How to convert a decimal base (10) to a negabinary base (-2)?

问题

I want to write a program to convert from decimal to negabinary.

I cannot figure out how to convert from decimal to negabinary.

I have no idea about how to find the rule and how it works.

Example: 7(base10)-->11011(base-2)

I just know it is 7 = (-2)^0*1 + (-2)^1*1 + (-2)^2*0 + (-2)^3*1 + (-2)^4*1.

回答1:

The algorithm is described in http://en.wikipedia.org/wiki/Negative_base#Calculation. Basically, you just pick the remainder as the positive base case and make sure the remainder is nonnegative and minimal.

 7 = -3*-2 + 1  (least significant digit)
-3 =  2*-2 + 1
 2 = -1*-2 + 0
-1 =  1*-2 + 1
 1 =  0*-2 + 1  (most significant digit)

回答2:

Just my two cents (C#):

public static int[] negaBynary(int value)
{
    List<int> result = new List<int> ();

    while (value != 0)
    {
        int remainder = value % -2;
        value = value / -2;

        if (remainder < 0)
        {
            remainder += 2;
            value += 1;
        }

        Console.WriteLine (remainder);
        result.Add(remainder);
    }

    return result.ToArray();
}

回答3:

There is a method (attributed to Librik/Szudzik/Schröppel) that is much more efficient:

uint64_t negabinary(int64_t num) {
    const uint64_t mask = 0xAAAAAAAAAAAAAAAA;
    return (mask + num) ^ mask;
}

The conversion method and its reverse are described in more detail in this answer.

回答4:

def neg2dec(arr):
    n = 0
    for i, num in enumerate(arr[::-1]):
        n+= ((-2)**i)*num
    return n

def dec2neg(num):
    if num == 0:
        digits = ['0']
    else:
        digits = []
        while num != 0:
            num, remainder = divmod(num, -2)
            if remainder < 0:
                num, remainder = num + 1, remainder + 2
            digits.append(str(remainder))
    return ''.join(digits[::-1])

来源：https://stackoverflow.com/questions/9330946/how-to-convert-a-decimal-base-10-to-a-negabinary-base-2

标签

algorithm

base

base-conversion