結果

問題 No.1113 二つの整数 / Two Integers
ユーザー Kiri8128
提出日時 2020-07-17 21:22:24
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 58 ms / 1,000 ms
コード長 2,287 bytes
コンパイル時間 531 ms
コンパイル使用メモリ 81,792 KB
実行使用メモリ 67,840 KB
最終ジャッジ日時 2024-11-30 06:20:17
合計ジャッジ時間 2,152 ms
ジャッジサーバーID
(参考情報)
judge2 / 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 * 15
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

def gcd(a, b):
while b: a, b = b, a % b
return a
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 7, 61] if n < 1<<32 else [2, 3, 5, 7, 11, 13, 17] if n < 1<<48 else [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
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 += i % 2 + (3 if i % 3 == 1 else 1)
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):
pf = primeFactor(N)
ret = [1]
for p in pf:
ret_prev = ret
ret = []
for i in range(pf[p]+1):
for r in ret_prev:
ret.append(r * (p ** i))
return sorted(ret)
A, B = map(int, input().split())
g = gcd(A, B)
print("Odd" if len(divisors(g)) % 2 else "Even")
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