結果

問題 No.1339 循環小数
ユーザー 👑 rin204
提出日時 2022-10-31 23:28:24
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 103 ms / 2,000 ms
コード長 2,141 bytes
コンパイル時間 441 ms
コンパイル使用メモリ 82,560 KB
実行使用メモリ 76,416 KB
最終ジャッジ日時 2024-07-08 11:03:01
合計ジャッジ時間 4,153 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
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 * 1
other AC * 36
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

from math import gcd
def isprime(n):
if n <= 1:
return False
elif n == 2:
return True
elif n % 2 == 0:
return False
A = [2, 325, 9375, 28178, 450775, 9780504, 1795265022]
s = 0
d = n - 1
while d % 2 == 0:
s += 1
d >>= 1
for a in A:
if a % n == 0:
return True
x = pow(a, d, n)
if x != 1:
for t in range(s):
if x == n - 1:
break
x = x * x % n
else:
return False
return True
def pollard(n):
if n % 2 == 0:
return 2
if isprime(n):
return n
f = lambda x:(x * x + 1) % n
step = 0
while 1:
step += 1
x = step
y = f(x)
while 1:
p = gcd(y - x + n, n)
if p == 0 or p == n:
break
if p != 1:
return p
x = f(x)
y = f(f(y))
def primefact(n):
if n == 1:
return []
p = pollard(n)
if p == n:
return [p]
left = primefact(p)
right = primefact(n // p)
left += right
return sorted(left)
def divisor_lst(n):
if n == 1:
return [1]
primes = primefact(n)
primes.append(primes[-1] + 1)
bef = primes[0]
cnt = 0
ret = [1]
for p in primes:
if p == bef:
cnt += 1
else:
times = bef
le = len(ret)
for _ in range(cnt):
for i in range(le):
ret.append(ret[i] * times)
times *= bef
bef = p
cnt = 1
ret.sort()
return ret
def solve(n):
while n % 2 == 0:
n //= 2
while n % 5 == 0:
n //= 5
if n == 1:
return 1
primes = set(primefact(n))
N = n
for p in primes:
n *= p - 1
n //= p
divs = divisor_lst(n)
for d in divs:
x = pow(10, d, N) - 1
if x == 0:
return d
for _ in range(int(input())):
ans = solve(int(input()))
print(ans)
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