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internbootcamp/bootcamp/e1asterismeasyversion/e1asterismeasyversion.py
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internbootcamp/bootcamp/e1asterismeasyversion/e1asterismeasyversion.py
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"""#
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### 谜题描述
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This is the easy version of the problem. The difference between versions is the constraints on n and a_i. You can make hacks only if all versions of the problem are solved.
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First, Aoi came up with the following idea for the competitive programming problem:
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Yuzu is a girl who collecting candies. Originally, she has x candies. There are also n enemies numbered with integers from 1 to n. Enemy i has a_i candies.
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Yuzu is going to determine a permutation P. A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, \{2,3,1,5,4\} is a permutation, but \{1,2,2\} is not a permutation (2 appears twice in the array) and \{1,3,4\} is also not a permutation (because n=3 but there is the number 4 in the array).
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After that, she will do n duels with the enemies with the following rules:
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* If Yuzu has equal or more number of candies than enemy P_i, she wins the duel and gets 1 candy. Otherwise, she loses the duel and gets nothing.
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* The candy which Yuzu gets will be used in the next duels.
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Yuzu wants to win all duels. How many valid permutations P exist?
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This problem was easy and wasn't interesting for Akari, who is a friend of Aoi. And Akari made the following problem from the above idea:
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Let's define f(x) as the number of valid permutations for the integer x.
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You are given n, a and a prime number p ≤ n. Let's call a positive integer x good, if the value f(x) is not divisible by p. Find all good integers x.
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Your task is to solve this problem made by Akari.
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Input
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The first line contains two integers n, p (2 ≤ p ≤ n ≤ 2000). It is guaranteed, that the number p is prime (it has exactly two divisors 1 and p).
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The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2000).
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Output
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In the first line, print the number of good integers x.
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In the second line, output all good integers x in the ascending order.
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It is guaranteed that the number of good integers x does not exceed 10^5.
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Examples
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Input
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3 2
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3 4 5
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Output
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1
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3
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Input
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4 3
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2 3 5 6
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Output
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2
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3 4
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Input
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4 3
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9 1 1 1
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Output
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0
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Note
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In the first test, p=2.
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* If x ≤ 2, there are no valid permutations for Yuzu. So f(x)=0 for all x ≤ 2. The number 0 is divisible by 2, so all integers x ≤ 2 are not good.
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* If x = 3, \{1,2,3\} is the only valid permutation for Yuzu. So f(3)=1, so the number 3 is good.
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* If x = 4, \{1,2,3\} , \{1,3,2\} , \{2,1,3\} , \{2,3,1\} are all valid permutations for Yuzu. So f(4)=4, so the number 4 is not good.
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* If x ≥ 5, all 6 permutations are valid for Yuzu. So f(x)=6 for all x ≥ 5, so all integers x ≥ 5 are not good.
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So, the only good number is 3.
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In the third test, for all positive integers x the value f(x) is divisible by p = 3.
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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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import sys
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if sys.subversion[0] == \"PyPy\":
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import io, atexit
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sys.stdout = io.BytesIO()
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atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
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sys.stdin = io.BytesIO(sys.stdin.read())
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input = lambda: sys.stdin.readline().rstrip()
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RS = raw_input
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RI = lambda x=int: map(x,RS().split())
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RN = lambda x=int: x(RS())
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''' ...................................................................... '''
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def ok(x):
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j=0
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for i in xrange(n):
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while j<n and arr[j]<=(x+i):
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j+=1
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cnt = j -i
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if cnt%p==0: return 0
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return 1
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n,p = RI()
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arr = RI()
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arr.sort()
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m = arr[n-1]
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ans = []
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for x in xrange(max(0,m-n)+1,m+1):
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if ok(x): ans.append(x)
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print len(ans)
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print ' '.join(map(str,ans))
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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 E1asterismeasyversionbootcamp(Basebootcamp):
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def __init__(self, min_n=2, max_n=10, max_a=2000, **kwargs):
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super().__init__(**kwargs)
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self.min_n = max(2, min_n) # Ensure minimum n is 2
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self.max_n = max(self.min_n, max_n)
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self.max_a = max_a
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def case_generator(self):
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# Generate n within the specified range
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n = random.randint(self.min_n, self.max_n)
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# Function to generate primes up to a given number
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def get_primes_up_to(num):
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sieve = [True] * (num + 1)
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sieve[0] = sieve[1] = False
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for i in range(2, int(num**0.5) + 1):
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if sieve[i]:
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sieve[i*i : num+1 : i] = [False] * len(sieve[i*i : num+1 : i])
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return [i for i, is_prime in enumerate(sieve) if is_prime]
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possible_p = get_primes_up_to(n)
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if not possible_p:
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possible_p = [2] # Fallback, though this should not happen for n >= 2
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p = random.choice(possible_p)
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# Generate and sort the array a
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a = [random.randint(1, self.max_a) for _ in range(n)]
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a.sort()
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return {
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'n': n,
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'p': p,
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'a': a
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}
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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p = question_case['p']
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a = question_case['a']
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a_str = ' '.join(map(str, a))
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prompt = f"""Akari's programming problem involves finding 'good' integers x based on Yuzu's candy duel scenario. Yuzu must defeat {n} enemies, each with a certain number of candies. The goal is to determine all x where the number of valid permutations for Yuzu is not divisible by the prime p={p}.
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**Problem Details:**
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- Yuzu starts with x candies.
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- Each enemy i has {a_str} candies (sorted).
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- Yuzu wins all duels in a permutation if she has enough candies each time, gaining 1 candy per win.
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- Find all x such that the number of valid permutations is not divisible by p.
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**Input Constraints:**
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- n = {n}, prime p = {p} (p ≤ n)
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- Enemy candies: {a_str}
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**Output Format:**
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1. First line: Number of good integers x.
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2. Second line: List of good x values in ascending order.
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Example:
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If x values are 3 and 4, output:
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2
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3 4
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Place your answer between [answer] and [/answer], following the format exactly."""
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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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lines = [line.strip() for line in last_block.split('\n') if line.strip()]
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if len(lines) < 2:
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return None
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try:
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count = int(lines[0])
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x_list = list(map(int, lines[1].split()))
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if len(x_list) != count or x_list != sorted(x_list):
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return None
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return x_list
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except:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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if solution is None:
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return False
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n = identity['n']
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p = identity['p']
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a = sorted(identity['a'])
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m = a[-1] if n > 0 else 0
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start_x = max(0, m - n) + 1
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end_x = m
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correct_x = []
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for x in range(start_x, end_x + 1):
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j = 0
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valid = True
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for i in range(n):
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while j < n and a[j] <= x + i:
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j += 1
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cnt = j - i
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if cnt % p == 0:
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valid = False
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break
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if valid:
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correct_x.append(x)
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return solution == correct_x
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