結果
| 問題 |
No.458 異なる素数の和
|
| ユーザー |
lam6er
|
| 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()
lam6er