結果
問題 |
No.1611 Minimum Multiple with Double Divisors
|
ユーザー |
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提出日時 | 2025-04-16 01:11:12 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 2,224 bytes |
コンパイル時間 | 439 ms |
コンパイル使用メモリ | 81,792 KB |
実行使用メモリ | 143,960 KB |
最終ジャッジ日時 | 2025-04-16 01:12:44 |
合計ジャッジ時間 | 13,956 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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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 math def sieve(n): sieve = [True] * (n + 1) sieve[0] = sieve[1] = False for i in range(2, int(math.sqrt(n)) + 1): if sieve[i]: sieve[i*i::i] = [False] * len(sieve[i*i::i]) primes = [i for i, is_p in enumerate(sieve) if is_p] return primes sieve_primes = sieve(10**6) 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 factorize(x, primes): factors = {} for p in primes: if p * p > x: break while x % p == 0: factors[p] = factors.get(p, 0) + 1 x //= p if x > 1: if is_prime(x): factors[x] = 1 else: i = 2 while i * i <= x: while x % i == 0: factors[i] = factors.get(i, 0) + 1 x //= i i += 1 if x > 1: factors[x] = 1 return factors def find_min_p(S, primes): for p in primes: if p not in S: return p candidate = 2 while True: if candidate not in S and is_prime(candidate): return candidate candidate += 1 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, sieve_primes) S = set(factors.keys()) p = find_min_p(S, sieve_primes) Y1 = X * p if factors: min_y2 = min(p**(a+1) for p, a in factors.items()) Y2 = X * min_y2 print(min(Y1, Y2)) else: print(Y1) if __name__ == "__main__": main()