C#

sb.ToString().Split(' ').ToList().Distinct().Count()

kanamala subin wrote:

i want to check the count of "hi" in the sb

so my final output would be like this.

 

Count is 2.

Dictionary<string, int> wordCount = new Dictionary<string,int>();
sb.ToString().Split(' ').ToList().ForEach(w=>
{
  if (wordCount.ContainsKey(w) ++wordCountCount[w];
  else wordCount[w]=1;
}

private static ulong[] fibValues = { 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...};

Private Function Fibonacci(ByVal n As Integer) As Double
    Return (1 / Math.Sqrt(5)) * ((1 + Math.Sqrt(5)) / 2) ^ n - (1 / Math.Sqrt(5)) * ((1 - Math.Sqrt(5)) / 2) ^ n
End Function

Private Function Fibonacci2(ByVal n As Integer) As Double
    ' Golden ration = 1.61803398874989484820458683436563811772030917980576286213544862270526046281890244970720720418939113748475
    Dim gold As Double = 1.6180339887498949
    Dim gold2 As Double = 0.6180339887498949
    Dim sqrt1over5 As Double = 0.44721359549995793
    Return sqrt1over5 * ((gold) ^ n - (gold2) ^ n)
End Function

Private Function GetFib3(ByVal n As Integer) As ULong
       Dim a As Int64 = 0
       Dim b As Int64 = 1

       If (n = 0) Then Return a
       If (n = 1) Then Return b

       For i As UInt64 = 2 To n
           a = a Xor b
           b = b Xor a
           a = a Xor b

           ' add
           b += a
       Next

       Return b
   End Function

Kenneth Haugland wrote:

Dim a As Int64 = 0
Dim b As Int64 = 1

Private Function GetFib3(ByVal n As Integer) As ULong
   Dim a As ULong = 0
   Dim b As ULong = 1

   If (n = 0) Then Return a
   If (n = 1) Then Return b

   For i As ULong = 2 To Convert.ToUInt64(n)
       a = a Xor b
       b = b Xor a
       a = a Xor b

       ' add
       b += a
   Next

   Return b
End Function

Private Function GetFib2(ByVal n As Integer) As Numerics.BigInteger
    Dim a As New Numerics.BigInteger
    a = 0
    Dim b As New Numerics.BigInteger
    b = 1

    If (n = 0) Then Return a
    If (n = 1) Then Return b

    For i As UInt64 = 2 To n
        a = a Xor b
        b = b Xor a
        a = a Xor b

        ' add
        b += a
    Next

    Return b
End Function

static ulong Fibonacci(uint n)
{
    ulong m00 = 1, m01 = 1, m10 = 1, m11 = 0;
    ulong x00 = 1, x01 = 1, x10 = 1, x11 = 0;

    while (n != 0)
    {
        if ((n & 1) != 0)
        {
            ulong t00 = m00 * x00 + m01 * x10;
            ulong t01 = m00 * x01 + m01 * x11;
            ulong t10 = m10 * x00 + m11 * x10;
            ulong t11 = m10 * x01 + m11 * x11;
            x00 = t00;
            x01 = t01;
            x10 = t10;
            x11 = t11;
        }

        ulong y00 = m00 * m00 + m01 * m10;
        ulong y01 = m00 * m01 + m01 * m11;
        ulong y10 = m10 * m00 + m11 * m10;
        ulong y11 = m10 * m01 + m11 * m11;
        m00 = y00;
        m01 = y01;
        m10 = y10;
        m11 = y11;

        n >>= 1;
    }

    return x01;
}

Max Holder wrote:

I tested the swap with a temp ulong too, it was a bit slower than the XOR swap.

harold aptroot wrote:

is that XOR-swap really faster than a normal swap?