結果

問題 No.385 カップ麺生活
ユーザー lam6er
提出日時 2025-03-20 20:49:45
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,360 bytes
コンパイル時間 247 ms
コンパイル使用メモリ 82,172 KB
実行使用メモリ 67,492 KB
最終ジャッジ日時 2025-03-20 20:49:48
合計ジャッジ時間 2,969 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 28 WA * 4
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

M = int(input())
N = int(input())
C = list(map(int, input().split()))
# Sieve of Eratosthenes to find primes up to M
def sieve(n):
if n < 2:
return [False] * (n + 1), []
is_prime = [True] * (n + 1)
is_prime[0], is_prime[1] = False, 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 is_prime, primes
is_prime, primes_list = sieve(M)
# DP to compute the maximum number of cups for each sum up to M
max_num = [-1] * (M + 1)
max_num[0] = 0 # Base case: 0 money buys 0 cups
for x in range(M + 1):
if max_num[x] == -1:
continue
for c in C:
next_x = x + c
if next_x > M:
continue
if max_num[next_x] < max_num[x] + 1:
max_num[next_x] = max_num[x] + 1
# Calculate the sum of valid k_p values
sum_k = 0
for p in primes_list:
required = M - p
if required >= 0 and max_num[required] > 0:
sum_k += max_num[required]
# Calculate the maximum x for the last non-reset purchase
max_x = 0
for x in range(M + 1):
r = M - x
if r < 0:
continue
if r == 0 or not is_prime[r]:
if max_num[x] != -1 and max_num[x] > max_x:
max_x = max_num[x]
print(sum_k + max_x)
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