# nCr

## Problem

The problem can be found here. The statement of this problem is extremely simple, in short, you need to calculate the ${n \choose r} \bmod M$ where M is 142857.

## Lucas Theorem

If 142857 is a prime number, then we can just use the Lucas theorem to solve it. However, $142857 = 3^3 \times 11 \times 13 \times 37$, that is why the problem is marked as an expert level problem in hackerrank.

The core idea of using Lucas theorem to calculate ${n \choose r} \bmod P$ where P is a prime number is to represent $n!$ as $a \times P^e$ where a is relative prime to P. To calculate the representation of $n!$, we can group $1 \times 2 \times \cdots \times n$ of P elements each, so it can be expressed as $(1 \times 2 \cdots \times (P -1)) \times P \times ((P+1)$ $\times$ $(P+2) \cdots \times (2P - 1)) \times$ $2P \cdots$, so that $1 \times 2 \times \cdots \times (P - 1)$ is relative prime to P, and each group mod P is of the same result.

And then use Wilson theorem which said that $(P - 1)! \equiv -1 (\bmod P)$. We can finally get the solution to calculate $n! = a \times P^e$. The python code as follow:

def fact_mod(n, p, facts):
"""
Type :: (Int, Int, [Int]) -> (Int, Int)
Suppose n! = a * p^e (mod p), then the function return (a mod p, e)
facts is i!(mod p) for 0 <= i < p, use Lucas theory

>>> facts = gen_fact_mod_prime(7)
>>> fact_mod(5, 7, facts)
(1, 0)

>>> fact_mod(15, 7, facts)
(2, 2)
"""
if (n == 0): return (1, 0)
(a, e) = fact_mod(n // p, p, facts)
e += n // p

if (n // p % 2 != 0): return (a * (p - facts[n % p]) % p, e)
return (a * facts[n % p] % p, e)


Notice the tricky in the last two line of the above code. Since $(P - 1)! = 1 (\bmod P)$, so if the number of group is even, then the result is just facts[n % p] % p, otherwise is -facts[n % p] % p, which is equivelent to (p - facts[n % p]) % p.

And then we can express ${n \choose r} = {n! \over {r! \times (n - r)!}}$, calculate $n! = a_1 \times P^{e_1}$, $k! = a_2 \times P^{e_2}$ and $(n - k)! = a_3 \times P^{e_3}$. Then if $e_1 > e_2 + e_3$, ${n \choose k}$ must be zero mod P, otherwise, the result should be a1 * modinv (a2 * a3, P) where modinv is the modulo inverse function.

def comb_mod(n, k, p):
"""
Type :: (Int, Int, Int) -> Int
Return C(n, k) mod p, p is a prime number.

>>> comb_mod(5, 3, 7)
3

>>> comb_mod(6, 2, 7)
1
"""

if n < 0 or k < 0 or n < k: return 0
facts = gen_fact_mod_prime(p)
a1, e1 = fact_mod(n, p, facts)
a2, e2 = fact_mod(k, p, facts)
a3, e3 = fact_mod(n - k, p, facts)
if (e1 > e2 + e3):
return 0
else:
return a1 * modinv(a2 * a3 % p, p) % p


## Generalize to this Problem

We can use very similar idea to calculate $n! \bmod P^a$ for $a\ge1$. Let $b = max \{i | n! \bmod P^i == 0, i \in \mathcal{N} \}$, then we need to calculate largest divisors of $n!$ relative prime to $P^a$.

This time we group $1 \times 2 \cdots \times n$ into group each has $P^a$ items. We first pre calculate the array facts[0..P^a], facts[i] is the largest divisors of $i!$ relative prime to $P^a$ mod $P^a$. Then we can use the same strategy when calculate $n! \bmod P^a$.

def comb_mod2(n, r, m, pa, facts1):
"""
Type :: (Int, Int, Int) -> Int
m is of form p^a, and n is very large
"""
p, a = pa

def n_fact_fact(n):
if n is 0 or n is 1:
return 1
elif n < m:
return facts1[n] * n_fact_fact(n // p) % m
else:
a = facts1[m - 1]
b = facts1[n % m]
c = n_fact_fact(n // p)
# print 'n = %d a = %d b = %d c = %d' % (n, a, b, c)
return pow(a, n // m, m) * b * c % m

def get_power(n, p):
ret = 0
while n > 0:
ret += n // p
n //= p
return ret
b = get_power(n, p) - get_power(r, p) - get_power(n - r, p)

if b >= a: return 0

m1 = n_fact_fact(n)
m2 = n_fact_fact(r)
m3 = n_fact_fact(n - r)

return (p ** b) * m1 * modinv_table[(m2, m)] * modinv_table[(m3, m)] % m


The above code is just to calculate ${n \choose r} \bmod P^a$. The function n_fact_fact is just to calculate the largest divisor of $n!$ mod $P^a$ which is relative to $P^a$. And get_power is to get the largest number of b such that ${n \choose r} \equiv 0 (\bmod P^b)$.

Finally, we can use Chinese Remainder Theorem to solve the problem. Since $142857 = 3^3 \times 11 \times 13 \times 37$, we can calculate ${n \choose r} \bmod 3^3$, ${n \choose r} \bmod 11$, ${n \choose r} \bmod 13$ and ${n \choose r} \bmod 37$, and the modulo numbers are relative prime to each other, so we can use it to finally get the answer.

You can find the whole solution in C++ and Python here.

11 April 2015