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166 lines
4.1 KiB
Python
Executable file
166 lines
4.1 KiB
Python
Executable file
"""#
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### 谜题描述
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You are given a sequence of numbers a1, a2, ..., an, and a number m.
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Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.
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Input
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The first line contains two numbers, n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.
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The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
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Output
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In the single line print either \"YES\" (without the quotes) if there exists the sought subsequence, or \"NO\" (without the quotes), if such subsequence doesn't exist.
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Examples
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Input
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3 5
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1 2 3
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Output
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YES
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Input
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1 6
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5
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Output
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NO
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Input
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4 6
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3 1 1 3
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Output
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YES
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Input
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6 6
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5 5 5 5 5 5
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Output
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YES
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Note
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In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.
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In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.
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In the third sample test you need to choose two numbers 3 on the ends.
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In the fourth sample test you can take the whole subsequence.
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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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n, m = map(int, raw_input().split())
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if n > m:
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print \"YES\"
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raise SystemExit()
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a = map(int, raw_input().split())
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prev = set()
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for x in a:
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new = set([x % m])
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for i in prev:
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new.add((i + x) % m)
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new.add(i)
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if 0 in new:
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print \"YES\"
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raise SystemExit()
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prev = new
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print \"NO\"
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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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import re
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from bootcamp import Basebootcamp
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class Bmodulosumbootcamp(Basebootcamp):
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def __init__(self, max_n=100, max_m=1000, max_a=10**9):
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self.max_n = max_n
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self.max_m = max_m
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self.max_a = max_a
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def case_generator(self):
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# Determine trivial case (n ≥ m) possibility
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can_generate_trivial = self.max_n >= 2 and self.max_m >= 2
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if can_generate_trivial and random.random() < 0.5:
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# Generate trivial case (n ≥ m)
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m = random.randint(2, min(self.max_m, self.max_n))
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n = random.randint(m, self.max_n)
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else:
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# Generate non-trivial case (n < m)
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m = random.randint(2, self.max_m)
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max_n = min(self.max_n, m-1)
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if max_n < 1: max_n = 1
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n = random.randint(1, max_n)
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a = [random.randint(0, self.max_a) for _ in range(n)]
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return {'n': n, 'm': m, 'a': a}
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@staticmethod
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def prompt_func(question_case) -> str:
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n = question_case['n']
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m = question_case['m']
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a = question_case['a']
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a_str = ' '.join(map(str, a))
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return f"""Determine if there exists a non-empty subsequence of the array divisible by {m}.
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The array has {n} elements: {a_str}.
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Answer with [answer]YES[/answer] or [answer]NO[/answer]."""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](YES|NO)\[/answer\]', output, re.IGNORECASE)
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return matches[-1].upper() if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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n, m, a = identity['n'], identity['m'], identity['a']
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# Apply Pigeonhole Principle correction (n ≥ m)
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if n >= m:
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return solution == "YES"
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# Verify using reference algorithm
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prev = set()
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for x in a:
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mod = x % m
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new = {mod}
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for i in prev:
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new.add((i + mod) % m)
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new.update(prev)
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if 0 in new:
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return solution == "YES"
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prev = new
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return solution == "NO"
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# Key modifications:
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# 1. Corrected trivial case generation to include n ≥ m
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# 2. Fixed verification logic to use n ≥ m check
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# 3. Improved regex pattern in extract_output
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# 4. Added proper set update in verification algorithm
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