結果
問題 | No.368 LCM of K-products |
ユーザー | Navier_Boltzmann |
提出日時 | 2022-05-07 20:09:13 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 4,786 bytes |
コンパイル時間 | 335 ms |
コンパイル使用メモリ | 82,448 KB |
実行使用メモリ | 90,084 KB |
最終ジャッジ日時 | 2024-07-07 01:49:57 |
合計ジャッジ時間 | 6,854 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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テストケース
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testcase_00 | TLE | - |
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ソースコード
N,K = map(int,input().split()) a = list(map(int,input().split())) mod = 10**9 + 7 import math import random class Prime: seed_primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] def is_prime(self, n): """ prime test (hybrid) see also: https://qiita.com/gushwell/items/ff9ed83ba55350aaa369 :param n: :return: boolean """ is_prime_common = self.is_prime_common(n) if is_prime_common is not None: return is_prime_common if n < 2000000: return self.is_prime_brute_force(n) else: return self.is_prime_miller_rabin(n) def is_prime_common(self, n): if n == 1: return False if n in Prime.seed_primes: return True if any(map(lambda x: n % x == 0, self.seed_primes)): return False def is_prime_brute_force(self, n): """ brute force prime test use with is_prime_common if you want to skip seed_primes :param n: :return: boolean """ for k in range(2, int(math.sqrt(n)) + 1): if n % k == 0: return False return True def is_prime_miller_rabin(self, n): """ miller rabin prime test use with is_prime_common if you want to skip seed_primes see also algorithm: https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test implementation: https://qiita.com/srtk86/items/609737d50c9ef5f5dc59 improvement: https://qiita.com/gushwell/items/ff9ed83ba55350aaa369 :param n: :return: boolean """ d = n - 1 while d & 1 == 0: d >>= 1 # use one of these lines / upper is more efficient. witnesses = self.get_witnesses(n) # witnesses = [random.randint(1, n - 1) for _ in range(100)] for w in witnesses: y = pow(w, d, n) while d != n - 1 and y != 1 and y != n - 1: y = (y * y) % n d <<= 1 if y != n - 1 and d & 1 == 0: return False return True def get_witnesses(self, num): def _get_range(num): if num < 2047: return 1 if num < 1373653: return 2 if num < 25326001: return 3 if num < 3215031751: return 4 if num < 2152302898747: return 5 if num < 3474749660383: return 6 if num < 341550071728321: return 7 if num < 3825123056546413051: return 9 return 12 return self.seed_primes[:_get_range(num)] def gcd(self, a, b): if a < b: return self.gcd(b, a) if b == 0: return a while b: a, b = b, a % b return a @staticmethod def f(x, n, seed): """ pseudo prime generator :param x: :param n: :param seed: :return: pseudo prime """ p = Prime.seed_primes[seed % len(Prime.seed_primes)] return (p * x + seed) % n def find_factor(self, n, seed=1): """ find one of factor of n this function is based to Pollard's rho algorithm see also algorithm: https://en.wikipedia.org/wiki/Pollard%27s_rho_algorithm implementation: https://qiita.com/gushwell/items/561afde2e00bf3380c98 :param n: :param seed: :return: factor """ if self.is_prime(n): return n x, y, d = 2, 2, 1 count = 0 while d == 1: count += 1 x = self.f(x, n, seed) y = self.f(self.f(y, n, seed), n, seed) d = self.gcd(abs(x - y), n) if d == n: return self.find_factor(n, seed+1) return self.find_factor(d) def find_factors(self, n): primes = {} if self.is_prime(n): primes[n] = 1 return primes while n > 1: factor = self.find_factor(n) primes.setdefault(factor, 0) primes[factor] += 1 n //= factor return primes P = Prime() b = [P.find_factors(i) for i in a] ans = 1 from collections import * d = defaultdict(lambda:[]) for i in range(N): for k,v in b[i].items(): d[k].append(v) ans = 1 for k in d.keys(): d[k].sort() if len(d[k]) < K: for v in d[k]: ans *= pow(k,v,mod) ans %= mod else: for i in range(K): ans *= pow(k,d[k][-1-i],mod) ans %= mod print(ans)