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