The Weird and The Wonderful

decimal tenE18 = 1;
for (int i = 0; i < 18; i++)
    {
    tenE18 *= 10;
    }

decimal d = 1E18M;

chevu wrote:

how can you say algo i gave takes more time to debug.. complexity of given algo is O(n)

OriginalGriff wrote:

Still, I think I will stick to the old-fashioned, boring way:

decimal d = 10E18M;

//
// TODO: Check the number is 10^18
//

decimal bigNumber = 1000000000000000000M;

using System;

namespace DecimalTest
{
    class Program
    {
        static void Main(string[] args)
        {
            decimal bigNumber = 1000000000000000000M;
            decimal bigNumber2 = 10E18M;

            Console.WriteLine("Is {0} the same as {1}?", bigNumber, bigNumber2);
            if (bigNumber == bigNumber2)
                Console.WriteLine("WTF???");
            else
                Console.WriteLine("No - Maths saved!");
         
            Console.ReadKey();
        }
    }
}

using System;

namespace DecimalTest
{
    class Program
    {
        static void Main(string[] args)
        {
            decimal divider; 
            // we need 10^18, but Math.Pow does not support decimal
            // types and decimal does not provide a power function
            divider = 10*10*10;
            divider = Decimal.Multiply(divider,10*10*10*10*10);
            divider = Decimal.Multiply(divider,10*10*10*10*10);
            divider = Decimal.Multiply(divider,10*10*10*10*10);
            Console.WriteLine(divider);

            decimal tenE18 = 10;
            for (int i = 0; i < 18; i++)
                tenE18 *= 10;
            Console.WriteLine(tenE18);
            Console.WriteLine(10E18M);

            Console.WriteLine(String.Format("{0:N}", Math.Pow(10d, 18d)));
         
            Console.ReadKey();
        }
    }
}

chevu wrote:

This one is correct code

double pow(long long a, long long b)
{
	if(b == 0)
		return 1.0;
	else if(b == 1)
		return a;
	else if(b%2 == 0)
		return pow(a*a,b/2);
	else
		return a* pow(a*a,b/2);
}

chevu wrote:

long long a

static decimal pow(decimal x, uint n)
{
    decimal result = 1;
    while (n > 0)
    {
        if ((n & 1) == 1)
        {
            result *= x;
        }
        n >>= 1;
        if (n == 0)
            break;   //not nice, but needed in case x*x overflow on the last step
        x *= x;
    }
    return result;
}

harold aptroot wrote:

(tested)