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internbootcamp/bootcamp/efixedpointremoval/efixedpointremoval.py

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+"""# 
+
+### 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;
+inline int re() {
+  int x = 0, f = 1;
+  char ch = getchar();
+  while (ch < '0' || ch > '9') {
+    if (ch == '-') f = -1;
+    ch = getchar();
+  }
+  while (ch >= '0' && ch <= '9') {
+    x = (x << 3) + (x << 1) + (ch ^ 48);
+    ch = getchar();
+  }
+  return x * f;
+}
+long long gcd(long long a, long long b) {
+  long long r;
+  while (b) {
+    r = a % b;
+    a = b;
+    b = r;
+  }
+  return a;
+}
+long long kk(long long a, long long b) {
+  long long sum = 1;
+  while (b) {
+    if (b) sum = sum * a;
+    a = a * a;
+    b >>= 1;
+  }
+  return sum;
+}
+int ly[300030 << 2], tree[300030 << 2], a[300030];
+void pdd(int l, int r, int tr) {
+  if (ly[tr]) {
+    ly[2 * tr] += ly[tr];
+    ly[2 * tr + 1] += ly[tr];
+    int mid = (l + r) >> 1;
+    tree[2 * tr] += (mid - l + 1) * ly[tr];
+    tree[2 * tr + 1] += (r - mid) * ly[tr];
+    ly[tr] = 0;
+  }
+}
+void pup(int tr) { tree[tr] = tree[2 * tr] + tree[2 * tr + 1]; }
+void upd(int l, int r, int L, int R, int tr) {
+  if (L <= l && r <= R) {
+    ly[tr] += 1;
+    tree[tr] += (r - l + 1);
+    return;
+  }
+  pdd(l, r, tr);
+  int mid = (l + r) >> 1;
+  if (L <= mid) upd(l, mid, L, R, 2 * tr);
+  if (R > mid) upd(mid + 1, r, L, R, 2 * tr + 1);
+  pup(tr);
+}
+int qry(int l, int r, int pos, int tr) {
+  if (l == r) return tree[tr];
+  pdd(l, r, tr);
+  int mid = (l + r) >> 1;
+  if (pos <= mid)
+    return qry(l, mid, pos, 2 * tr);
+  else
+    return qry(mid + 1, r, pos, 2 * tr + 1);
+}
+vector<pair<int, int>> v[300030];
+int e[300030];
+int main() {
+  int n = re(), q = re();
+  for (int i = 1; i <= n; i++) {
+    a[i] = re();
+    a[i] = i - a[i];
+    if (a[i] < 0) a[i] = 0x3f3f3f3f;
+  }
+  for (int i = 1; i <= q; i++) {
+    int u = re(), r = re();
+    r = n - r;
+    v[r].push_back({u, i});
+  }
+  for (int i = 1; i <= n; i++) {
+    int l = 1, r = i;
+    int ans = -1;
+    while (l <= r) {
+      int mid = (l + r) >> 1;
+      if (qry(1, n, mid, 1) >= a[i]) {
+        ans = mid;
+        l = mid + 1;
+      } else
+        r = mid - 1;
+    }
+    if (ans != -1) {
+      upd(1, n, 1, ans, 1);
+    }
+    for (int j = 0; j < v[i].size(); j++) {
+      int u = v[i][j].first;
+      e[v[i][j].second] = qry(1, n, u + 1, 1);
+    }
+  }
+  for (int i = 1; i <= q; i++) cout << e[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
+import random
+from bootcamp import Basebootcamp
+
+class Efixedpointremovalbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        self.params = params
+        self.n = params.get('n', 5)
+        self.q = params.get('q', 2)
+        self.seed = params.get('seed', None)
+        if self.seed is not None:
+            random.seed(self.seed)
+
+    def case_generator(self):
+        n = self.n
+        q = self.q
+
+        a = [random.randint(1, n) for _ in range(n)]
+        
+        queries = []
+        for _ in range(q):
+            while True:
+                x = random.randint(0, n - 1)
+                max_y = n - 1 - x
+                if max_y < 0:
+                    continue
+                y = random.randint(0, max_y)
+                if x + y < n:
+                    queries.append((x, y))
+                    break
+        
+        answers = self.solve_case(n, q, a, queries)
+        
+        return {
+            'n': n,
+            'q': q,
+            'a': a,
+            'queries': queries,
+            'answers': answers
+        }
+
+    @staticmethod
+    def solve_case(n, q, a_list, queries):
+        a = [0] * (n + 2)
+        for i in range(1, n + 1):
+            ai = a_list[i - 1]
+            a_i_val = i - ai
+            a[i] = a_i_val if a_i_val >= 0 else float('inf')
+        
+        v = [[] for _ in range(n + 2)]
+        for idx, (x, y) in enumerate(queries):
+            r = n - y
+            v[r].append((x, idx))
+        
+        st = Efixedpointremovalbootcamp.SegmentTree(n)
+        answers = [0] * len(queries)
+        
+        for i in range(1, n + 1):
+            current_a = a[i]
+            if current_a == float('inf'):
+                ans = -1
+            else:
+                l, r, ans = 1, i, -1
+                while l <= r:
+                    mid = (l + r) // 2
+                    val = st.point_query(mid)
+                    if val >= current_a:
+                        ans = mid
+                        l = mid + 1
+                    else:
+                        r = mid - 1
+            if ans != -1:
+                st.range_add(1, ans, 1)
+            
+            for (x, query_idx) in v[i]:
+                pos = x + 1
+                val = st.point_query(pos) if pos <= n else 0
+                answers[query_idx] = val
+        
+        return answers
+
+    class SegmentTree:
+        def __init__(self, size):
+            self.n = size
+            self.tree = [0] * (4 * size)
+            self.lazy = [0] * (4 * size)
+
+        def _push_down(self, node, l, r):
+            if self.lazy[node]:
+                mid = (l + r) // 2
+                left, right = 2 * node, 2 * node + 1
+                self.tree[left] += self.lazy[node] * (mid - l + 1)
+                self.lazy[left] += self.lazy[node]
+                self.tree[right] += self.lazy[node] * (r - mid)
+                self.lazy[right] += self.lazy[node]
+                self.lazy[node] = 0
+
+        def _range_add(self, node, l, r, L, R, val):
+            if R < l or L > r:
+                return
+            if L <= l and r <= R:
+                self.tree[node] += val * (r - l + 1)
+                self.lazy[node] += val
+                return
+            self._push_down(node, l, r)
+            mid = (l + r) // 2
+            self._range_add(2*node, l, mid, L, R, val)
+            self._range_add(2*node+1, mid+1, r, L, R, val)
+            self.tree[node] = self.tree[2*node] + self.tree[2*node+1]
+
+        def _point_query(self, node, l, r, pos):
+            if l == r:
+                return self.tree[node]
+            self._push_down(node, l, r)
+            mid = (l + r) // 2
+            if pos <= mid:
+                return self._point_query(2*node, l, mid, pos)
+            else:
+                return self._point_query(2*node+1, mid+1, r, pos)
+
+        def range_add(self, L, R, val):
+            self._range_add(1, 1, self.n, L, R, val)
+
+        def point_query(self, pos):
+            return self._point_query(1, 1, self.n, pos)
+
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        problem_desc = """\
+给定一个由n个正整数构成的数组a。每个查询要求将前x个元素和后y个元素替换为n+1（无法移除），计算处理后数组的最大可移除元素数目。
+
+规则：
+1. 每次操作可以选择一个索引i，使得当前a_i == i，并移除该元素。
+2. 移除后数组的后续元素索引依次前移。
+3. 权重是最多能移除的元素数量。
+
+输入格式：
+- 第1行：n 和 q（数组长度，查询数量）
+- 第2行：a_1 a_2 ... a_n
+- 接下来q行：每行两个整数x和y
+
+输出格式：
+- 每个查询的答案占一行，置于[answer]和[/answer]标签内。
+
+示例输入：
+5 2
+1 4 1 2 4
+0 0
+1 0
+
+示例输出：
+2
+0
+"""
+        current_input = f"{question_case['n']} {question_case['q']}\n"
+        current_input += ' '.join(map(str, question_case['a'])) + '\n'
+        current_input += '\n'.join(f"{x} {y}" for x, y in question_case['queries'])
+        prompt = f"{problem_desc}\n当前输入数据为：\n{current_input}\n请输出每个查询的结果，每行一个，放在[answer]标签内。"
+        return prompt
+
+    @staticmethod
+    def extract_output(output):
+        import re
+        answer_blocks = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not answer_blocks:
+            return None
+        last_block = answer_blocks[-1].strip()
+        answers = []
+        for line in last_block.split('\n'):
+            line = line.strip()
+            if line:
+                try:
+                    answers.append(int(line))
+                except ValueError:
+                    continue
+        return answers if answers else None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        return solution == identity['answers']