mirror of
https://github.com/InternLM/InternBootcamp.git
synced 2026-04-19 12:58:04 +00:00
266 lines
8.3 KiB
Python
Executable file
266 lines
8.3 KiB
Python
Executable file
"""#
|
||
|
||
### 谜题描述
|
||
Grigory loves strings. Recently he found a metal strip on a loft. The strip had length n and consisted of letters \"V\" and \"K\". Unfortunately, rust has eaten some of the letters so that it's now impossible to understand which letter was written.
|
||
|
||
Grigory couldn't understand for a long time what these letters remind him of, so he became interested in the following question: if we put a letter \"V\" or \"K\" on each unreadable position, which values can the period of the resulting string be equal to?
|
||
|
||
A period of a string is such an integer d from 1 to the length of the string that if we put the string shifted by d positions to the right on itself, then all overlapping letters coincide. For example, 3 and 5 are periods of \"VKKVK\".
|
||
|
||
Input
|
||
|
||
There are several (at least one) test cases in the input. The first line contains single integer — the number of test cases.
|
||
|
||
There is an empty line before each test case. Each test case is described in two lines: the first line contains single integer n (1 ≤ n ≤ 5·105) — the length of the string, the second line contains the string of length n, consisting of letters \"V\", \"K\" and characters \"?\". The latter means the letter on its position is unreadable.
|
||
|
||
It is guaranteed that the sum of lengths among all test cases doesn't exceed 5·105.
|
||
|
||
For hacks you can only use tests with one test case.
|
||
|
||
Output
|
||
|
||
For each test case print two lines. In the first line print the number of possible periods after we replace each unreadable letter with \"V\" or \"K\". In the next line print all these values in increasing order.
|
||
|
||
Example
|
||
|
||
Input
|
||
|
||
3
|
||
|
||
5
|
||
V??VK
|
||
|
||
6
|
||
??????
|
||
|
||
4
|
||
?VK?
|
||
|
||
|
||
Output
|
||
|
||
2
|
||
3 5
|
||
6
|
||
1 2 3 4 5 6
|
||
3
|
||
2 3 4
|
||
|
||
Note
|
||
|
||
In the first test case from example we can obtain, for example, \"VKKVK\", which has periods 3 and 5.
|
||
|
||
In the second test case we can obtain \"VVVVVV\" which has all periods from 1 to 6.
|
||
|
||
In the third test case string \"KVKV\" has periods 2 and 4, and string \"KVKK\" has periods 3 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;
|
||
const int N = (1 << 20);
|
||
const double pi = acos(-1);
|
||
inline int read() {
|
||
int x = 0, f = 1;
|
||
char ch = getchar();
|
||
while (ch < '0' || ch > '9') {
|
||
if (ch == '-') f = -1;
|
||
ch = getchar();
|
||
}
|
||
while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar();
|
||
return x * f;
|
||
}
|
||
struct CP {
|
||
double x, y;
|
||
CP(double xx = 0, double yy = 0) { x = xx, y = yy; }
|
||
};
|
||
CP operator+(CP aa, CP bb) { return CP(aa.x + bb.x, aa.y + bb.y); }
|
||
CP operator-(CP aa, CP bb) { return CP(aa.x - bb.x, aa.y - bb.y); }
|
||
CP operator*(CP aa, CP bb) {
|
||
return CP(aa.x * bb.x - aa.y * bb.y, aa.x * bb.y + aa.y * bb.x);
|
||
}
|
||
CP operator*(CP aa, double bb) { return CP(aa.x * bb, aa.y * bb); }
|
||
int pp[N], lim, Lim;
|
||
CP Pow[N], iPow[N];
|
||
void up(int x) {
|
||
for (lim = 1; lim <= x; lim <<= 1)
|
||
;
|
||
}
|
||
void revlim() {
|
||
for (int i = 0, iE = lim - 1; i <= iE; i++)
|
||
pp[i] = ((pp[i >> 1] >> 1) | ((i & 1) * (lim >> 1)));
|
||
}
|
||
void init(int x) {
|
||
up(x), Lim = lim;
|
||
Pow[0] = iPow[0] = CP(1, 0);
|
||
for (int i = 1, iE = Lim - 1; i <= iE; i++)
|
||
Pow[i] = iPow[Lim - i] = CP(cos(2 * pi / Lim * i), sin(2 * pi / Lim * i));
|
||
}
|
||
void FFT(CP *f, int flag) {
|
||
for (int i = 0, iE = lim - 1; i <= iE; i++)
|
||
if (pp[i] < i) swap(f[pp[i]], f[i]);
|
||
for (int i = 2; i <= lim; i <<= 1)
|
||
for (int j = 0, l = (i >> 1), ch = Lim / i; j < lim; j += i) {
|
||
for (int k = j, now = 0; k < j + l; k++) {
|
||
CP pa = f[k], pb = f[k + l] * (flag == 1 ? Pow[now] : iPow[now]);
|
||
f[k] = pa + pb, f[k + l] = pa - pb, now += ch;
|
||
}
|
||
}
|
||
if (flag == -1)
|
||
for (int i = 0, iE = lim - 1; i <= iE; i++) f[i].x /= lim, f[i].y /= lim;
|
||
}
|
||
int n, m, a[N], f[N], tot, g[N];
|
||
CP A[N], AA[N], AAA[N];
|
||
CP B[N], BB[N], BBB[N];
|
||
char s[N];
|
||
void Main() {
|
||
n = read(), scanf(\"%s\", s);
|
||
up(n << 1), revlim();
|
||
for (int i = 0, iE = lim - 1; i <= iE; i++)
|
||
A[i] = B[i] = AA[i] = BB[i] = AAA[i] = BBB[i] = CP(0, 0);
|
||
for (int i = 0, iE = n - 1; i <= iE; i++) {
|
||
int o;
|
||
if (s[i] == '?') o = 0;
|
||
if (s[i] == 'V') o = 1;
|
||
if (s[i] == 'K') o = 2;
|
||
B[n - 1 - i].x = A[i].x = o;
|
||
BB[n - 1 - i].x = AA[i].x = o * o;
|
||
BBB[n - 1 - i].x = AAA[i].x = o * o * o;
|
||
;
|
||
}
|
||
FFT(A, 1), FFT(AA, 1), FFT(AAA, 1);
|
||
FFT(B, 1), FFT(BB, 1), FFT(BBB, 1);
|
||
for (int i = 0, iE = lim - 1; i <= iE; i++)
|
||
A[i] = A[i] * BBB[i] + AAA[i] * B[i] - AA[i] * BB[i] * 2;
|
||
FFT(A, -1);
|
||
for (int i = 0, iE = n - 1; i <= iE; i++) f[i] = A[n - 1 - i].x;
|
||
tot = 0;
|
||
for (int i = 1, iE = n; i <= iE; i++) {
|
||
bool ok = 1;
|
||
for (int j = 1, jE = (n - 1) / i; j <= jE; j++)
|
||
if (f[i * j]) ok = 0;
|
||
if (ok) g[++tot] = i;
|
||
}
|
||
printf(\"%d\n\", tot);
|
||
for (int i = 1, iE = tot; i <= iE; i++) printf(\"%d \", g[i]);
|
||
puts(\"\");
|
||
}
|
||
int main() {
|
||
init(1e6);
|
||
int T = read();
|
||
while (T--) Main();
|
||
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
|
||
from bootcamp import Basebootcamp
|
||
|
||
class Erustystringbootcamp(Basebootcamp):
|
||
def __init__(self, max_length=5, max_questions=3):
|
||
self.max_length = max_length
|
||
self.max_questions = max_questions
|
||
|
||
def case_generator(self):
|
||
n = random.randint(1, self.max_length)
|
||
pattern_type = random.choice(['all_v', 'all_k', 'alternating', 'random'])
|
||
original = []
|
||
if pattern_type == 'all_v':
|
||
original = ['V'] * n
|
||
elif pattern_type == 'all_k':
|
||
original = ['K'] * n
|
||
elif pattern_type == 'alternating':
|
||
original = ['V' if i%2 == 0 else 'K' for i in range(n)]
|
||
else:
|
||
original = [random.choice(['V', 'K']) for _ in range(n)]
|
||
|
||
num_q = random.randint(0, min(self.max_questions, n))
|
||
q_indices = random.sample(range(n), k=num_q)
|
||
s = original.copy()
|
||
for i in q_indices:
|
||
s[i] = '?'
|
||
s_str = ''.join(s)
|
||
|
||
correct_periods = self.calculate_possible_periods(s_str)
|
||
correct_periods.sort()
|
||
return {
|
||
'n': n,
|
||
's': s_str,
|
||
'correct_periods': correct_periods
|
||
}
|
||
|
||
def calculate_possible_periods(self, s):
|
||
n = len(s)
|
||
valid_periods = []
|
||
|
||
for d in range(1, n+1):
|
||
valid = True
|
||
for r in range(d): # Check each residue group
|
||
has_v = False
|
||
has_k = False
|
||
# Check all positions in this residue group
|
||
for pos in range(r, n, d):
|
||
char = s[pos]
|
||
if char == 'V':
|
||
has_v = True
|
||
elif char == 'K':
|
||
has_k = True
|
||
# Conflict detected in this group
|
||
if has_v and has_k:
|
||
valid = False
|
||
break
|
||
if not valid:
|
||
break
|
||
if valid:
|
||
valid_periods.append(d)
|
||
|
||
return valid_periods
|
||
|
||
@staticmethod
|
||
def prompt_func(question_case) -> str:
|
||
n = question_case['n']
|
||
s = question_case['s']
|
||
prompt = f"""You are a programming competition participant. Solve the following string period puzzle.
|
||
|
||
Problem Description:
|
||
A string's period is an integer d (1 ≤ d ≤ n) where for all positions i, characters at i and i+d are equal (where i+d < n). Determine all possible periods after replacing '?' with 'V' or 'K'.
|
||
|
||
Input:
|
||
- A string of length {n}: {s}
|
||
|
||
Output Format:
|
||
Two lines:
|
||
1. Number of valid periods
|
||
2. Sorted valid periods separated by spaces
|
||
|
||
Enclose your answer between [answer] and [/answer]. Example:
|
||
[answer]
|
||
3
|
||
2 3 4
|
||
[/answer]"""
|
||
return prompt
|
||
|
||
@staticmethod
|
||
def extract_output(output):
|
||
start = output.rfind('[answer]')
|
||
end = output.rfind('[/answer]')
|
||
if start == -1 or end == -1 or start >= end:
|
||
return None
|
||
|
||
content = output[start+8:end].strip().split('\n')
|
||
try:
|
||
if len(content) < 2:
|
||
return None
|
||
periods = list(map(int, content[1].strip().split()))
|
||
return periods
|
||
except:
|
||
return None
|
||
|
||
@classmethod
|
||
def _verify_correction(cls, solution, identity):
|
||
return solution == identity['correct_periods']
|