mirror of
https://github.com/InternLM/InternBootcamp.git
synced 2026-04-19 12:58:04 +00:00
init-commit
This commit is contained in:
lilinyang
commit
18a552597a
3461 changed files with 1150579 additions and 0 deletions
251
internbootcamp/bootcamp/dpickingstrings/dpickingstrings.py
Executable file
251
internbootcamp/bootcamp/dpickingstrings/dpickingstrings.py
Executable file
|
|
@ -0,0 +1,251 @@
|
|||
"""#
|
||||
|
||||
### 谜题描述
|
||||
Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times:
|
||||
|
||||
* A <image> BC
|
||||
* B <image> AC
|
||||
* C <image> AB
|
||||
* AAA <image> empty string
|
||||
|
||||
|
||||
|
||||
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.
|
||||
|
||||
Input
|
||||
|
||||
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.
|
||||
|
||||
The third line contains the number of queries Q (1 ≤ Q ≤ 105).
|
||||
|
||||
The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?
|
||||
|
||||
Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.
|
||||
|
||||
It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.
|
||||
|
||||
Output
|
||||
|
||||
Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.
|
||||
|
||||
Example
|
||||
|
||||
Input
|
||||
|
||||
AABCCBAAB
|
||||
ABCB
|
||||
5
|
||||
1 3 1 2
|
||||
2 2 2 4
|
||||
7 9 1 1
|
||||
3 4 2 3
|
||||
4 5 1 3
|
||||
|
||||
|
||||
Output
|
||||
|
||||
10011
|
||||
|
||||
Note
|
||||
|
||||
In the first query we can achieve the result, for instance, by using transitions <image>.
|
||||
|
||||
The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
|
||||
|
||||
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>
|
||||
#pragma GCC target(\"pclmul\")
|
||||
using ul = std::uint32_t;
|
||||
using ull = std::uint64_t;
|
||||
using li = std::int32_t;
|
||||
using ll = std::int64_t;
|
||||
using uss = std::uint8_t;
|
||||
const ul maxn = 1e5;
|
||||
char str[maxn + 2];
|
||||
ul scb[maxn + 1];
|
||||
ul sca[maxn + 1];
|
||||
ul tcb[maxn + 1];
|
||||
ul tca[maxn + 1];
|
||||
ul q;
|
||||
int main() {
|
||||
std::scanf(\"%s\", str + 1);
|
||||
for (ul i = 1; str[i]; ++i) {
|
||||
if (str[i] == 'A') {
|
||||
sca[i] = sca[i - 1] + 1;
|
||||
scb[i] = scb[i - 1];
|
||||
} else {
|
||||
scb[i] = scb[i - 1] + 1;
|
||||
}
|
||||
}
|
||||
std::scanf(\"%s\", str + 1);
|
||||
for (ul i = 1; str[i]; ++i) {
|
||||
if (str[i] == 'A') {
|
||||
tca[i] = tca[i - 1] + 1;
|
||||
tcb[i] = tcb[i - 1];
|
||||
} else {
|
||||
tcb[i] = tcb[i - 1] + 1;
|
||||
}
|
||||
}
|
||||
std::scanf(\"%u\", &q);
|
||||
for (ul i = 1; i <= q; ++i) {
|
||||
ul a, b, c, d;
|
||||
std::scanf(\"%u%u%u%u\", &a, &b, &c, &d);
|
||||
ul sb = scb[b] - scb[a - 1];
|
||||
ul tb = tcb[d] - tcb[c - 1];
|
||||
ul sa = std::min(sca[b], b - a + 1);
|
||||
ul ta = std::min(tca[d], d - c + 1);
|
||||
if ((sb ^ tb) & 1) {
|
||||
std::putchar('0');
|
||||
continue;
|
||||
}
|
||||
if (sa < ta) {
|
||||
std::putchar('0');
|
||||
continue;
|
||||
}
|
||||
if ((sa - ta) % 3 == 0 && sb > tb) {
|
||||
std::putchar('0');
|
||||
continue;
|
||||
}
|
||||
if ((sa - ta) % 3 != 0 && sb + 2 > tb) {
|
||||
std::putchar('0');
|
||||
continue;
|
||||
}
|
||||
if (sb == 0 && tb != 0 && sa == ta) {
|
||||
std::putchar('0');
|
||||
continue;
|
||||
}
|
||||
std::putchar('1');
|
||||
}
|
||||
std::putchar('\n');
|
||||
return 0;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
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
|
||||
import re
|
||||
from bootcamp import Basebootcamp
|
||||
|
||||
def compute_prefixes(s):
|
||||
sca = [0] * (len(s) + 1)
|
||||
scb = [0] * (len(s) + 1)
|
||||
for i in range(1, len(s) + 1):
|
||||
char = s[i-1]
|
||||
if char == 'A':
|
||||
sca[i] = sca[i-1] + 1
|
||||
scb[i] = scb[i-1]
|
||||
else:
|
||||
sca[i] = sca[i-1]
|
||||
scb[i] = scb[i-1] + 1
|
||||
return sca, scb
|
||||
|
||||
def compute_query_result(sca, scb, tca, tcb, a, b, c, d):
|
||||
sb = scb[b] - scb[a-1]
|
||||
sa = min(sca[b], b - a + 1)
|
||||
tb = tcb[d] - tcb[c-1]
|
||||
ta = min(tca[d], d - c + 1)
|
||||
|
||||
if (sb ^ tb) & 1:
|
||||
return '0'
|
||||
if sa < ta:
|
||||
return '0'
|
||||
if (sa - ta) % 3 == 0:
|
||||
if sb > tb:
|
||||
return '0'
|
||||
else:
|
||||
if (sb + 2) > tb:
|
||||
return '0'
|
||||
if sb == 0 and tb != 0 and sa == ta:
|
||||
return '0'
|
||||
return '1'
|
||||
|
||||
class Dpickingstringsbootcamp(Basebootcamp):
|
||||
def __init__(self, max_s_length=10, max_t_length=10, max_queries=5, max_attempts=10):
|
||||
self.max_s_length = max_s_length
|
||||
self.max_t_length = max_t_length
|
||||
self.max_queries = max_queries
|
||||
self.max_attempts = max_attempts
|
||||
|
||||
def case_generator(self):
|
||||
for _ in range(self.max_attempts):
|
||||
s = ''.join(random.choices(['A', 'B', 'C'], k=random.randint(1, self.max_s_length)))
|
||||
t = ''.join(random.choices(['A', 'B', 'C'], k=random.randint(1, self.max_t_length)))
|
||||
sca, scb = compute_prefixes(s)
|
||||
tca, tcb = compute_prefixes(t)
|
||||
a, b = 1, len(s)
|
||||
c, d = 1, len(t)
|
||||
if compute_query_result(sca, scb, tca, tcb, a, b, c, d) == '1':
|
||||
queries = [[a, b, c, d]]
|
||||
for _ in range(self.max_queries - 1):
|
||||
a_q = random.randint(1, len(s))
|
||||
b_q = random.randint(a_q, len(s))
|
||||
c_q = random.randint(1, len(t))
|
||||
d_q = random.randint(c_q, len(t))
|
||||
queries.append([a_q, b_q, c_q, d_q])
|
||||
return {
|
||||
'S': s,
|
||||
'T': t,
|
||||
'queries': queries
|
||||
}
|
||||
# Fallback if no valid case found
|
||||
s = ''.join(random.choices(['A', 'B', 'C'], k=random.randint(1, self.max_s_length)))
|
||||
t = ''.join(random.choices(['A', 'B', 'C'], k=random.randint(1, self.max_t_length)))
|
||||
queries = []
|
||||
for _ in range(self.max_queries):
|
||||
a = random.randint(1, len(s))
|
||||
b = random.randint(a, len(s))
|
||||
c = random.randint(1, len(t))
|
||||
d = random.randint(c, len(t))
|
||||
queries.append([a, b, c, d])
|
||||
return {
|
||||
'S': s,
|
||||
'T': t,
|
||||
'queries': queries
|
||||
}
|
||||
|
||||
@staticmethod
|
||||
def prompt_func(question_case):
|
||||
s = question_case['S']
|
||||
t = question_case['T']
|
||||
queries = question_case['queries']
|
||||
prompt = f"""Alice有一个字符串S，内容为：{s}。Bob可以对该字符串的任意子串应用以下转换规则（次数不限、顺序不限）：
|
||||
|
||||
- 将单个'A'替换为'BC'；
|
||||
- 将单个'B'替换为'AC'；
|
||||
- 将单个'C'替换为'AB'；
|
||||
- 将连续的三个'A'（即"AAA"）替换为空字符串（即删除）。
|
||||
|
||||
现有目标字符串T，内容为：{t}。你需要处理{len(queries)}个查询，每个查询询问是否存在一种可能，通过应用上述规则，将S的某个子串转换为T的对应子串。
|
||||
|
||||
每个查询的参数为四个整数a、b、c、d，表示：源子串是S的第a到第b个字符（包含两端，按1-based索引），目标子串是T的第c到第d个字符。要求判断是否可以进行这样的转换。
|
||||
|
||||
请依次对每个查询给出“是”（用'1'表示）或“否”（用'0'表示）的答案，并将所有答案按顺序组成一个连续的字符串，无需分隔符。
|
||||
|
||||
请将最终答案放置在[answer]和[/answer]标签之间。例如：[answer]101[/answer]。"""
|
||||
return prompt
|
||||
|
||||
@staticmethod
|
||||
def extract_output(output):
|
||||
matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
|
||||
if not matches:
|
||||
return None
|
||||
return matches[-1].strip()
|
||||
|
||||
@classmethod
|
||||
def _verify_correction(cls, solution, identity):
|
||||
s = identity['S']
|
||||
t = identity['T']
|
||||
queries = identity['queries']
|
||||
sca, scb = compute_prefixes(s)
|
||||
tca, tcb = compute_prefixes(t)
|
||||
expected = []
|
||||
for a, b, c, d in queries:
|
||||
res = compute_query_result(sca, scb, tca, tcb, a, b, c, d)
|
||||
expected.append(res)
|
||||
expected_str = ''.join(expected)
|
||||
return solution == expected_str
|
||||
Loading…
Add table
Add a link
Reference in a new issue