結果

問題 No.1200 お菓子配り-3
ユーザー tpyneriver
提出日時 2020-08-28 22:11:49
言語 PyPy3
(7.3.15)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 2,655 bytes
コンパイル時間 380 ms
コンパイル使用メモリ 82,260 KB
実行使用メモリ 80,048 KB
最終ジャッジ日時 2024-11-14 15:23:06
合計ジャッジ時間 6,886 ms
ジャッジサーバーID
(参考情報)
judge2 / judge1
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 2
other AC * 28 WA * 3
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
readline = sys.stdin.readline
from itertools import product
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]
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
S = int(readline())
Ans = [None]*S
for qu in range(S):
X, Y = map(int, readline().split())
if X == Y:
Ans[qu] = X - 1
continue
if X < Y:
X, Y = Y, X
pk, pv = map(list, zip(*list(primeFactor(X+Y).items())))
ans = 0
for k in product(*[range(v+1) for v in pv]):
res = 1
for i in range(len(pv)):
res = res*pow(pk[i], k[i])
if res - 1 <= 0:
continue
if res == 2:
continue
if (X-Y)%(res-2):
continue
l, r = (X+Y)//res, (X-Y)//(res-2)
if (l+r)%2:
continue
if 0 <= l+r and 0 <= l-r:
ans += 1
#print(res-1, (l+r)//2, (l-r)//2)
Ans[qu] = ans
print('\n'.join(map(str, Ans)))
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0