和值v。现在有{q}个查询，每个查询给出系数a和b。要求对每个查询找出最大子序列值。规则如下：
+1. 子序列保持原顺序
+2. 第一个球的贡献为v_i × b
+3. 后续球如果颜色与前一个相同，贡献为v_i × a，否则为v_i × b
+
+输入数据：
+n = {n}, q = {q}
+v数组: {v_str}
+c数组: {c_str}
+查询列表:
+{queries}
+
+请计算每个查询的最大价值，并按顺序将结果用换行分隔放在[answer]标签内，例如：
+[answer]
+结果1
+结果2
+[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        import re
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        answers = []
+        for line in matches[-1].strip().split('\n'):
+            line = line.strip()
+            if line:
+                try:
+                    answers.append(int(line))
+                except:
+                    continue
+        return answers if answers else None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        expected = identity['expected_outputs']
+        return isinstance(solution, list) and len(solution) == len(expected) and all(s == e for s, e in zip(solution, expected))