mirror of
https://github.com/InternLM/InternBootcamp.git
synced 2026-04-25 17:10:49 +00:00
368 lines
9.8 KiB
Python
Executable file
368 lines
9.8 KiB
Python
Executable file
"""#
|
||
|
||
### 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']
|