結果
問題 |
No.577 Prime Powerful Numbers
|
ユーザー |
![]() |
提出日時 | 2025-03-31 17:33:16 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,588 bytes |
コンパイル時間 | 160 ms |
コンパイル使用メモリ | 82,056 KB |
実行使用メモリ | 142,180 KB |
最終ジャッジ日時 | 2025-03-31 17:33:55 |
合計ジャッジ時間 | 6,592 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | -- * 1 |
other | TLE * 1 -- * 9 |
ソースコード
import math def generate_s_list(): sieve_size = 10**6 sieve = [True] * (sieve_size + 1) sieve[0] = sieve[1] = False for i in range(2, int(math.sqrt(sieve_size)) + 1): if sieve[i]: sieve[i*i : sieve_size+1 : i] = [False] * len(sieve[i*i : sieve_size+1 : i]) primes = [i for i, is_p in enumerate(sieve) if is_p] s_set = set() for p in primes: a = 1 while True: power = p ** a if power > sieve_size: break s_set.add(power) a += 1 return sorted(s_set) s_list = generate_s_list() def is_prime(n): if n <= 1: return False elif n <= 3: return True elif n % 2 == 0: return False d = n - 1 s = 0 while d % 2 == 0: d //= 2 s += 1 bases = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37] for a in bases: 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 is_prime_power(x): if x < 2: return False if is_prime(x): return True max_k = x.bit_length() for k in range(max_k, 1, -1): low = 2 high = x found = False while low <= high: mid = (low + high) // 2 m = pow(mid, k) if m == x: if is_prime(mid): return True else: break elif m < x: low = mid + 1 else: high = mid - 1 return False small_primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71] Q = int(input()) for _ in range(Q): N = int(input()) found = False # Step 1: Check precomputed prime powers for s in s_list: if s > N - 2: continue rem = N - s if rem >= 2 and is_prime_power(rem): found = True break if found: print("Yes") continue # Step 2: Check small primes' powers for p in small_primes: a = 1 while True: current = pow(p, a) if current > N - 2: break rem = N - current if rem >= 2 and is_prime_power(rem): found = True break a += 1 if found: break print("Yes" if found else "No")