結果

問題 No.2497 GCD of LCMs
ユーザー mkawa2
提出日時 2023-10-08 16:29:12
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 354 ms / 2,000 ms
コード長 2,934 bytes
コンパイル時間 332 ms
コンパイル使用メモリ 82,520 KB
実行使用メモリ 79,200 KB
最終ジャッジ日時 2024-07-26 18:08:31
合計ジャッジ時間 3,944 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 * 14
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
# sys.setrecursionlimit(200005)
# sys.set_int_max_str_digits(200005)
int1 = lambda x: int(x)-1
pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
# inf = -1-(-1 << 31)
inf = -1-(-1 << 63)
# md = 10**9+7
md = 998244353
from collections import Counter
class Sieve:
def __init__(self, n):
self.plist = [2]
min_prime_factor = [2, 0]*(n//2+1)
for x in range(3, n+1, 2):
if min_prime_factor[x] == 0:
min_prime_factor[x] = x
self.plist.append(x)
if x**2 > n: continue
for y in range(x**2, n+1, 2*x):
if min_prime_factor[y] == 0:
min_prime_factor[y] = x
self.min_prime_factor = min_prime_factor
def isprime(self, x):
return self.min_prime_factor[x] == x
def pf(self, x):
pp, ee = [], []
while x > 1:
mpf = self.min_prime_factor[x]
if pp and mpf == pp[-1]:
ee[-1] += 1
else:
pp.append(mpf)
ee.append(1)
x //= mpf
return pp, ee
# unsorted
def factor(self, a):
ff = [1]
pp, ee = self.pf(a)
for p, e in zip(pp, ee):
ff, gg = [], ff
w = p
for _ in range(e):
for f in gg: ff.append(f*w)
w *= p
ff += gg
return ff
sv = Sieve(31596)
primes = sv.plist
def prime_factorization(a):
res = Counter()
for p in primes:
while a%p == 0:
a //= p
res[p] += 1
if a > 1:
res[a] = 1
return res
from heapq import *
def dijkstra(p, root=0):
n = len(to)
dist = [inf]*n
dist[root] = cnt[root][p]
hp = [(dist[root], root)]
while hp:
d, u = heappop(hp)
if d > dist[u]: continue
for v in to[u]:
nd = max(d,cnt[v][p])
if dist[v] <= nd: continue
dist[v] = nd
heappush(hp, (nd, v))
return dist
n,m=LI()
aa=LI()
to=[[] for _ in range(n)]
for _ in range(m):
u,v=LI1()
to[u].append(v)
to[v].append(u)
cnt=[]
pp=set()
for a in aa:
c=prime_factorization(a)
for p in c:pp.add(p)
cnt.append(c)
dp=[1]*n
for p in pp:
dist=dijkstra(p,0)
for u in range(n):
dp[u]*=pow(p,dist[u],md)
dp[u]%=md
print(*dp,sep="\n")
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