计算二进制的二项式系数

Question

有没有一种计算二项式系数并将结果保留为二进制的快速算法。例如，(10 选择 8) = 101101。我不需要将我的结果转换为以 10 为基数，而是我想将我的结果存储为二进制字符串。我在重新发明轮子之前问过这个问题。

Answer 1

我找到了最有效的方法。它正在执行二项式系数的质因数分解，然后转换为二进制。我添加了非常快速的代码来查找二项式系数的质因数分解。如果您喜欢那种东西，它被称为 Kummer's theorem and you can use this online calculator to verify your results. The factorization algorithm is from this Jstor paper. This is a Haskell 版本。请注意，您需要先自己生成一个素数列表。然后测试单个素数。还有算术基本定理在起作用

//Note primeNumber is always less than n
//Inputs n,k,primeNumber 
//Output: e (the exponent of prime number)
//Note: 0 means primeNumber is not a factor of this binomial coefficient
//Example: (n=10,k=3,primeNumber=3), e = 1
//So if you had a list of primes 2,3,5 then for each you get e=3,e=1,e=1. (10,3) = 2^3 *3^1 * 5 ^1 
int BinomialFactorization(int n, int k, int primeNumber)
{
    int a = 0;
    int b = 0;
    int exponent = 0;
    int r = 0;
    
    //Finds smaller value between n and n-k since (n choose k) == (n choose n-k)
    //Algorithm only works when k < n/2 
    if(k > (n/2))
    {   
        k = n - k; 
    }
    //Speeds up according to paper
    if(primeNumber > n - k)
    {
        return 1;
    }
    //Speeds up according to paper
    if(primeNumber > n/2)
    {
        printf("%d", 0);
        return 0;
    }
    if(primeNumber * primeNumber > n)
    {
        if(n % primeNumber < k % primeNumber)
        {   
            return 1;
        }
        else
        {   //Saw this on online calculator
            return 0;
        }
    }
    //Changing base algorithm 
    while(n > 0)
    {
        a = n % primeNumber;
        n = n / primeNumber;
        b = k % primeNumber + r;
        k = k / primeNumber;
        if(a < b)
        {
            exponent = exponent + 1;
            r = 1;
        }
        else
        {
            r = 0;
        }
    }
    return exponent;    
}

计算二进制的二项式系数

Calculate binomial coefficient in binary

c

math

numerical-methods

factorial

binomial-coefficients