The algorithm for obtaining values ​​from the sum of a given number from an array

Question

I do not know to search or google, so I ask him here. I have an array of integers with a fixed size and it is with this logic.

sample [1,2,4,8,16,32]

Now I am assigned a number, for example 26. And I will find numbers the sum of which will be this number, in this case [2,8,16]

for the number 20 it will be [4.16]

for 40 it is [8.32]

and for 63 all these numbers [1,2,4,8,16,32]

What is the correct algorithm for this?

I know for sure that there is always a continuation, that the number doubles from the previous value. as well as only numbers from a given array are summed with a given number, and each number will be used only once or not

#, ints int ints, int, , .

Answer 1

, - base-2, , shift.

:

private IEnumerable<int> FindBits(int value)
{
    // check for bits.
    for (int i = 0; i < 32; i++)
    {
        // shift 1 by i
        var bitVal = 1 << i;   // you could use (int)Math.Pow(2, i); instead
        // check if the value contains that bit.
        if ((value & bitVal) == bitVal)
            // yep, it did.
            yield return bitVal;
    }
}

, ienumerable. ( )

---

:

// find the bits.
var res = FindBits(40).ToArray();

// format it using the string.join
var str = $"[{string.Join(",", res)}]";

// present the results
Console.WriteLine(str);

[8,32]

---

:

                          counter
00000001 =   1     = 1 << 0
00000010 =   2     = 1 << 1 
00000100 =   4     = 1 << 2
00001000 =   8     = 1 << 3
00010000 =  16     = 1 << 4
00100000 =  32     = 1 << 5
01000000 =  64     = 1 << 6
10000000 = 128     = 1 << 7

, , for, .

---

:

, FindBits :

private Func<int, IEnumerable<int>> FindBits = (int value) => Enumerable
    .Range(0, 31)
    .Select(i => 2 << i).Where(i => (value & i) == i);

/.

Answer 2

,

 ( 1    2    4    8    16 ... ) = (2^0  2^1  2^2  2^3  2^4 ... )

, . , , - 10 .

:

public List<int> dec_to_bin(int num)
{
    List<int> return_list = new List<int>();
    int index = 0; 
    int remainder = num;
    int bit = 0;

    while (remainder > 0)
    {
        bit = remainder % 2;

        if (bit == 1 )
        {
            return_list.Add((int)Math.Pow(2, index));
        }

        remainder = remainder / 2; 
        index = index + 1;
   }

   return return_list;
}

, .

public List<int> dec_to_bin(int num)
{
    List<int> return_list = new List<int>();
    int value = 1; 

    while( value < num )
    {
         if( (value & num) == value )
         {
             return_list.Add(value);
         }
         value = value * 2; 
    }
    return return_list;
}

Answer 3

" ints", , 2.

    private int[] count (int num)
    {

        int factor = 0;
        List<int> facts = new List<int>();
        while (num > 0)
        {
            int counter = 0;
            int div = num;
            int remainder = 0;
            while (remainder == 0)
            {
                remainder = div % 2;
                div = div / 2;
                counter++;
            }
            factor = 1;
            for (int i = 1; i < counter; i++) 
                factor *= 2;
            num = num - factor;
            facts.Add(factor);
        }

        return (facts.ToArray());
    }