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internbootcamp/bootcamp/cmikeandgcdproblem/cmikeandgcdproblem.py

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+"""# 
+
+### 谜题描述
+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>. 
+
+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.
+
+<image> is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).
+
+Input
+
+The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.
+
+The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.
+
+Output
+
+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.
+
+If the answer was \"YES\", output the minimal number of moves needed to make sequence A beautiful.
+
+Examples
+
+Input
+
+2
+1 1
+
+
+Output
+
+YES
+1
+
+
+Input
+
+3
+6 2 4
+
+
+Output
+
+YES
+0
+
+
+Input
+
+2
+1 3
+
+
+Output
+
+YES
+1
+
+Note
+
+In the first example you can simply make one move to obtain sequence [0, 2] with <image>.
+
+In the second example the gcd of the sequence is already greater than 1. 
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+from fractions import gcd
+
+n = int(raw_input())
+p = [int(x) for x in raw_input().split()]
+
+g = p[0]
+ans = 0
+cnt = 0
+
+for x in p:
+    g = gcd(x,g)
+    if(x %2 == 0):
+        ans += (cnt/2) + 2*(cnt%2)
+        cnt = 0
+    else: cnt += 1
+
+ans += (cnt/2) + 2*(cnt%2)
+
+print \"YES\"
+if g > 1:
+    print 0
+else:
+    print ans
+```
+
+
+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 re
+import random
+from math import gcd
+from functools import reduce
+from bootcamp import Basebootcamp
+
+class Cmikeandgcdproblembootcamp(Basebootcamp):
+    def __init__(self, n_min=2, n_max=10**5, max_value=10**9):
+        self.n_min = n_min
+        self.n_max = n_max
+        self.max_value = max_value
+    
+    def case_generator(self):
+        n = random.randint(self.n_min, self.n_max)
+        # 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
+        if random.choice([True, False]):
+            # 生成初始gcd>1的情况
+            d = random.randint(2, 5)
+            a = [d * random.randint(1, self.max_value//d) for _ in range(n)]
+        else:
+            # 生成初始gcd=1的数组（必须通过操作解决）
+            while True:
+                # 生成至少两个互质数确保整体gcd=1
+                a = []
+                # 生成前两个互质数
+                x, y = random.sample(range(1, self.max_value//2 +1), 2)
+                while gcd(x, y) != 1:
+                    x, y = random.sample(range(1, self.max_value//2 +1), 2)
+                a.extend([x, y])
+                # 生成其余元素
+                for _ in range(n-2):
+                    a.append(random.randint(1, self.max_value))
+                # 验证整体gcd
+                if reduce(gcd, a) == 1:
+                    break
+        return {'n': n, 'a': a}
+    
+    @staticmethod
+    def prompt_func(question_case):
+        a_str = ' '.join(map(str, question_case['a']))
+        return f"""给定长度为{question_case['n']}的数列: {a_str}
+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]。"""
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\]\s*(.*?)\s*\[/answer\]', output, re.DOTALL | re.IGNORECASE)
+        if not matches:
+            return None
+        content = matches[-1].strip().upper()
+        lines = [line.strip() for line in content.split('\n')]
+        if not lines:
+            return None
+        if lines[0] == 'YES':
+            if len(lines) < 2:
+                return None
+            try:
+                steps = int(lines[1])
+                return ('YES', steps)
+            except ValueError:
+                return None
+        elif lines[0] == 'NO':
+            return ('NO', None)
+        return None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        a = identity['a']
+        current_gcd = reduce(gcd, a)
+        
+        # 初始gcd>1的验证
+        if current_gcd > 1:
+            return solution == ('YES', 0)
+        
+        # 必须通过操作的情况
+        cnt = ans = 0
+        for x in a:
+            if x % 2 == 0:
+                ans += (cnt // 2) + 2 * (cnt % 2)
+                cnt = 0
+            else:
+                cnt += 1
+        ans += (cnt // 2) + 2 * (cnt % 2)
+        return solution == ('YES', ans)