結果

問題 No.2211 Frequency Table of GCD
ユーザー タコイモタコイモ
提出日時 2023-02-10 21:56:17
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 973 ms / 2,000 ms
コード長 4,325 bytes
コンパイル時間 170 ms
コンパイル使用メモリ 82,440 KB
実行使用メモリ 108,012 KB
最終ジャッジ日時 2024-07-07 17:59:19
合計ジャッジ時間 17,068 ms
ジャッジサーバーID
(参考情報)
judge4 / judge2
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 * 3
other AC * 26
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)
def divisors_pf(pf):
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)
def isPrime(n):
if n in {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}: return 1
if n <= 100: return 0
for i in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]:
if n % i == 0: return 0
return isPrimeMR(n)
def findPrime(n):
if n <= 2: return 2
i = n | 1
while 1:
if isPrime(i): return i
i += 2
def findNttFriendlyPrime(n, k, m=1):
a = (n >> k) + 1
i = (a << k) + 1
while 1:
if (i - 1) % m == 0:
if isPrime(i):
g = primitiveRoot(i)
ig = pow(g, i - 2, i)
return (i, g, ig) # p, g, invg
i += 1 << k
import time
import sys
#sys.setrecursionlimit(500000)
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def TI(): return tuple(map(int,sys.stdin.readline().rstrip().split()))
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip())
#for i, pi in enumerate(p):
from collections import defaultdict,deque
import bisect
import itertools
dic = defaultdict(int)
def make_divisors(n):
lower_divisors , upper_divisors = [], []
i = 1
while i*i <= n:
if n % i == 0:
lower_divisors.append(i)
if i != n // i:
upper_divisors.append(n//i)
i += 1
return lower_divisors + upper_divisors[::-1]
d = deque()
N,M = MI()
A = LI()
mod = 998244353
#
S = [0]*(M) #x
for i in A:
s = divisors(i)#make_divisors(i)
for j in s:
S[j-1] += 1
ans = [0]*(M)
for i in range(M-1,-1,-1):
ans[i] = (ans[i]+pow(2,S[i],mod)-1)%mod
s = divisors(i+1)#make_divisors(i+1)
for j in s:
if i+1 == j:
continue
ans[j-1] -= ans[i]
for i in ans:
print(i)
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