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368 lines
9.8 KiB
Python
Executable file
368 lines
9.8 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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inline int re() {
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int x = 0, f = 1;
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char ch = getchar();
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while (ch < '0' || ch > '9') {
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if (ch == '-') f = -1;
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ch = getchar();
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}
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while (ch >= '0' && ch <= '9') {
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x = (x << 3) + (x << 1) + (ch ^ 48);
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ch = getchar();
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}
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return x * f;
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}
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long long gcd(long long a, long long b) {
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long long r;
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while (b) {
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r = a % b;
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a = b;
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b = r;
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}
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return a;
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}
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long long kk(long long a, long long b) {
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long long sum = 1;
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while (b) {
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if (b) sum = sum * a;
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a = a * a;
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b >>= 1;
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}
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return sum;
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}
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int ly[300030 << 2], tree[300030 << 2], a[300030];
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void pdd(int l, int r, int tr) {
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if (ly[tr]) {
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ly[2 * tr] += ly[tr];
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ly[2 * tr + 1] += ly[tr];
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int mid = (l + r) >> 1;
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tree[2 * tr] += (mid - l + 1) * ly[tr];
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tree[2 * tr + 1] += (r - mid) * ly[tr];
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ly[tr] = 0;
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}
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}
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void pup(int tr) { tree[tr] = tree[2 * tr] + tree[2 * tr + 1]; }
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void upd(int l, int r, int L, int R, int tr) {
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if (L <= l && r <= R) {
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ly[tr] += 1;
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tree[tr] += (r - l + 1);
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return;
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}
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pdd(l, r, tr);
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int mid = (l + r) >> 1;
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if (L <= mid) upd(l, mid, L, R, 2 * tr);
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if (R > mid) upd(mid + 1, r, L, R, 2 * tr + 1);
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pup(tr);
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}
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int qry(int l, int r, int pos, int tr) {
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if (l == r) return tree[tr];
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pdd(l, r, tr);
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int mid = (l + r) >> 1;
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if (pos <= mid)
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return qry(l, mid, pos, 2 * tr);
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else
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return qry(mid + 1, r, pos, 2 * tr + 1);
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}
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vector<pair<int, int>> v[300030];
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int e[300030];
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int main() {
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int n = re(), q = re();
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for (int i = 1; i <= n; i++) {
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a[i] = re();
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a[i] = i - a[i];
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if (a[i] < 0) a[i] = 0x3f3f3f3f;
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}
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for (int i = 1; i <= q; i++) {
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int u = re(), r = re();
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r = n - r;
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v[r].push_back({u, i});
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}
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for (int i = 1; i <= n; i++) {
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int l = 1, r = i;
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int ans = -1;
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while (l <= r) {
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int mid = (l + r) >> 1;
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if (qry(1, n, mid, 1) >= a[i]) {
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ans = mid;
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l = mid + 1;
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} else
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r = mid - 1;
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}
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if (ans != -1) {
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upd(1, n, 1, ans, 1);
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}
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for (int j = 0; j < v[i].size(); j++) {
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int u = v[i][j].first;
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e[v[i][j].second] = qry(1, n, u + 1, 1);
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}
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}
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for (int i = 1; i <= q; i++) cout << e[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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import random
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from bootcamp import Basebootcamp
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class Efixedpointremovalbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.params = params
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self.n = params.get('n', 5)
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self.q = params.get('q', 2)
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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 = self.n
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q = self.q
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a = [random.randint(1, n) for _ in range(n)]
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queries = []
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for _ in range(q):
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while True:
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x = random.randint(0, n - 1)
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max_y = n - 1 - x
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if max_y < 0:
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continue
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y = random.randint(0, max_y)
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if x + y < n:
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queries.append((x, y))
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break
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answers = self.solve_case(n, q, a, queries)
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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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'queries': queries,
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'answers': answers
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}
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@staticmethod
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def solve_case(n, q, a_list, queries):
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a = [0] * (n + 2)
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for i in range(1, n + 1):
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ai = a_list[i - 1]
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a_i_val = i - ai
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a[i] = a_i_val if a_i_val >= 0 else float('inf')
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v = [[] for _ in range(n + 2)]
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for idx, (x, y) in enumerate(queries):
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r = n - y
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v[r].append((x, idx))
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st = Efixedpointremovalbootcamp.SegmentTree(n)
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answers = [0] * len(queries)
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for i in range(1, n + 1):
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current_a = a[i]
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if current_a == float('inf'):
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ans = -1
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else:
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l, r, ans = 1, i, -1
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while l <= r:
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mid = (l + r) // 2
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val = st.point_query(mid)
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if val >= current_a:
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ans = mid
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l = mid + 1
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else:
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r = mid - 1
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if ans != -1:
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st.range_add(1, ans, 1)
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for (x, query_idx) in v[i]:
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pos = x + 1
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val = st.point_query(pos) if pos <= n else 0
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answers[query_idx] = val
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return answers
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class SegmentTree:
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def __init__(self, size):
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self.n = size
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self.tree = [0] * (4 * size)
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self.lazy = [0] * (4 * size)
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def _push_down(self, node, l, r):
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if self.lazy[node]:
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mid = (l + r) // 2
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left, right = 2 * node, 2 * node + 1
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self.tree[left] += self.lazy[node] * (mid - l + 1)
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self.lazy[left] += self.lazy[node]
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self.tree[right] += self.lazy[node] * (r - mid)
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self.lazy[right] += self.lazy[node]
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self.lazy[node] = 0
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def _range_add(self, node, l, r, L, R, val):
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if R < l or L > r:
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return
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if L <= l and r <= R:
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self.tree[node] += val * (r - l + 1)
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self.lazy[node] += val
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return
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self._push_down(node, l, r)
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mid = (l + r) // 2
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self._range_add(2*node, l, mid, L, R, val)
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self._range_add(2*node+1, mid+1, r, L, R, val)
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self.tree[node] = self.tree[2*node] + self.tree[2*node+1]
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def _point_query(self, node, l, r, pos):
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if l == r:
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return self.tree[node]
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self._push_down(node, l, r)
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mid = (l + r) // 2
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if pos <= mid:
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return self._point_query(2*node, l, mid, pos)
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else:
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return self._point_query(2*node+1, mid+1, r, pos)
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def range_add(self, L, R, val):
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self._range_add(1, 1, self.n, L, R, val)
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def point_query(self, pos):
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return self._point_query(1, 1, self.n, pos)
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@staticmethod
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def prompt_func(question_case) -> str:
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problem_desc = """\
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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。
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规则：
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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。
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2. 移除后数组的后续元素索引依次前移。
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3. 权重是最多能移除的元素数量。
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输入格式：
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- 第1行：n 和 q（数组长度，查询数量）
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- 第2行：a_1 a_2 ... a_n
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- 接下来q行：每行两个整数x和y
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输出格式：
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- 每个查询的答案占一行，置于[answer]和[/answer]标签内。
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示例输入：
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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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示例输出：
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2
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0
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"""
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current_input = f"{question_case['n']} {question_case['q']}\n"
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current_input += ' '.join(map(str, question_case['a'])) + '\n'
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current_input += '\n'.join(f"{x} {y}" for x, y in question_case['queries'])
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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。"
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return prompt
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@staticmethod
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def extract_output(output):
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import re
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answer_blocks = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not answer_blocks:
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return None
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last_block = answer_blocks[-1].strip()
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answers = []
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for line in last_block.split('\n'):
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line = line.strip()
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if line:
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try:
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answers.append(int(line))
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except ValueError:
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continue
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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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return solution == identity['answers']
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