Finding binomial coefficient for large n and k modulo m

问题

I want to compute nCk mod m with following constraints:

n<=10^18

k<=10^5

m=10^9+7

I have read this article:

Calculating Binomial Coefficient (nCk) for large n & k

But here value of m is 1009. Hence using Lucas theorem, we need only to calculate 1009*1009 different values of aCb where a,b<=1009

How to do it with above constraints. I cannot make a array of O(m*k) space complexity with given constraints.

Help!

回答1:

Just use the fact that

(n, k) = n! / k! / (n - k)! = n*(n-1)*...*(n-k+1)/[k*(k-1)*...*1]

so you actually have just 2*k=2*10^5 factors. For the inverse of a number you can use suggestion of kfx since your m is prime.

回答2:

First, you don't need to pre-compute and store all the possible aCb values! they can be computed per case.

Second, for the special case when (k < m) and (n < m^2), the Lucas theorem easily reduces to the following result:

(n choose k) mod m = ((n mod m) choose k) mod m

then since (n mod m) < 10^9+7 you can simply use the code proposed by @kfx.

回答3:

The binominal coefficient of (n, k) is calculated by the formula:

(n, k) = n! / k! / (n - k)!

To make this work for large numbers n and k modulo m observe that:

Factorial of a number modulo m can be calculated step-by-step, in each step taking the result % m. However, this will be far too slow with n up to 10^18. So there are faster methods where the complexity is bounded by the modulo, and you can use some of those.
The division (a / b) mod m is equal to (a * b^-1) mod m, where b^-1 is the inverse of b modulo m (that is, (b * b^-1 = 1) mod m).

This means that:

(n, k) mod m = (n! * (k!)^-1 * ((n - k)!)^-1) mod m

The inverse of a number can be efficiently found using the Extended Euclidean algorithm. Assuming you have the factorial calculation sorted out, the rest of the algorithm is straightforward, just watch out for integer overflows on multiplication. Here's reference code that works up to n=10^9. To handle for larger numbers the factorial computation should be replaced with a more efficient algorithm and the code should be slightly adapted to avoid integer overflows, but the main idea will remain the same:

#define MOD 1000000007

// Extended Euclidean algorithm
int xGCD(int a, int b, int &x, int &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }

    int x1, y1, gcd = xGCD(b, a % b, x1, y1);
    x = y1;
    y = x1 - (long long)(a / b) * y1;
    return gcd;
}

// factorial of n modulo MOD
int modfact(int n) {
    int result = 1;
    while (n > 1) {
        result = (long long)result * n % MOD;
        n -= 1;
    }
    return result;
}

// multiply a and b modulo MOD
int modmult(int a, int b) {
    return (long long)a * b % MOD;
}

// inverse of a modulo MOD
int inverse(int a) {
    int x, y;
    xGCD(a, MOD, x, y);
    return x;
}

// binomial coefficient nCk modulo MOD
int bc(int n, int k)
{
    return modmult(modmult(modfact(n), inverse(modfact(k))), inverse(modfact(n - k)));
}

回答4:

We want to compute nCk (mod p). I'll handle when 0 <= k <= p-2, because Lucas's theorem handles the rest.

Wilson's theorem states that for prime p, (p-1)! = -1 (mod p), or equivalently (p-2)! = 1 (mod p) (by division).

By division: (k!)^(-1) = (p-2)!/(k!) = (p-2)(p-3)...(k+1) (mod p)

Thus, the binomial coefficient is n!/(k!(n-k)!) = n(n-1)...(n-k+1)/(k!) = n(n-1)...(n-k+1)(p-2)(p-3)...(k+1) (mod p)

Voila. You don't have to do any inverse computations or anything like that. It's also fairly easy to code. A couple optimizations to consider: (1) you can replace (p-2)(p-3)... with (-2)(-3)...; (2) nCk is symmetric in the sense that nCk = nC(n-k) so choose the half that requires you to do less computations.

来源：https://stackoverflow.com/questions/35226781/finding-binomial-coefficient-for-large-n-and-k-modulo-m

标签

c++

algorithm

modulus

modular-arithmetic

binomial-coefficients