結果
問題 | No.12 限定された素数 |
ユーザー |
|
提出日時 | 2024-02-29 00:35:28 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 393 ms / 5,000 ms |
コード長 | 2,882 bytes |
コンパイル時間 | 348 ms |
コンパイル使用メモリ | 82,228 KB |
実行使用メモリ | 141,428 KB |
最終ジャッジ日時 | 2024-09-29 12:27:51 |
合計ジャッジ時間 | 9,645 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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ファイルパターン | 結果 |
---|---|
other | AC * 26 |
ソースコード
import sysimport mathimport bisectfrom heapq import heapify, heappop, heappushfrom collections import deque, defaultdict, Counterfrom functools import lru_cachefrom itertools import accumulate, combinations, permutations, productsys.setrecursionlimit(1000000)MOD = 10 ** 9 + 7MOD99 = 998244353input = lambda: sys.stdin.readline().strip()NI = lambda: int(input())NMI = lambda: map(int, input().split())NLI = lambda: list(NMI())SI = lambda: input()SMI = lambda: input().split()SLI = lambda: list(SMI())EI = lambda m: [NLI() for _ in range(m)]# 高速エラストテネス sieve[n]はnの最小の素因数def make_prime_table(n):sieve = list(range(n + 1))sieve[0] = -1sieve[1] = -1for i in range(4, n + 1, 2):sieve[i] = 2for i in range(3, int(n ** 0.5) + 1, 2):if sieve[i] != i:continuefor j in range(i * i, n + 1, i * 2):if sieve[j] == j:sieve[j] = ireturn sieveprime_table = make_prime_table(5000001)# 素数列挙primes = [p for i, p in enumerate(prime_table) if i == p]# 素因数分解 上のprime_tableと組み合わせて使うdef prime_factorize(n):result = []while n != 1:p = prime_table[n]e = 0while n % p == 0:n //= pe += 1result.append((p, e))return result# Nの素因数分解を辞書で返す(単体)def prime_fact(n):root = int(n**0.5) + 1prime_dict = {}for i in range(2, root):cnt = 0while n % i == 0:cnt += 1n = n // iif cnt:prime_dict[i] = cntif n != 1:prime_dict[n] = 1return prime_dict# 約数列挙(単体)def divisors(x):res = set()for i in range(1, int(x**0.5) + 2):if x % i == 0:res.add(i)res.add(x//i)return resdef main():N = NI()A = set(SLI())if len(A) == 10:print(4999999)returnPN = len(primes)ans = -1l = 0r = 0# print(A)while l < PN and r < PN:S = set()r = lK = 0L = 0for i in range(l, PN):p = str(primes[i])ok = Truefor sp in p:if sp not in A:ok = Falseif ok:S |= set(list(p))else:r = ibreakif S == A and l < r:if l == 0:K = 1else:K = primes[l-1] + 1if r == PN:L = 5000000else:L = primes[r] - 1ans = max(ans, L-K)l = relif l < r:l = relif l == r:l += 1print(ans)if __name__ == "__main__":main()