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179 lines
5.1 KiB
Python
Executable file
179 lines
5.1 KiB
Python
Executable file
"""#
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### 谜题描述
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Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e. <image>.
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Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.
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<image> is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).
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Input
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The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.
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The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.
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Output
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Output on the first line \"YES\" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and \"NO\" (without quotes) otherwise.
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If the answer was \"YES\", output the minimal number of moves needed to make sequence A beautiful.
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Examples
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Input
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2
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1 1
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Output
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YES
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1
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Input
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3
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6 2 4
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Output
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YES
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0
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Input
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2
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1 3
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Output
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YES
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1
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Note
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In the first example you can simply make one move to obtain sequence [0, 2] with <image>.
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In the second example the gcd of the sequence is already greater than 1.
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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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from fractions import gcd
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n = int(raw_input())
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p = [int(x) for x in raw_input().split()]
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g = p[0]
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ans = 0
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cnt = 0
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for x in p:
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g = gcd(x,g)
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if(x %2 == 0):
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ans += (cnt/2) + 2*(cnt%2)
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cnt = 0
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else: cnt += 1
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ans += (cnt/2) + 2*(cnt%2)
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print \"YES\"
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if g > 1:
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print 0
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else:
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print 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 re
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import random
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from math import gcd
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from functools import reduce
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from bootcamp import Basebootcamp
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class Cmikeandgcdproblembootcamp(Basebootcamp):
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def __init__(self, n_min=2, n_max=10**5, max_value=10**9):
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self.n_min = n_min
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self.n_max = n_max
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self.max_value = max_value
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def case_generator(self):
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n = random.randint(self.n_min, self.n_max)
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# 决定生成类型：50%概率生成可立即解决的案例，50%需要计算的案例
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if random.choice([True, False]):
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# 生成初始gcd>1的情况
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d = random.randint(2, 5)
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a = [d * random.randint(1, self.max_value//d) for _ in range(n)]
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else:
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# 生成初始gcd=1的数组（必须通过操作解决）
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while True:
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# 生成至少两个互质数确保整体gcd=1
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a = []
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# 生成前两个互质数
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x, y = random.sample(range(1, self.max_value//2 +1), 2)
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while gcd(x, y) != 1:
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x, y = random.sample(range(1, self.max_value//2 +1), 2)
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a.extend([x, y])
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# 生成其余元素
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for _ in range(n-2):
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a.append(random.randint(1, self.max_value))
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# 验证整体gcd
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if reduce(gcd, a) == 1:
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break
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return {'n': n, 'a': a}
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@staticmethod
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def prompt_func(question_case):
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a_str = ' '.join(map(str, question_case['a']))
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return f"""给定长度为{question_case['n']}的数列: {a_str}
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请你判断是否可以通过替换相邻两数为它们的差和和（操作次数最少），使得数列所有元素的最大公约数大于1。若可以，输出YES并给出最少操作数，否则输出NO。答案格式：[answer]YES\nX[/answer] 或 [answer]NO[/answer]。"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\]\s*(.*?)\s*\[/answer\]', output, re.DOTALL | re.IGNORECASE)
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if not matches:
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return None
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content = matches[-1].strip().upper()
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lines = [line.strip() for line in content.split('\n')]
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if not lines:
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return None
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if lines[0] == 'YES':
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if len(lines) < 2:
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return None
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try:
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steps = int(lines[1])
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return ('YES', steps)
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except ValueError:
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return None
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elif lines[0] == 'NO':
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return ('NO', None)
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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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a = identity['a']
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current_gcd = reduce(gcd, a)
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# 初始gcd>1的验证
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if current_gcd > 1:
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return solution == ('YES', 0)
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# 必须通过操作的情况
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cnt = ans = 0
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for x in a:
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if x % 2 == 0:
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ans += (cnt // 2) + 2 * (cnt % 2)
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cnt = 0
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else:
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cnt += 1
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ans += (cnt // 2) + 2 * (cnt % 2)
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return solution == ('YES', ans)
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