結果
問題 |
No.458 異なる素数の和
|
ユーザー |
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提出日時 | 2025-03-31 17:30:11 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,670 bytes |
コンパイル時間 | 313 ms |
コンパイル使用メモリ | 82,460 KB |
実行使用メモリ | 60,852 KB |
最終ジャッジ日時 | 2025-03-31 17:31:14 |
合計ジャッジ時間 | 2,914 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 27 WA * 1 |
ソースコード
def sieve(n): is_prime = [True] * (n + 1) is_prime[0] = is_prime[1] = False for i in range(2, int(n ** 0.5) + 1): if is_prime[i]: for j in range(i * i, n + 1, i): is_prime[j] = False primes = [i for i, prime in enumerate(is_prime) if prime] return primes, is_prime def main(): N = int(input().strip()) if N < 2: print(-1) return # Generate primes up to a large enough value (2*1e4) primes_list, is_prime = sieve(20000) sum_primes = [] primes_used = [] current_sum = 0 for p in primes_list: if current_sum + p > N: break current_sum += p sum_primes.append(current_sum) primes_used.append(p) # Check for all possible k from maximum possible down to 1 for k in range(len(sum_primes), 0, -1): sum_k = sum_primes[k-1] if sum_k == N: print(k) return d = N - sum_k # Check condition a: d is a prime and larger than last prime in current sum if d > primes_used[k-1] and is_prime[d]: print(k + 1) return # Check condition b: find if there is a prime[i] such that prime[i] + d is a prime larger than current last prime max_prime_in_k = primes_used[k-1] for i in range(k): q = primes_used[i] + d if q > max_prime_in_k and is_prime[q]: print(k) return # Check if N itself is a prime (when k=1 is possible) if is_prime[N]: print(1) return print(-1) if __name__ == "__main__": main()