对于UInt64，价值太大或太小。

Question

My codes are like below. I just want to try to convert number to binary and summarize binary numbers as if they are decimal ones, eg the desired outcome for 3 is 22 :

1 ->  1
2 -> 10
3 -> 11
-------
     22 == 1 + 10 + 11     

But number array is growing and the code blowing :)

static void Main(string[] args)
{
    long deger = 1;

    for (int i = 0; i < 1000000; i++)
    {
        deger *= (deger + 1);
        int result = solve(deger);
        Console.WriteLine(result);
    }
}

public static int solve(long a)
{
    ulong[] lastValue = new ulong[a];
    for (int i = 1; i < a; i++)
    {
        var binary = Convert.ToString(i, 2);
        lastValue[i] = Convert.ToUInt64(binary);// this part gives an error
    }

    var result =  lastValue.Aggregate((t, c) => t + c);

    return (int)result;
}

Answer 1

Look at UInt64.MaxValue and UInt64.MinValue .

They defined as 18446744073709551615 and 0 corresponding.

Answer 2

Well, UInt64 is not large enough ; you can try either BigInteger or you may sum up string s:

private string MyBinarySum(string left, string right) {
  var x = left
    .PadLeft(Math.Max(left.Length, right.Length) + 1, '0')
    .Reverse()
    .Select(c => c - '0')
    .ToArray();

  var y = right
    .PadLeft(Math.Max(left.Length, right.Length) + 1, '0')
    .Reverse().Select(c => c - '0')
    .ToArray();

  StringBuilder sb = new StringBuilder(left.Length);

  int shift = 0;

  for (int i = 0; i < x.Length; ++i) {
    int v = x[i] + y[i] + shift;

    shift = v / 2;
    v = v % 2;

    sb.Append((char)('0' + v));
  }

  return String.Concat(sb.ToString().TrimEnd('0').Reverse());
}

Then, with a help of Linq

   var result = Enumerable
    .Range(0, 1000000)
    .Select(item => Convert.ToString(item, 2))
    .Aggregate((sum, item) => MyBinarySum(sum, item));

   Console.Write(result);

Outcome:

111010001101010010010101110011011100000

which is beyond UInt64.MaxValue == 18446744073709551615