init-commit

internbootcamp/bootcamp/cfixedpointremoval/cfixedpointremoval.py

@@ -0,0 +1,272 @@
+"""# 
+
+### 谜题描述
+Let a_1, …, a_n be an array of n positive integers. In one operation, you can choose an index i such that a_i = i, and remove a_i from the array (after the removal, the remaining parts are concatenated).
+
+The weight of a is defined as the maximum number of elements you can remove.
+
+You must answer q independent queries (x, y): after replacing the x first elements of a and the y last elements of a by n+1 (making them impossible to remove), what would be the weight of a?
+
+Input
+
+The first line contains two integers n and q (1 ≤ n, q ≤ 3 ⋅ 10^5) — the length of the array and the number of queries.
+
+The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — elements of the array.
+
+The i-th of the next q lines contains two integers x and y (x, y ≥ 0 and x+y < n).
+
+Output
+
+Print q lines, i-th line should contain a single integer — the answer to the i-th query.
+
+Examples
+
+Input
+
+
+13 5
+2 2 3 9 5 4 6 5 7 8 3 11 13
+3 1
+0 0
+2 4
+5 0
+0 12
+
+
+Output
+
+
+5
+11
+6
+1
+0
+
+
+Input
+
+
+5 2
+1 4 1 2 4
+0 0
+1 0
+
+
+Output
+
+
+2
+0
+
+Note
+
+Explanation of the first query:
+
+After making first x = 3 and last y = 1 elements impossible to remove, a becomes [×, ×, ×, 9, 5, 4, 6, 5, 7, 8, 3, 11, ×] (we represent 14 as × for clarity).
+
+Here is a strategy that removes 5 elements (the element removed is colored in red):
+
+  * [×, ×, ×, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, ×] 
+  * [×, ×, ×, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, ×] 
+  * [×, ×, ×, 9, 4, \color{red}{6}, 5, 7, 8, 3, ×] 
+  * [×, ×, ×, 9, 4, 5, 7, \color{red}{8}, 3, ×] 
+  * [×, ×, ×, 9, 4, 5, \color{red}{7}, 3, ×] 
+  * [×, ×, ×, 9, 4, 5, 3, ×] (final state) 
+
+
+
+It is impossible to remove more than 5 elements, hence the weight is 5.
+
+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 N = 3e5 + 1;
+int n, q, a[N], ans[N], it[4 * N], b[N];
+vector<pair<pair<int, int>, int>> v;
+bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
+  return (x.first.second < y.first.second);
+}
+void upd(int id, int l, int r, int pos, int val) {
+  if (pos < l || r < pos) return;
+  if (l == r) {
+    it[id] += val;
+    return;
+  }
+  int mid = (l + r) / 2;
+  upd(2 * id, l, mid, pos, val);
+  upd(2 * id + 1, mid + 1, r, pos, val);
+  it[id] = it[2 * id] + it[2 * id + 1];
+}
+int get_pos(int id, int l, int r, int pos) {
+  if (l == r) return l;
+  int mid = (l + r) / 2;
+  if (it[2 * id] >= pos)
+    return get_pos(2 * id, l, mid, pos);
+  else
+    return get_pos(2 * id + 1, mid + 1, r, pos - it[2 * id]);
+}
+int get_sum(int id, int l, int r, int l1, int r1) {
+  if (r1 < l || r < l1) return 0;
+  if (l1 <= l && r <= r1) return it[id];
+  int mid = (l + r) / 2;
+  return get_sum(2 * id, l, mid, l1, r1) +
+         get_sum(2 * id + 1, mid + 1, r, l1, r1);
+}
+int main() {
+  ios_base::sync_with_stdio(false);
+  cin.tie();
+  cin >> n >> q;
+  for (int i = 1; i <= n; i++) {
+    cin >> a[i];
+  }
+  for (int i = 1; i <= q; i++) {
+    int l, r;
+    cin >> l >> r;
+    v.push_back({{l, r}, i});
+  }
+  sort(v.begin(), v.end(), cmp);
+  int cnt = 0;
+  for (int i = 1; i <= n; i++) {
+    if (a[i] > i) continue;
+    int tmp = i - a[i];
+    if (tmp <= cnt) {
+      if (tmp == 0) {
+        b[i] = i;
+        upd(1, 1, n, i, 1);
+      } else {
+        tmp = cnt - tmp + 1;
+        b[i] = get_pos(1, 1, n, tmp);
+        upd(1, 1, n, b[i], 1);
+      }
+      cnt++;
+    }
+  }
+  int cur = n + 1;
+  for (int i = 0; i < q; i++) {
+    int tmp = n - v[i].first.second + 1;
+    for (int j = tmp; j < cur; j++) {
+      if (b[j] == 0) continue;
+      upd(1, 1, n, b[j], -1);
+    }
+    ans[v[i].second] = get_sum(1, 1, n, v[i].first.first + 1, n);
+    cur = tmp;
+  }
+  for (int i = 1; i <= q; i++) cout << ans[i] << endl;
+}
+```
+
+
+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
+from bootcamp import Basebootcamp
+import random
+from bisect import bisect_right
+
+class Cfixedpointremovalbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        self.n_min = params.get('n_min', 5)
+        self.n_max = params.get('n_max', 15)
+        self.q_max = params.get('q_max', 5)
+        self.seed = params.get('seed', None)
+        if self.seed is not None:
+            random.seed(self.seed)
+    
+    def case_generator(self):
+        n = random.randint(self.n_min, self.n_max)
+        a = [random.randint(1, n) for _ in range(n)]
+        
+        # 预计算有效元素位置（改进算法）
+        valid = []
+        for i in range(n):
+            pos = i + 1  # 1-based index
+            if a[i] == pos:
+                valid.append(pos)
+            elif a[i] < pos:
+                required = pos - a[i]
+                if required <= len(valid):
+                    start = len(valid) - required
+                    if all(valid[j] >= a[i] for j in range(start, len(valid))):
+                        valid = valid[:start] + [pos]
+
+        # 生成查询并计算答案
+        q = random.randint(1, self.q_max)
+        queries = []
+        answers = []
+        for _ in range(q):
+            while True:
+                x = random.randint(0, n//2)
+                y = random.randint(0, (n-1)-x)
+                if x + y < n:
+                    break
+            
+            l = x + 1
+            r = n - y
+            # 二分查找有效范围内元素数量
+            left = bisect_right(valid, l-1)
+            right_idx = bisect_right(valid, r)
+            answers.append(right_idx - left)
+            queries.append((x, y))
+
+        return {
+            'n': n,
+            'q': q,
+            'a': a,
+            'valid': valid,
+            'queries': queries,
+            'answers': answers
+        }
+    
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        n = question_case['n']
+        q = question_case['q']
+        a = question_case['a']
+        queries = question_case['queries']
+        return f"""## 数组权重谜题
+
+给定长度为{n}的数组：{a}
+
+**规则**：
+1. 每次移除一个元素a_i（满足a_i = 当前索引）
+2. 移除后数组重新索引
+3. 权重是最大可移除元素数量
+
+**查询操作**：
+对于每个查询(x,y):
+- 前x个元素设置为{n+1}
+- 后y个元素设置为{n+1}
+- 计算修改后的数组权重
+
+**输入格式**：
+{q}个查询，每个查询x和y满足x+y<{n}
+
+**输出格式**：
+将每个查询的答案按顺序放在[answer]标签内，每个答案占一行：
+
+示例：
+[answer]
+3
+0
+5
+[/answer]"""
+    
+    @staticmethod
+    def extract_output(output):
+        import re
+        answer_block = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL)
+        if not answer_block:
+            return None
+        answers = []
+        for line in answer_block[-1].strip().split('\n'):
+            line = line.strip()
+            if line and line.isdigit():
+                answers.append(int(line))
+        return answers if answers else None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        expected = identity['answers']
+        return isinstance(solution, list) and solution == expected