結果

問題 No.1339 循環小数
ユーザー chineristAC
提出日時 2021-01-15 21:40:18
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 83 ms / 2,000 ms
コード長 2,787 bytes
コンパイル時間 288 ms
コンパイル使用メモリ 82,328 KB
実行使用メモリ 76,732 KB
最終ジャッジ日時 2024-11-26 13:41:11
合計ジャッジ時間 3,996 ms
ジャッジサーバーID
(参考情報)
judge3 / judge1
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

import sys
from math import gcd
input = sys.stdin.readline
def euler_phi(n):
res = n
for x in range(2, int(n**.5)+1):
if n % x == 0:
res = res // x * (x-1)
while n % x == 0:
n //= x
if n!=1:
res = (res//n) * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
prime = primeFactor(n)
res = [1]
for p in prime:
new = []
for a in res:
for j in range(prime[p]+1):
new.append(a*p**j)
res = new
return res
ans = []
import random
for _ in range(int(input())):
N = int(input())
while N%2==0:
N //= 2
while N%5==0:
N //= 5
primef = primeFactor(N)
phi = N
for p in primef:
phi *= (p-1)
phi //= p
divi = divisors(phi)
#print(divi)
for d in divi:
if pow(10,d,N)==1:
ans.append(d)
break
if N==1:
ans.append(1)
print(*ans,sep="\n")
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