結果
| 問題 | 
                            No.2682 Visible Divisible
                             | 
                    
| コンテスト | |
| ユーザー | 
                             | 
                    
| 提出日時 | 2024-03-21 00:47:50 | 
| 言語 | PyPy3  (7.3.15)  | 
                    
| 結果 | 
                             
                                TLE
                                 
                             
                            
                         | 
                    
| 実行時間 | - | 
| コード長 | 4,740 bytes | 
| コンパイル時間 | 325 ms | 
| コンパイル使用メモリ | 81,920 KB | 
| 実行使用メモリ | 120,056 KB | 
| 最終ジャッジ日時 | 2024-09-30 09:51:19 | 
| 合計ジャッジ時間 | 5,270 ms | 
| 
                            ジャッジサーバーID (参考情報)  | 
                        judge5 / judge3 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | -- * 3 | 
| other | AC * 2 TLE * 1 -- * 11 | 
ソースコード
# https://qiita.com/daikw/items/f48d6ac374255763463d
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
prime = Prime()
N, K = map(int, input().split())
A = list(map(int, input().split()))
facts = prime.find_factors(K)
d = {}
p = facts.keys()
for key in facts.keys():
    d[key] = 0
for i in range(N):
    for key in p:
        cnt = 0
        while A[i] % key == 0:
            cnt += 1
            A[i] //= key
        d[key] = max(d[key], cnt)
for key in p:
    if d[key] < facts[key]:
        exit(print("No"))
print("Yes")