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internbootcamp/bootcamp/conebasedarithmetic/conebasedarithmetic.py

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+"""# 
+
+### 谜题描述
+Prof. Vasechkin wants to represent positive integer n as a sum of addends, where each addends is an integer number containing only 1s. For example, he can represent 121 as 121=111+11+–1. Help him to find the least number of digits 1 in such sum.
+
+Input
+
+The first line of the input contains integer n (1 ≤ n < 1015).
+
+Output
+
+Print expected minimal number of digits 1.
+
+Examples
+
+Input
+
+121
+
+
+Output
+
+6
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+def solve(n,p):
+	aparitii = n/a[p];
+	if(n%a[p]==0):
+		return aparitii*p;
+	sol1 = aparitii * p + solve(n%a[p],p-1);
+	sol2 = (aparitii+1) * p + solve(a[p]*(aparitii+1)-n,p-1);
+	return min(sol1,sol2);
+		
+
+a = [];
+n = int(input());
+
+a.append(0);
+for i in range(1,18):
+	a.append(a[i-1]*10+1);
+
+print solve(n,16)
+```
+
+
+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 functools import lru_cache
+from bootcamp import Basebootcamp
+
+class Conebasedarithmeticbootcamp(Basebootcamp):
+    def __init__(self, min_n=1, max_n=10**15-1):
+        self.min_n = min_n
+        self.max_n = max_n
+        # 预处理生成1~17位的全1数
+        self.a = [0]
+        for i in range(1, 17+1):
+            self.a.append(self.a[-1] * 10 + 1)
+    
+    def case_generator(self):
+        # 生成三类测试用例：全1数、接近全1数的边缘值、随机数
+        rand = random.random()
+        if rand < 0.2:  # 全1数
+            digits = random.randint(1, 17)
+            n = self.a[digits]
+        elif rand < 0.4:  # 全1数±1
+            digits = random.randint(2, 16)
+            base = self.a[digits]
+            n = base + random.choice([-1, 1])
+        else:  # 普通随机数
+            n = random.randint(self.min_n, min(self.max_n, 10**17))
+        return {'n': n}
+    
+    @staticmethod
+    def prompt_func(question_case):
+        n = question_case['n']
+        return (
+            f"将整数 {n} 表示为全1数字的加减组合，找到最少需要使用的1的总个数。\n"
+            "规则说明：\n"
+            "1. 每个加数/减数必须是由连续1组成的整数（如11, 111等）\n"
+            "2. 允许使用加减法组合，例如：121 = 111 + 11 - 1\n"
+            "3. 要求最终解中使用的1字符总数最少\n\n"
+            "请给出最少需要的1的总个数，并置于[answer]标签内，例如：[answer]6[/answer]"
+        )
+    
+    @staticmethod
+    def extract_output(output):
+        import re
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        try:
+            return int(matches[-1].strip()) if matches else None
+        except:
+            return None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        try:
+            n = identity['n']
+            instance = cls()  # 创建带预计算数据的实例
+            return int(solution) == instance.calculate_min_ones(n)
+        except:
+            return False
+
+    @lru_cache(maxsize=None)
+    def calculate_min_ones(self, n):
+        def dfs(n, p):
+            if p == 0:
+                return float('inf') if n !=0 else 0
+            
+            current = self.a[p]
+            if current == 0:
+                return dfs(n, p-1)
+            
+            quotient, remainder = divmod(n, current)
+            
+            if remainder == 0:
+                return quotient * p
+            
+            # 正向处理方案
+            option1 = quotient * p + dfs(remainder, p-1)
+            # 溢出处理方案（多用一个current）
+            option2 = (quotient + 1) * p + dfs(current*(quotient+1)-n, p-1)
+            
+            return min(option1, option2)
+        
+        return dfs(n, p=len(self.a)-1)  # 从最大位数开始处理