結果

問題 No.575 n! / m / m / m...
ユーザー mkawa2
提出日時 2021-11-25 16:21:47
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 60 ms / 2,000 ms
コード長 1,528 bytes
コンパイル時間 162 ms
コンパイル使用メモリ 82,176 KB
実行使用メモリ 58,624 KB
最終ジャッジ日時 2024-06-28 08:43:43
合計ジャッジ時間 2,773 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
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ファイルパターン 結果
sample AC * 3
other AC * 23
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)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
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 = 18446744073709551615
inf = 4294967295
# md = 10**9+7
md = 998244353
def prime_factorization(a):
    pp, ee = [], []
    if a & 1 == 0:
        pp += [2]
        ee += [0]
        while a & 1 == 0:
            a >>= 1
            ee[-1] += 1
    p = 3
    while p**2 <= a:
        if a%p == 0:
            pp += [p]
            ee += [0]
            while a%p == 0:
                a //= p
                ee[-1] += 1
        p += 2
    if a > 1:
        pp += [a]
        ee += [1]
    return pp, ee
from math import *
n, m = LI()
pp, ee = prime_factorization(m)
ptoe = {p: e for p, e in zip(pp, ee)}
mn = inf
for p, e in zip(pp, ee):
    p2 = p
    s = 0
    while p2 <= n:
        s += n//p2
        p2 *= p
    mn = min(mn, s//e)
s = 0
pp, ee = prime_factorization(n)
for p, e in zip(pp, ee):
    if p in ptoe:
        e -= ptoe[p]*mn
    s += e*log10(p)
l = lgamma(n+1)/log(10)-log10(m)*mn
d = floor(l)
p = 10**(l-d)
print(f"{p:.5f}e{d}")
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