結果
問題 |
No.1611 Minimum Multiple with Double Divisors
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ユーザー |
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提出日時 | 2025-04-16 01:13:09 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
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実行時間 | - |
コード長 | 2,364 bytes |
コンパイル時間 | 569 ms |
コンパイル使用メモリ | 82,572 KB |
実行使用メモリ | 106,588 KB |
最終ジャッジ日時 | 2025-04-16 01:15:08 |
合計ジャッジ時間 | 23,979 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | -- * 2 |
other | AC * 1 WA * 10 TLE * 1 -- * 25 |
ソースコード
import sys import random def is_prime(n): if n < 2: return False for p in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]: if n % p == 0: return n == p d = n - 1 s = 0 while d % 2 == 0: d //= 2 s += 1 for a in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]: if a >= n: continue x = pow(a, d, n) if x == 1 or x == n - 1: continue for _ in range(s - 1): x = pow(x, 2, n) if x == n - 1: break else: return False return True def pollards_rho(n): if n % 2 == 0: return 2 if n % 3 == 0: return 3 if n % 5 == 0: return 5 while True: c = random.randint(1, n - 1) f = lambda x: (pow(x, 2, n) + c) % n x, y, d = 2, 2, 1 while d == 1: x = f(x) y = f(f(y)) d = gcd(abs(x - y), n) if d != n: return d def gcd(a, b): while b: a, b = b, a % b return a def factor(n): factors = [] def _factor(n): if n == 1: return if is_prime(n): factors.append(n) return d = pollards_rho(n) _factor(d) _factor(n // d) _factor(n) return factors def factorize(n): if n == 1: return {} factors = factor(n) res = {} for p in factors: res[p] = res.get(p, 0) + 1 return res def find_min_p(primes_set): candidate = 2 while True: if candidate not in primes_set: if is_prime(candidate): return candidate if candidate == 2: candidate = 3 else: candidate += 2 def main(): input = sys.stdin.read().split() T = int(input[0]) cases = list(map(int, input[1:T+1])) for X in cases: if X == 1: print(2) continue factors = factorize(X) primes = list(factors.keys()) primes_set = set(primes) min_p = find_min_p(primes_set) Y1 = X * min_p min_Y = Y1 for p in primes: exponent = factors[p] new_Y = X * (p ** (exponent + 1)) if new_Y < min_Y: min_Y = new_Y print(min_Y) if __name__ == "__main__": main()