mirror of
https://github.com/InternLM/InternBootcamp.git
synced 2026-04-26 17:13:14 +00:00
init-commit
This commit is contained in:
lilinyang
commit
18a552597a
3461 changed files with 1150579 additions and 0 deletions
305
internbootcamp/bootcamp/cgreedyshopping/cgreedyshopping.py
Executable file
305
internbootcamp/bootcamp/cgreedyshopping/cgreedyshopping.py
Executable file
|
|
@ -0,0 +1,305 @@
|
|||
"""#
|
||||
|
||||
### 谜题描述
|
||||
You are given an array a_1, a_2, …, a_n of integers. This array is non-increasing.
|
||||
|
||||
Let's consider a line with n shops. The shops are numbered with integers from 1 to n from left to right. The cost of a meal in the i-th shop is equal to a_i.
|
||||
|
||||
You should process q queries of two types:
|
||||
|
||||
* 1 x y: for each shop 1 ≤ i ≤ x set a_{i} = max(a_{i}, y).
|
||||
* 2 x y: let's consider a hungry man with y money. He visits the shops from x-th shop to n-th and if he can buy a meal in the current shop he buys one item of it. Find how many meals he will purchase. The man can buy a meal in the shop i if he has at least a_i money, and after it his money decreases by a_i.
|
||||
|
||||
Input
|
||||
|
||||
The first line contains two integers n, q (1 ≤ n, q ≤ 2 ⋅ 10^5).
|
||||
|
||||
The second line contains n integers a_{1},a_{2}, …, a_{n} (1 ≤ a_{i} ≤ 10^9) — the costs of the meals. It is guaranteed, that a_1 ≥ a_2 ≥ … ≥ a_n.
|
||||
|
||||
Each of the next q lines contains three integers t, x, y (1 ≤ t ≤ 2, 1≤ x ≤ n, 1 ≤ y ≤ 10^9), each describing the next query.
|
||||
|
||||
It is guaranteed that there exists at least one query of type 2.
|
||||
|
||||
Output
|
||||
|
||||
For each query of type 2 output the answer on the new line.
|
||||
|
||||
Example
|
||||
|
||||
Input
|
||||
|
||||
|
||||
10 6
|
||||
10 10 10 6 6 5 5 5 3 1
|
||||
2 3 50
|
||||
2 4 10
|
||||
1 3 10
|
||||
2 2 36
|
||||
1 4 7
|
||||
2 2 17
|
||||
|
||||
|
||||
Output
|
||||
|
||||
|
||||
8
|
||||
3
|
||||
6
|
||||
2
|
||||
|
||||
Note
|
||||
|
||||
In the first query a hungry man will buy meals in all shops from 3 to 10.
|
||||
|
||||
In the second query a hungry man will buy meals in shops 4, 9, and 10.
|
||||
|
||||
After the third query the array a_1, a_2, …, a_n of costs won't change and will be \{10, 10, 10, 6, 6, 5, 5, 5, 3, 1\}.
|
||||
|
||||
In the fourth query a hungry man will buy meals in shops 2, 3, 4, 5, 9, and 10.
|
||||
|
||||
After the fifth query the array a of costs will be \{10, 10, 10, 7, 6, 5, 5, 5, 3, 1\}.
|
||||
|
||||
In the sixth query a hungry man will buy meals in shops 2 and 4.
|
||||
|
||||
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;
|
||||
template <class T>
|
||||
using vc = vector<T>;
|
||||
template <class T>
|
||||
using vvc = vc<vc<T>>;
|
||||
template <class T>
|
||||
void mkuni(vector<T> &v) {
|
||||
sort(v.begin(), v.end());
|
||||
v.erase(unique(v.begin(), v.end()), v.end());
|
||||
}
|
||||
long long rand_int(long long l, long long r) {
|
||||
static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
|
||||
return uniform_int_distribution<long long>(l, r)(gen);
|
||||
}
|
||||
template <class T>
|
||||
void print(T x, int suc = 1) {
|
||||
cout << x;
|
||||
if (suc == 1)
|
||||
cout << '\n';
|
||||
else
|
||||
cout << ' ';
|
||||
}
|
||||
template <class T>
|
||||
void print(const vector<T> &v, int suc = 1) {
|
||||
for (int i = 0; i < v.size(); i++)
|
||||
print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
|
||||
}
|
||||
const int N = 3e5 + 10;
|
||||
struct Tree {
|
||||
long long l, r, lazy, sum, mi, ma;
|
||||
} tree[N << 2];
|
||||
void push_up(int rt) {
|
||||
tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
|
||||
tree[rt].ma = tree[rt << 1].ma;
|
||||
tree[rt].mi = tree[rt << 1 | 1].mi;
|
||||
}
|
||||
void build(int l, int r, int rt, vector<int> &a) {
|
||||
tree[rt].l = l, tree[rt].r = r, tree[rt].lazy = 0;
|
||||
if (l == r) {
|
||||
tree[rt].sum = tree[rt].mi = tree[rt].ma = a[l];
|
||||
return;
|
||||
}
|
||||
int mid = l + r >> 1;
|
||||
build(l, mid, rt << 1, a);
|
||||
build(mid + 1, r, rt << 1 | 1, a);
|
||||
push_up(rt);
|
||||
}
|
||||
void push_down(int rt) {
|
||||
if (tree[rt].lazy) {
|
||||
int x = tree[rt].lazy, l = tree[rt].l, r = tree[rt].r;
|
||||
tree[rt].lazy = 0;
|
||||
tree[rt << 1].sum = 1ll * (tree[rt << 1].r - tree[rt << 1].l + 1) * x;
|
||||
tree[rt << 1].mi = tree[rt << 1].ma = x;
|
||||
tree[rt << 1].lazy = x;
|
||||
tree[rt << 1 | 1].sum =
|
||||
1ll * (tree[rt << 1 | 1].r - tree[rt << 1 | 1].l + 1) * x;
|
||||
tree[rt << 1 | 1].mi = tree[rt << 1 | 1].ma = x;
|
||||
tree[rt << 1 | 1].lazy = x;
|
||||
}
|
||||
}
|
||||
void update_range(int L, int R, long long Y, int rt) {
|
||||
int l = tree[rt].l, r = tree[rt].r;
|
||||
if (tree[rt].mi >= Y || l > R) return;
|
||||
if (tree[rt].ma <= Y && r <= R) {
|
||||
tree[rt].sum = 1ll * (r - l + 1) * Y;
|
||||
tree[rt].mi = tree[rt].ma = Y;
|
||||
tree[rt].lazy = Y;
|
||||
return;
|
||||
}
|
||||
push_down(rt);
|
||||
update_range(L, R, Y, rt << 1);
|
||||
update_range(L, R, Y, rt << 1 | 1);
|
||||
push_up(rt);
|
||||
}
|
||||
int query_range(int L, int R, int rt, long long &Y) {
|
||||
int l = tree[rt].l, r = tree[rt].r;
|
||||
if (tree[rt].mi > Y || r < L) return 0;
|
||||
if (tree[rt].sum <= Y && l >= L) {
|
||||
Y -= tree[rt].sum;
|
||||
return r - l + 1;
|
||||
}
|
||||
push_down(rt);
|
||||
long long res = 0;
|
||||
res += query_range(L, R, rt << 1, Y);
|
||||
res += query_range(L, R, rt << 1 | 1, Y);
|
||||
return res;
|
||||
}
|
||||
int main() {
|
||||
ios::sync_with_stdio(false);
|
||||
cin.tie(nullptr);
|
||||
int n, q;
|
||||
cin >> n >> q;
|
||||
vector<int> a(n + 1);
|
||||
for (int i = 1; i <= n; ++i) cin >> a[i];
|
||||
build(1, n, 1, a);
|
||||
while (q--) {
|
||||
long long x, y, op;
|
||||
cin >> op >> x >> y;
|
||||
if (op == 1) {
|
||||
update_range(1, x, y, 1);
|
||||
} else
|
||||
cout << query_range(x, n, 1, y) << '\n';
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
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 re
|
||||
import random
|
||||
from itertools import accumulate
|
||||
from bootcamp import Basebootcamp
|
||||
|
||||
class Cgreedyshoppingbootcamp(Basebootcamp):
|
||||
def __init__(self, **params):
|
||||
self.n = params.get('n', 10)
|
||||
self.q = params.get('q', 5)
|
||||
self.n = max(1, self.n) # 确保最小为1
|
||||
self.q = max(1, self.q) # 确保至少一个查询
|
||||
|
||||
def case_generator(self):
|
||||
# 生成非递增数组（优化版）
|
||||
a = [random.randint(1, 10**9)]
|
||||
for _ in range(1, self.n):
|
||||
a.append(random.randint(1, a[-1]))
|
||||
|
||||
# 生成查询列表并确保类型2存在
|
||||
queries = []
|
||||
type2_indices = []
|
||||
for i in range(self.q):
|
||||
t = random.choices([1, 2], weights=[0.4, 0.6])[0] # 增加类型2概率
|
||||
x = random.randint(1, self.n)
|
||||
y = random.randint(1, 10**9)
|
||||
queries.append([t, x, y]) # 统一使用列表存储
|
||||
if t == 2:
|
||||
type2_indices.append(i)
|
||||
|
||||
# 确保至少一个类型2查询
|
||||
if not type2_indices:
|
||||
queries[-1] = [2, random.randint(1, self.n), random.randint(1, 10**9)]
|
||||
type2_indices = [self.q-1]
|
||||
|
||||
# 预处理答案（优化模拟）
|
||||
current_a = a.copy()
|
||||
answers = []
|
||||
for op in queries:
|
||||
t, x, y = op
|
||||
if t == 1:
|
||||
# 使用二分查找确定有效更新范围
|
||||
left = 0
|
||||
right = x-1 # 转换为0-based索引
|
||||
update_pos = next((i for i in range(x) if current_a[i] < y), None)
|
||||
if update_pos is not None:
|
||||
current_a[update_pos:x] = [max(y, val) for val in current_a[update_pos:x]]
|
||||
else:
|
||||
# 使用累积和优化计算
|
||||
prefix = list(accumulate(current_a[x-1:]))
|
||||
money = y
|
||||
count = 0
|
||||
for s in prefix:
|
||||
if s > money:
|
||||
break
|
||||
count += 1
|
||||
money -= s - (prefix[count-2] if count>1 else 0)
|
||||
answers.append(count)
|
||||
|
||||
return {
|
||||
'n': self.n,
|
||||
'q': self.q,
|
||||
'initial_array': a,
|
||||
'queries': queries, # 统一使用列表存储
|
||||
'answers': answers
|
||||
}
|
||||
|
||||
@staticmethod
|
||||
def prompt_func(question_case) -> str:
|
||||
input_lines = [
|
||||
f"{question_case['n']} {question_case['q']}",
|
||||
' '.join(map(str, question_case['initial_array']))
|
||||
]
|
||||
for op in question_case['queries']:
|
||||
input_lines.append(f"{op[0]} {op[1]} {op[2]}")
|
||||
|
||||
return f"""你正在处理餐馆消费查询系统，需要处理两种操作类型：
|
||||
|
||||
**规则详解**
|
||||
1. 类型1 (1 x y)：将前x个餐馆的餐费更新为原值和y的较大值
|
||||
2. 类型2 (2 x y)：顾客从第x个餐馆开始向后消费，直到余额不足
|
||||
|
||||
**输入格式**
|
||||
{" ".join(input_lines[:2])}
|
||||
{chr(10).join(input_lines[2:])}
|
||||
|
||||
**答案格式要求**
|
||||
请将所有类型2查询的答案按顺序排列在[answer]标签内，例如：
|
||||
[answer]
|
||||
3
|
||||
5
|
||||
2
|
||||
[/answer]"""
|
||||
|
||||
@staticmethod
|
||||
def extract_output(output):
|
||||
matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
|
||||
if not matches:
|
||||
return None
|
||||
answers = []
|
||||
for line in matches[-1].strip().split('\n'):
|
||||
if line.strip().isdigit():
|
||||
answers.append(int(line.strip()))
|
||||
return answers or None
|
||||
|
||||
@classmethod
|
||||
def _verify_correction(cls, solution, identity):
|
||||
return solution == identity['answers']
|
||||
|
||||
# 验证示例
|
||||
if __name__ == "__main__":
|
||||
bootcamp = Cgreedyshoppingbootcamp(n=10, q=6)
|
||||
case = bootcamp.case_generator()
|
||||
print("Generated Case:")
|
||||
print(f"Initial array: {case['initial_array']}")
|
||||
print(f"Queries: {case['queries']}")
|
||||
print(f"Expected answers: {case['answers']}")
|
||||
|
||||
prompt = Cgreedyshoppingbootcamp.prompt_func(case)
|
||||
print("\nGenerated Prompt:")
|
||||
print(prompt)
|
||||
|
||||
# 模拟模型回答
|
||||
response = f"[answer]\n" + "\n".join(map(str, case['answers'])) + "\n[/answer]"
|
||||
extracted = Cgreedyshoppingbootcamp.extract_output(response)
|
||||
print("\nExtracted Answer:", extracted)
|
||||
|
||||
score = Cgreedyshoppingbootcamp.verify_score(response, case)
|
||||
print("Validation Score:", score)
|
||||
Loading…
Add table
Add a link
Reference in a new issue