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252 lines
6.1 KiB
Python
Executable file
252 lines
6.1 KiB
Python
Executable file
"""#
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### 谜题描述
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Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.
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Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.
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Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?
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Input
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The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).
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Output
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Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.
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If there are many possible integers k giving the same value of the least common multiple, print the smallest one.
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Examples
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Input
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6 10
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Output
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2
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Input
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21 31
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Output
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9
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Input
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5 10
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Output
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0
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Note
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In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
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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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import random
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import time
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seed = int(time.time())
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random.seed(seed)
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def gcd(a, b):
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if b > a:
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return gcd(b, a)
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while b > 0:
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a, b = b, a % b
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return a
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def lcm(a, b):
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return (a*b)//gcd(a, b)
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def solve_slow(a, b):
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best = lcm(a, b)
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best_k = 0
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for k in range(1, 20**2):
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cand = lcm(a+k, b+k)
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if cand < best:
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best = cand
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best_k = k
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return best_k
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def solve_slow2(a, b):
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if a > b:
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return solve_slow2(b, a)
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best = lcm(a, b)
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best_k = 0
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for m in range(2, a+1):
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if (a-b) % m == 0:
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print 'm:', m
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break
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return 0
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def solve(a, b):
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if a > b:
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return solve(b, a)
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elif a == b:
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return 0
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delta = b - a
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best = lcm(a, b)
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best_k = 0
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for fac in range(1, int(delta**.5)+1):
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if delta % fac != 0:
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continue
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k = fac - (a % fac)
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candidate = lcm(a+k, b+k)
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if candidate < best:
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best = candidate
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best_k = k
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invfac = delta//fac
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k = invfac - (a % invfac)
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candidate = lcm(a+k, b+k)
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if candidate < best:
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best = candidate
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best_k = k
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return best_k
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def test():
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print solve(random.randint(1, 10**9), random.randint(1, 10**9))
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return
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for a in range(1, 100):
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for b in range(1, 100):
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actual = solve(a, b)
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expected = solve_slow(a, b)
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if actual != expected:
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print 'death on a=%d b=%d: expected %d, got %d' % (a, b, expected, actual)
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return
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def main():
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xs = [int(x) for x in raw_input().strip().split()]
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print solve(*xs)
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if '__main__'==__name__:
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main()
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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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from bootcamp import Basebootcamp
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import random
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import re
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class Cnekodoesmathsbootcamp(Basebootcamp):
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def __init__(self, min_val=1, max_val=10**9):
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if min_val < 1:
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raise ValueError("min_val must be ≥1")
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if max_val < min_val:
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raise ValueError("max_val must be ≥ min_val")
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self.min_val = min_val
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self.max_val = max_val
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@staticmethod
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def gcd(a, b):
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while b:
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a, b = b, a % b
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return a
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@staticmethod
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def lcm(a, b):
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return a * b // Cnekodoesmathsbootcamp.gcd(a, b)
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@staticmethod
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def solve(a, b):
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if a == b:
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return 0
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if a > b:
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a, b = b, a
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delta = b - a
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best_k = 0
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best_lcm = Cnekodoesmathsbootcamp.lcm(a, b)
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factors = set()
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for fac in range(1, int(delta**0.5) + 1):
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if delta % fac == 0:
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factors.update({fac, delta//fac})
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for current_fac in sorted(factors, reverse=True):
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k = (current_fac - a % current_fac) % current_fac
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new_a = a + k
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new_b = b + k
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current_lcm = Cnekodoesmathsbootcamp.lcm(new_a, new_b)
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if (current_lcm < best_lcm) or (current_lcm == best_lcm and k < best_k):
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best_lcm = current_lcm
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best_k = k
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return best_k
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def case_generator(self):
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if random.random() < 0.2:
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case_type = random.choice([
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('equal', 1, 1),
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('small_diff', 5, 10),
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('prime_diff', 2, 5),
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('large_diff', 10**9-100, 10**9)
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])
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a, b = {
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'equal': (lambda: (x:=random.randint(1,10**9), x)),
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'small_diff': lambda: (random.randint(1,100), random.randint(1,100)+5),
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'prime_diff': lambda: (random.choice([2,3,5,7,11]), random.choice([2,3,5,7,11])+2),
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'large_diff': lambda: (10**9 - random.randint(1,1000), 10**9)
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}[case_type[0]]()
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else:
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a = random.randint(self.min_val, self.max_val)
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b = random.randint(self.min_val, self.max_val)
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return {
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'a': a,
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'b': b,
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'correct_k': self.solve(a, b)
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}
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@staticmethod
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def prompt_func(question_case):
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a = question_case['a']
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b = question_case['b']
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return f"""Neko遇到了一个数论问题：给定两个正整数a和b，找到最小非负整数k，使得a+k和b+k的最小公倍数最小。
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**问题参数**
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- a = {a}
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- b = {b}
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**求解要求**
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1. 如果存在多个k能得到相同的最小LCM，返回最小的k值
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2. 答案必须是非负整数
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3. 请将最终答案放在[answer]和[/answer]标签之间
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**示例格式**
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如果正确答案是0，应写：[answer]0[/answer]"""
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@staticmethod
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def extract_output(output):
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answers = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output)
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return int(answers[-1]) if answers else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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return solution == identity['correct_k']
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