"0830 commoncode"의 두 판 사이의 차이
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$$ \frac{(a+b+c+\cdots+z)!}{a!b!c!\cdots z!} \pmod p \text{ where } p \text{ is a prime}$$ | $$ \frac{(a+b+c+\cdots+z)!}{a!b!c!\cdots z!} \pmod p \text{ where } p \text{ is a prime}$$ | ||
| − | Above value is always integer. Refer [https://math.stackexchange.com/questions/2158/division-of-factorials the proofs]. The key observation is that the product of \(n\) consecutive integers is divisible by \(n!\).<ref> This can be proved by induction. [https://math.stackexchange.com/a/2171/64186]. And maybe possible | + | Above value is always integer. Refer [https://math.stackexchange.com/questions/2158/division-of-factorials the proofs]. The key observation is that the product of \(n\) consecutive integers is divisible by \(n!\).<ref> This can be proved by induction. [https://math.stackexchange.com/a/2171/64186]. And maybe possible by group theory</ref> |
<pre> | <pre> | ||
p = int(1e9)+7 #example. this is a prime number | p = int(1e9)+7 #example. this is a prime number | ||
2017년 8월 30일 (수) 22:47 기준 최신판
$$ \frac{(a+b+c+\cdots+z)!}{a!b!c!\cdots z!} \pmod p \text{ where } p \text{ is a prime}$$
Above value is always integer. Refer the proofs. The key observation is that the product of \(n\) consecutive integers is divisible by \(n!\).[1]
p = int(1e9)+7 #example. this is a prime number
def anmodp(a, n): # a^n mod p
if n==1:
return a%p
tmp = anmodp(a, n/2)
if n%2==1: #odd
return (a*(tmp**2))%p
else:
return (tmp**2)%p
def fac(n): # n! mod p
k = 1
for i in range(2, n+1):
k *= i
k %= p
return k
def invfac(n): # 1/(n!) (mod p)
return anmodp(fac(n), p-2) #euler totient
def comb(x): # input = int array
s = sum(x)
ans = fac(s)
for i in x:
ans *= invfac(i)
return ans % p