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130 lines
3.8 KiB
Python
Executable file
130 lines
3.8 KiB
Python
Executable file
"""#
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### 谜题描述
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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.
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Input
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The first line of the input contains integer n (1 ≤ n < 1015).
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Output
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Print expected minimal number of digits 1.
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Examples
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Input
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121
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Output
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6
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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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def solve(n,p):
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aparitii = n/a[p];
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if(n%a[p]==0):
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return aparitii*p;
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sol1 = aparitii * p + solve(n%a[p],p-1);
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sol2 = (aparitii+1) * p + solve(a[p]*(aparitii+1)-n,p-1);
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return min(sol1,sol2);
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a = [];
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n = int(input());
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a.append(0);
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for i in range(1,18):
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a.append(a[i-1]*10+1);
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print solve(n,16)
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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 functools import lru_cache
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from bootcamp import Basebootcamp
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class Conebasedarithmeticbootcamp(Basebootcamp):
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def __init__(self, min_n=1, max_n=10**15-1):
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self.min_n = min_n
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self.max_n = max_n
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# 预处理生成1~17位的全1数
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self.a = [0]
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for i in range(1, 17+1):
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self.a.append(self.a[-1] * 10 + 1)
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def case_generator(self):
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# 生成三类测试用例：全1数、接近全1数的边缘值、随机数
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rand = random.random()
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if rand < 0.2: # 全1数
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digits = random.randint(1, 17)
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n = self.a[digits]
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elif rand < 0.4: # 全1数±1
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digits = random.randint(2, 16)
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base = self.a[digits]
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n = base + random.choice([-1, 1])
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else: # 普通随机数
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n = random.randint(self.min_n, min(self.max_n, 10**17))
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return {'n': n}
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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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return (
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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"
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"规则说明：\n"
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"1. 每个加数/减数必须是由连续1组成的整数（如11, 111等）\n"
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"2. 允许使用加减法组合，例如：121 = 111 + 11 - 1\n"
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"3. 要求最终解中使用的1字符总数最少\n\n"
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"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]"
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)
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@staticmethod
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def extract_output(output):
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import re
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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try:
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return int(matches[-1].strip()) if matches else None
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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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try:
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n = identity['n']
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instance = cls() # 创建带预计算数据的实例
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return int(solution) == instance.calculate_min_ones(n)
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except:
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return False
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@lru_cache(maxsize=None)
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def calculate_min_ones(self, n):
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def dfs(n, p):
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if p == 0:
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return float('inf') if n !=0 else 0
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current = self.a[p]
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if current == 0:
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return dfs(n, p-1)
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quotient, remainder = divmod(n, current)
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if remainder == 0:
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return quotient * p
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# 正向处理方案
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option1 = quotient * p + dfs(remainder, p-1)
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# 溢出处理方案（多用一个current）
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option2 = (quotient + 1) * p + dfs(current*(quotient+1)-n, p-1)
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return min(option1, option2)
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return dfs(n, p=len(self.a)-1) # 从最大位数开始处理
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