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226 lines
6 KiB
Python
Executable file
226 lines
6 KiB
Python
Executable file
"""#
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### 谜题描述
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Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English letters, so he writes any number only as a decimal number, it means that instead of the letter A he will write the number 10. Thus, by converting the number 475 from decimal to hexadecimal system, he gets 11311 (475 = 1·162 + 13·161 + 11·160). Alexander lived calmly until he tried to convert the number back to the decimal number system.
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Alexander remembers that he worked with little numbers so he asks to find the minimum decimal number so that by converting it to the system with the base n he will get the number k.
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Input
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The first line contains the integer n (2 ≤ n ≤ 109). The second line contains the integer k (0 ≤ k < 1060), it is guaranteed that the number k contains no more than 60 symbols. All digits in the second line are strictly less than n.
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Alexander guarantees that the answer exists and does not exceed 1018.
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The number k doesn't contain leading zeros.
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Output
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Print the number x (0 ≤ x ≤ 1018) — the answer to the problem.
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Examples
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Input
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13
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12
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Output
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12
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Input
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16
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11311
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Output
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475
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Input
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20
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999
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Output
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3789
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Input
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17
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2016
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Output
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594
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Note
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In the first example 12 could be obtained by converting two numbers to the system with base 13: 12 = 12·130 or 15 = 1·131 + 2·130.
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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 = int(input())
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k = str(input())
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INF = 10**19
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pown = []
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tr = len(str(n))
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dp = [range(0,70) for i in range(0,70)]
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for i in range(0,70) :
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for j in range(0,70) :
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dp[i][j] = (-1,-1)
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pown.append(1)
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for i in range(1,70) :
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if pown[i-1] == INF :
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pown.append(INF)
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continue
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x = pown[i-1]
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kk = x*n
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if kk > INF :
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kk = INF
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pown.append(kk)
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def check(f) :
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if len(f) > tr :
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return INF
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if int(f) >= n :
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return INF
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return int(f)
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def DP(x,y) :
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if dp[x][y] != (-1,-1) :
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return dp[x][y]
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if x > y :
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dp[x][y] = (0,0)
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return dp[x][y]
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ret = (INF,0)
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for i in range(x,y) :
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a = DP(x,i)
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b = DP(i+1,y)
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pp = pown[b[1]]
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if pp == INF :
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continue
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p = b[0]+(a[0]* pp)
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if p < ret[0] :
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ret = (p,a[1]+b[1])
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f =\"\"
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for i in range(x,y+1) :
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f += k[i]
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e = check(f)
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#print x,y,ret,e
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dp[x][y] = ret
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if e < dp[x][y][0] :
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dp[x][y] = (e,1)
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return dp[x][y]
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print DP(0,len(k)-1)[0]
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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 bootcamp import Basebootcamp
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class Dabilitytoconvertbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.min_n = params.get('min_n', 2)
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self.max_n = params.get('max_n', 100) # 测试时采用较小范围
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self.max_k_length = params.get('max_k_length', 10)
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self.min_k_length = params.get('min_k_length', 1)
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def case_generator(self):
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for _ in range(1000):
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n = random.randint(self.min_n, self.max_n)
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allowed_digits = list(map(str, range(min(n, 10))))
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if not allowed_digits:
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allowed_digits = ['0']
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# 生成有效k_str（每个字符严格<n）
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k_length = random.randint(self.min_k_length, self.max_k_length)
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k_chars = [random.choice(allowed_digits) for _ in range(k_length)]
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k_str = ''.join(k_chars).lstrip('0') or '0'
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if len(k_str) > 60:
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continue
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# 计算正确答案
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try:
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expected_x = self.calculate_min_x(n, k_str)
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if expected_x is not None and 0 <= expected_x <= 1e18:
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return {'n': n, 'k': k_str, 'expected_x': expected_x}
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except Exception as e:
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continue
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# 保底用例
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return {'n': 10, 'k': '0', 'expected_x': 0}
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@staticmethod
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def prompt_func(case):
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return f"""Alexander将十进制数转换为n进制时，用十进制数字代替字母。例如，当n=16时，475转换为'11311'。现在给定n={case['n']}和k={case['k']}，请找出最小的十进制数x，使得x转换为n进制的Alexander表示正好是k。答案请写在[answer]和[/answer]之间。"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](\d+)\[/answer\]', output)
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return int(matches[-1]) if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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return solution == identity['expected_x']
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def calculate_min_x(self, n, k_str):
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INF = 10**19
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length = len(k_str)
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if length == 0:
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return 0
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# 预处理n的位数阈值
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tr = len(str(n))
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# 初始化权值数组
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pown = [1] * 70
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for i in range(1, 70):
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pown[i] = pown[i-1] * n if pown[i-1] <= INF // n else INF
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# DP表：dp[i][j] = (min_value, digits_count)
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dp = [[(INF, 0) for _ in range(length)] for __ in range(length)]
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# 填充DP表
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for l in range(1, length+1):
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for i in range(length - l + 1):
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j = i + l - 1
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current_str = k_str[i:j+1]
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# 候选1：整个子串作为单个数字
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if len(current_str) <= tr:
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num = int(current_str)
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if num < n and num < dp[i][j][0]:
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dp[i][j] = (num, 1)
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# 候选2：分割子串
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for mid in range(i, j):
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left_val, left_len = dp[i][mid]
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right_val, right_len = dp[mid+1][j]
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if right_len >= len(pown) or pown[right_len] == INF:
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continue
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combined = left_val * pown[right_len] + right_val
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if combined < dp[i][j][0] and combined <= INF:
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dp[i][j] = (combined, left_len + right_len)
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return dp[0][length-1][0] if dp[0][length-1][0] <= 1e18 else None
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