結果

問題 No.2365 Present of good number
ユーザー lloyzlloyz
提出日時 2023-06-30 22:37:27
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,706 bytes
コンパイル時間 148 ms
コンパイル使用メモリ 82,036 KB
実行使用メモリ 67,080 KB
最終ジャッジ日時 2024-07-07 10:28:15
合計ジャッジ時間 3,398 ms
ジャッジサーバーID
(参考情報)
judge1 / 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 * 2
other AC * 31 WA * 8
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ソースコード

diff #
プレゼンテーションモードにする

from collections import defaultdict
IsPrime = [True for _ in range(10**5 + 1)]
IsPrime[0] = IsPrime[1] = False
Primes = []
for i in range(2, 10**5 + 1):
if IsPrime[i]:
Primes.append(i)
for j in range(i * i, 10**5 + 1, i):
IsPrime[j] = False
n, k = map(int, input().split())
mod = 10**9 + 7
P = defaultdict(int)
seen = set()
for prime in Primes:
while n % prime == 0:
seen.add(prime)
P[prime] += 1
n //= prime
C = [(p, c) for p, c in P.items()]
while not (len(seen) <= 2 and (2 in seen or 3 in seen)) and k > 0:
k -= 1
P = defaultdict(int)
seen = set()
for p, c in C:
p += 1
if IsPrime[p]:
seen.add(p)
P[p] += c
P[p] %= mod - 1
else:
for prime in Primes:
while p % prime == 0:
seen.add(prime)
P[prime] += c
P[prime] %= mod - 1
p //= prime
if p == 1:
break
C = [(p, c) for p, c in P.items()]
if k > 0:
q, r = divmod(k, 2)
P = defaultdict(int)
for p, c in C:
if p == 2:
if r == 0:
P[2] += c * pow(2, q, mod - 1)
P[2] %= mod - 1
else:
P[3] += c * pow(2, q, mod - 1)
P[3] %= mod - 1
else:
if r == 0:
P[3] += c * pow(2, q, mod - 1)
P[3] %= mod - 1
else:
P[2] += c * pow(2, q + 1, mod - 1)
P[2] %= mod - 1
C = [(p, c) for p, c in P.items()]
ans = 1
for p, c in C:
ans *= pow(p, c, mod)
ans %= mod
print(ans)
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