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Calculate 2^(n) where 0<n<10000

So, this is my problem to solve:
I want to calculate 2^(n) where 0 < n< 10000
I am representing each element of array as a space where 4digit number should be "living" and if extra digit appears, I am replacing it to the next element of this array.
The principle I am using looks like this:
The code I am using is the following:
static string NotEfficient(int power)
{
if (power < 0)
throw new Exception("Power shouldn't be negative");
if (power == 0)
return "1";
if (power == 1)
return "2";
int[] A = new int[3750];
int current4Digit = 0;
//at first 2 is written in first element of array
A[current4Digit] = 2;
int currentPower = 1;
while (currentPower < power)
{
//multiply every 4digit by 2
for (int i = 0; i <= current4Digit; i++)
{
A[i] *= 2;
}
currentPower++;
//checking every 4digit if it
//contains 5 digit and if yes remove and
//put it in next 4digit
for (int i = 0; i <= current4Digit; i++)
{
if (A[i] / 10000 > 0)
{
int more = A[i] / 10000;
A[i] = A[i] % 10000;
A[i + 1] += more;
//if new digit should be opened
if (i + 1 > current4Digit)
{
current4Digit++;
}
}
}
}
//getting data from array to generate answer
string answer = "";
for (int i = current4Digit; i >= 0; i--)
{
answer += A[i].ToString() + ",";
}
return answer;
}
The problem I have is that it doesn't display correctly the number, which contains 0 in reality. for example 2 ^ (50) = 1 125 899 906 842 624 and with my algorithm I get 1 125 899 96 842 624 (0 is missing). This isn't only for 50...
This happens when I have the following situation for example:
How I can make this algorithm better?

Use BigInteger, which is already included in .Net Core or available in the System.Runtime.Numerics Nuget Package.
static string Efficient(int power)
{
var result = BigInteger.Pow(2, power);
return result.ToString(CultureInfo.InvariantCulture);
}
On my machine, NotEfficient takes roughly 80ms, where Efficient takes 0.3ms. You should be able to manipulate that string (if I'm understanding your problem statement correctly):
static string InsertCommas(string value)
{
var sb = new StringBuilder(value);
for (var i = value.Length - 4; i > 0; i -= 4)
{
sb.Insert(i, ',');
}
return sb.ToString();
}

One way to resolve this is to pad your 4-digit numbers with leading zeroes if they are less than four digits by using the PadLeft method:
answer += A[i].ToString().PadLeft(4, '0') + ",";
And then you can use the TrimStart method to remove any leading zeros from the final result:
return answer.TrimStart('0');

How to do the loop and solve the following

I'm supposed to code a program that writes out a division just like in school.
Example:
13:3=4.333333333333
13
1
10
10
10....
So my approach was:
Solve the division then get the solution in a List.
Then question if the first number (in this case 1) is divisible by 3.
If not put it down and add the second number and so on...
I managed to do this the first time. It's sloppy but works. The problem is that it only works with numbers that when divided get to have a decimal in it.
Exapmle:
123:13
This is the first code:
do
{
for (int number = 1; number <= divNum; number++)
if (number % divisor == 0) countH++;
for (int i = 0; i < count; i++)
Console.Write(" ");
if ((c = divNum % divisor ) < divisor )
{
Console.WriteLine(" " + ((divNum- (countH * divisor ))) * 10);
}
else Console.WriteLine(" " + (divNum- (countH * divisor )));
c = divNum % divisor ;
if (c < divisor )
{
divNum = c * 10;
}
count++; countH = 0;
} while ((divNum >= divisor ) && (count < x));
Any ideas or help? Sorry if this is a bad question.
************ added
Try of a better explanation:
1 cant be divided by 13, so it goes down, we get the 2 down and try 12 divided by 13, still nothing so we get the 3 down and try 123:13, 13 goes 9 times in 123 so we have 123-9*13 = 6 the six goes down we write 9 in the result. We try 6:13 not going so we drop a 0 next to 6. Next we try 60:13, 13 goes 4 times so 60-4*13 = 8, we get the 8 down. And so on..
123:13=9.46153....
123
60
80
20
70
50
....

Something like this should work. Not the fastest solution most likely, but should do the job.
var number = 123;
var b = 12;
int quotient;
double remainder = number;
var x = 10;
do
{
quotient = (int)Math.Floor(remainder / b);
remainder = remainder - (quotient * b);
for (int i = 0; i < count; i++)
Console.Write(" ");
remainder *= 10;
Console.WriteLine(" " + remainder);
count++;
} while ((remainder > 0) && (count < x));

Get sum of each digit below n as long

This is the code I have but it's to slow, any way to do it faster..
the number range I'm hitting is 123456789 but I can't get it below 15 seconds, and I need it to get below 5 seconds..
long num = 0;
for (long i = 0; i <= n; i++)
{
num = num + GetSumOfDigits(i);
}
static long GetSumOfDigits(long n)
{
long num2 = 0;
long num3 = n;
long r = 0;
while (num3 != 0)
{
r = num3 % 10;
num3 = num3 / 10;
num2 = num2 + r;
}
return num2;
}
sum =(n(n+1))/2 is not giving me the results I need, not calculating properly..
For N = 12 the sum is 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+(1+2)= 51.
I need to do this with a formula instead of a loop..
I've got about 15 tests to run through each under 6 seconds..
with parallel I got one test from 15seconds to 4-8 seconds..
Just to show you the test I'm doing this is the hard one..
[Test]
public void When123456789_Then4366712385()
{
Assert.AreEqual(4366712385, TwistedSum.Solution(123456789));
}
On my computer I can run all the tests under 5 seconds..
Look at the photo..
With DineMartine Answer I got these results:

Your algorithm complexity is N log(N). I have found a better algorithm with complexity log(N). The idea is to iterate on the number of digits which is :
log10(n) = ln(n)/ln(10) = O(log(n)).
The demonstration of this algorithm involves a lot of combinatorial calculus. So I choose not to write it here.
Here is the code :
public static long Resolve(long input)
{
var n = (long)Math.Log10(input);
var tenPow = (long)Math.Pow(10, n);
var rest = input;
var result = 0L;
for (; n > 0; n--)
{
var dn = rest / tenPow;
rest = rest - dn * tenPow;
tenPow = tenPow / 10;
result += dn * (rest + 1) + dn * 45 * n * tenPow + dn * (dn-1) * tenPow * 5 ;
}
result += rest * (rest + 1) / 2;
return result;
}
Now you would solve the problem in a fraction of second.
The idea is to write the input as a list of digit :
Assuming the solution is given by a function f, we are looking for g a recursive expression of f over n :
Actually g can be written as follow :
If you find h, the problem would be practically solved.

A little bit convoluted but gets the time down to practicaly zero:
private static long getSumOfSumOfDigitsBelow(long num)
{
if (num == 0)
return 0;
// 1 -> 1 ; 12 -> 10; 123 -> 100; 321 -> 100, ...
int pow10 = (int)Math.Pow(10, Math.Floor(Math.Log10(num)));
long firstDigit = num / pow10;
long sum = 0;
var sum999 = getSumOfSumOfDigitsBelow(pow10 - 1);
var sumRest = getSumOfSumOfDigitsBelow(num % pow10);
sum += (firstDigit - 1)*(firstDigit - 0)/2*pow10 + firstDigit*sum999;
sum += firstDigit*(num%pow10 + 1) + sumRest;
return sum;
}
getSumOfSumOfDigitsBelow(123456789) -> 4366712385 (80us)
getSumOfSumOfDigitsBelow(9223372036854775807) -> 6885105964130132360 (500us - unverified)
The trick is to avoid to compute the same answer again and again. e.g. 33:
your approach:
sum = 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+(1+2)+ ... +(3+2)+(3+3)
my approach:
sum = 10*(0 + (1+2+3+4+5+6+7+8+9)) +
10*(1 + (1+2+3+4+5+6+7+8+9)) +
10*(2 + (1+2+3+4+5+6+7+8+9)) +
3*(3 + (1 + 2 + 3))
The (1+2+3+4+5+6+7+8+9)-part have to be calculated only once. The loop of 0..firstDigit-1 can be avoided by the n(n-1)/2-trick. I hope this makes sense.
The complexity is O(2^N) with N counting the number of digits. This looks very bad but is fast enough for your threshold of 5s even for long-max. It may be possible to transform this algorithm in something running in O(n) by calling getSumOfSumOfDigitsBelow() only once but it would look much more complex.
First step of optimization: look at your algorithm ;)
Comming back to this problem after the answer of DineMartine :
To further optimize the algorithm, the sum999-part can be replaced by an explicit formula. Lets take some number 9999...9=10^k-1 into the code and replace accordingly:
sum(10^k-1) = (9 - 1)*(9 - 0)/2*pow10 + 9*sum999 + 9*(num%pow10 + 1) + sumRest
sum(10^k-1) = 36*pow10 + 9*sum999 + 9*(num%pow10 + 1) + sumRest
sum999 and sumRest are the same for the numbers of type 10^k:
sum(10^k-1) = 36*pow10 + 10*sum(10^(k-1)-1) + 9*(num%pow10 + 1)
sum(10^k-1) = 36*pow10 + 10*sum(10^(k-1)-1) + 9*((10^k-1)%pow10 + 1)
sum(10^k-1) = 36*pow10 + 10*sum(10^(k-1)-1) + 9*pow10
sum(10^k-1) = 45*pow10 + 10*sum(10^(k-1)-1)
We have a definition of sum(10^k-1) and know sum(9)=45. And we get:
sum(10^k-1) = 45*k*10^k
The updated code:
private static long getSumOfSumOfDigitsBelow(long num)
{
if (num == 0)
return 0;
long N = (int) Math.Floor(Math.Log10(num));
int pow10 = (int)Math.Pow(10, N);
long firstDigit = num / pow10;
long sum = (firstDigit - 1)*firstDigit/2*pow10
+ firstDigit* 45 * N * pow10 / 10
+ firstDigit*(num%pow10 + 1)
+ getSumOfSumOfDigitsBelow(num % pow10);
return sum;
}
This is the same algorithm as the one from DineMartine but expressed in a recursive fashion (I've compared both implementations and yes I'm sure it is ;) ). The runtime goes down to practically zero and the time complexity is O(N) counting the number of digits or O(long(N)) taking the value.

If you have multiple processors (or cores) in your system, you can speed it up quite a lot by doing the calculations in parallel.
The following code demonstrates (it's a compilable console app).
The output when I try it on my system (4 cores with hyperthreading) is as follows for a release build:
x86 version:
Serial took: 00:00:14.6890714
Parallel took: 00:00:03.5324480
Linq took: 00:00:04.4480217
Fast Parallel took: 00:00:01.6371894
x64 version:
Serial took: 00:00:05.1424354
Parallel took: 00:00:00.9860272
Linq took: 00:00:02.6912356
Fast Parallel took: 00:00:00.4154711
Note that the parallel version is around 4 times faster. Also note that the x64 version is MUCH faster (due to the use of long in the calculations).
The code uses Parallel.ForEach along with a Partitioner to split the range of number into sensible regions for the number of processors available. It also uses Interlocked.Add() to quickly add the numbers with efficient locking.
I've also added another method where you need to pre-calculate the sums for the numbers between 0 and 1000. You should only need to pre-calculate the sums once for each run of the program. See FastGetSumOfDigits().
Using FastGetSumOfDigits() more than doubles the previous fastest time on my PC. You can increase the value of SUMS_SIZE to a larger multiple of 10 to increase the speed still further, at the expense of space. Increasing it to 10000 on my PC decreased the time to ~0.3s
(The sums array only needs to be a short array, to save space. It doesn't need a larger type.)
using System;
using System.Collections.Concurrent;
using System.Diagnostics;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Threading;
using System.Threading.Tasks;
namespace Demo
{
internal class Program
{
public static void Main()
{
long n = 123456789;
Stopwatch sw = Stopwatch.StartNew();
long num = 0;
for (long i = 0; i <= n; i++)
num = num + GetSumOfDigits(i);
Console.WriteLine("Serial took: " + sw.Elapsed);
Console.WriteLine(num);
sw.Restart();
num = 0;
var rangePartitioner = Partitioner.Create(0, n + 1);
Parallel.ForEach(rangePartitioner, (range, loopState) =>
{
long subtotal = 0;
for (long i = range.Item1; i < range.Item2; i++)
subtotal += GetSumOfDigits(i);
Interlocked.Add(ref num, subtotal);
});
Console.WriteLine("Parallel took: " + sw.Elapsed);
Console.WriteLine(num);
sw.Restart();
num = Enumerable.Range(1, 123456789).AsParallel().Select(i => GetSumOfDigits(i)).Sum();
Console.WriteLine("Linq took: " + sw.Elapsed);
Console.WriteLine(num);
sw.Restart();
initSums();
num = 0;
Parallel.ForEach(rangePartitioner, (range, loopState) =>
{
long subtotal = 0;
for (long i = range.Item1; i < range.Item2; i++)
subtotal += FastGetSumOfDigits(i);
Interlocked.Add(ref num, subtotal);
});
Console.WriteLine("Fast Parallel took: " + sw.Elapsed);
Console.WriteLine(num);
}
private static void initSums()
{
for (int i = 0; i < SUMS_SIZE; ++i)
sums[i] = (short)GetSumOfDigits(i);
}
[MethodImpl(MethodImplOptions.AggressiveInlining)]
private static long GetSumOfDigits(long n)
{
long sum = 0;
while (n != 0)
{
sum += n%10;
n /= 10;
}
return sum;
}
[MethodImpl(MethodImplOptions.AggressiveInlining)]
private static long FastGetSumOfDigits(long n)
{
long sum = 0;
while (n != 0)
{
sum += sums[n % SUMS_SIZE];
n /= SUMS_SIZE;
}
return sum;
}
static short[] sums = new short[SUMS_SIZE];
private const int SUMS_SIZE = 1000;
}
}

To increase performance you could calculate the sum starting from the highest number.
Let r=n%10 + 1. Calculate the sum for the last r numbers.
Then we note that if n ends with a 9, then the total sum can be calculated as 10 * sum(n/10) + (n+1)/10 * 45. The first term is the sum of all digits but the last, and the second term is the sum of the last digit.
The function to calculate the total sum becomes:
static long GetSumDigitFrom1toN(long n)
{
long num2 = 0;
long i;
long r = n%10 + 1;
if (n <= 0)
{
return 0;
}
for (i = 0; i < r; i++)
{
num2 += GetSumOfDigits(n - i);
}
// The magic number 45 is the sum of 1 to 9.
return num2 + 10 * GetSumDigitFrom1toN(n/10 - 1) + (n/10) * 45;
}
Test run:
GetSumDigitFrom1toN(12L): 51
GetSumDigitFrom1toN(123456789L): 4366712385
The time complexity is O(log n).

Sum of digits for 0..99999999 is 10000000 * 8 * (0 + 1 + 2 + ... + 9).
Then calculating the rest (100000000..123456789) using the loop might be fast enough.
For N = 12:
Sum of digits for 0..9 is 1 * 1 * 45. Then use your loop for 10, 11, 12.
For N = 123:
Sum of digits for 0..99 is 10 * 2 * 45. Then use your loop for 100..123.
You see the pattern?

A different approach you can try:
Convert the number to a string, then to a Char array
Sum the ASCII codes for all the chars, minus the code for 0
Example code:
long num = 123456789;
var numChars = num.ToString().ToCharArray();
var zeroCode = Convert.ToByte('0');
var sum = numChars.Sum(ch => Convert.ToByte(ch) - zeroCode);

Finding a number based on how many loops of the Collatz Conjecture it takes to reach 1

Basically, I want to write an algorithm to find out which number takes 500 iterations to reach 1. I tried some variations but couldn't get it right.
Here is my code so far:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace sequence4
{
class Program
{
static void Main(string[] args)
{
long startingNumber = 1;
long count = 0;
while (count != 500)
{
startingNumber = startingNumber * 2;
count++;
startingNumber = startingNumber / 3 - 1;
count++;
}
Console.WriteLine(count);
Console.WriteLine(startingNumber);
}
}
}
EDIT: Updated version of the code
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace sequence4
{
class Program
{
static void Main(string[] args)
{
int number = 2;
int count = 0;
while (count != 500)
{
if (number % 2 == 0)
{
number = 2 * number;
count++;
}
if (number % 2 != 0)
{
number = (number / 3) - 1;
count++;
}
}
Console.WriteLine(number);
Console.WriteLine(count);
}
}
}

Collatz Conjecture's example is:
Consider, we have a number 7 and we need to reach 1 using the Collatz Conjecture
7 is odd so we use the algorithm 3(7) + 1 = 22
22 is even so we use 22/2 = 11
11 is odd so we use the algorithm 3(11) + 1 = 34
34 is even so we use the algorithm 34/2 = 17
17 is odd so we use the algorithm 3(17) + 1 = 52
52 is even so we use the algorithm 52/2 = 26
26 is even so we use the algorithm 26/2 = 13
13 is odd so we use the algorithm 13(3) + 1 = 40
40 is even so we use the algorithm 40/2 = 20
20 is even so we use the algorithm 20/2 = 10
10 is even so we use the algorithm 10/2 = 5
5 is odd so we use the algorithm 5(3) + 1 = 16
16 is even so we use the algorithm 16/2 = 8
8 is even so we use the algorithm 8/2 = 4
4 is even so we use the algorithm 4/2 = 2
2 is even so we use the algorithm 2/2 = 1
For Odd Number: x = 3n + 1
For Even Number: x = n / 2
We have applied the algorithm 16 times for the number 7 and we got to 1. So, 16 is the cycle length.
Now, if we take above example, we need to move reverse from bottom line to 500 times upward.
For reverse iterations, we use:
For Odd Number: x = (n - 1) / 3
For Even Number: x = n * 2
Now, programmatically, implement as:
using System;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
double output = 1;
const int iterations = 500;
for (var i = 1; i <= iterations; i++)
{
output = GetOutput(output);
Console.WriteLine("Number after {0} iterations is: {1}", i, output);
}
Console.WriteLine("Required Number is: {0}", output);
VerifyResult(output, iterations);
Console.ReadKey();
}
private static double GetOutput(double input)
{
if (input == 1)
{
return 2;
}
var output = (input - 1) / 3;
return output % 1 == 0 && output % 2 != 0 && output > 3 ? output : input * 2;
}
//To verify the above results we need this method
private static void VerifyResult(double output, int iterations)
{
//-------------------------VERIFICATION-----------------------
Console.WriteLine("Press any key to check iterations in reverse");
Console.ReadKey();
Console.WriteLine("Running validation process ...");
var n = output;
var max = n;
var count = 0;
Console.WriteLine("{0} (starting number in Collatz Sequence)", n);
while (n > 1)
{
n = n % 2 == 0 ? n / 2 : 3 * n + 1;
count++;
if (n > max) max = n;
Console.WriteLine(n);
}
if (count == iterations) //match here iterations and outputs
{
Console.WriteLine("\n\nCONGRATULATION! Verification results matched. :-)\n\n");
Console.WriteLine("There are {0} cycle length in the sequence", count);
Console.WriteLine("The largest number in the sequence is {0}", output);
Console.WriteLine("\n\n-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-");
Console.WriteLine("\n\nREQUIRED NUMBER: {0}\n\n", output);
Console.WriteLine("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-\n");
Console.WriteLine("\nPress any key to exit");
}
else
{
Console.WriteLine("Oops... Verification results are not matching. :-(");
}
}
}
}
Example's Source: Algorithm guidance with 3n+1 Conjecture

Quickest Method to Reverse in String in C#.net

I'm currently writing a quick solution for Euler Problem #4 where one must find the largest palindromic number from the product of two 3-digit numbers.
To identify if a number is palindromic, you would obviously compare a reverse of the number with the original.
Since C# doesn't have a built in String.Reverse() method, what is the quickest way to reverse a string?
I will be testing all the suggested solution in a loop with 100,000,000 iterations. The correct answer will be given to the person who submitted the fastest solution.
I will be testing the solution in a C#.Net 3.5 console application

Wouldn't reversing the number be faster?
// unchecked code, don't kill me if it doesn't even compile.
ulong Reverse(ulong number) {
ulong result = 0;
while (number > 0) {
ulong digit = number % 10;
result = result * 10 + digit;
number /= 10;
}
return result;
}

A you want to compare a number with its reverse it may be faster to reverse the number using division rather than converting it to a string. I still need to test the speed of it.
private static int Reverse(int num) {
int res = 0;
while (num > 0) {
int rm ;
num = Math.DivRem(num, 10, out rm);
res = res * 10 + rm;
}
return res;
}
EDIT:
DivRem was about 1% faster than division and module in my computer.
A speed optimization is exit if the last digit is 0:
private static int Reverse(int num) {
int res = 0;
int rm;
num = Math.DivRem(num, 10, out rm);
//Some magic value or return false, see below.
if (rm == 0) return -1 ;
res = res * 10 + rm;
while (num > 0) {
num = Math.DivRem(num, 10, out rm);
res = res * 10 + rm;
}
return res ;
}
Making the method return a bool was slightly slower than comparing to a bool in a loop in my computer, but I don't understand why. Please test in your computer.
Multiplication and bit-shifing should be faster than division but probably are not precise enough. EDIT: using long seems be precise enough.
private static int FastReverse(int num) {
int res = 0;
int q = (int)((214748365L * num) >> 31);
int rm = num - 10 * q;
num = q;
if (rm == 0) return -1;
res = res * 10 + rm;
while (num > 0) {
q = (int)((214748365L * num) >> 31);
rm = num - 10 * q;
num = q;
res = res * 10 + rm;
}
return res;
}
(214748365L * num) >> 31 is equal to i / 10 until 1,073,741,829 where 1 / 10 gives 107374182 and the multiplication + binary shifting gives 107374183.

I think it might be faster to do the comparison in-place. If you reverse the string, you've got to:
Instantiate a new string object (or StringBuffer object)
Copy the data (in reverse) from the first string to the new string
Do your comparison.
If you perform the comparison in place, you do only the last step. An even then, your comparison is only half the string (or half - 0.5, in the event of an odd number of characters). Something like the following should work:
static bool IsPalindromic(string s){
int len = s.Length;
int half = len-- >> 1;
for(int i = 0; i < half; i++)
if(s[i] != s[len - i])
return false;
return true;
}
EDIT:
Although this answers the OP's question, the solutions offered by ggf31416 and configurator solve the OP's real need about 30% faster, by my tests. configurator's solution is a tiny bit faster than ggf31416's, if you convert it to a static method and use ints instead of ulongs (but much slower, otherwise).
Incidentally, running through these examples to solve the problem the OP mentions (finding the largest palindromic product of any two three-digit numbers) with the simple (perhaps naïve) loop below:
for(int i = 100; i < 1000; i++)
for(int j = i; j < 1000; j++) // calculations where j < i would be redundant
...
yields the following results on my machine:
IsPalindromic(product.ToString()) took 0.3064174 seconds.
ggf31416Reverse(product) == product took 0.1933994 seconds.
configuratorReverse(product) == product took 0.1872061 seconds.
Each produces the correct result of 913 * 993 = 906609.

Performance: Fastest string reversing algorithms... (final results)

string test = "ABC";
string reversed = new String(test.ToCharArray().Reverse().ToArray());

public static String Reverse(string input) {
var length = input.Length;
var buffer = new char[length];
for ( var i= 0; i < input.Length; i++ ) {
buffer[i] = input[(length-i)-1];
}
return new String(buffer);
}
EDIT: Doh! Forgot to halve the length for perf :)

The fastest way I have found to reverse a string in C# is with the following code. It's faster reading in 32bits at a time instead of a char's length of 16bits.
In debug mode, it is faster until you get to about 93 characters. Anything longer than that Array.Reverse() is faster. Using a release build and running outside of the IDE, this method will blow Array.Reverse() out of the water at any string length.
char[] MyCharArray = MyString.ToCharArray();
UIntStringReverse(ref MyCharArray); //Code to reverse is below.
string ReversedString = new string(MyCharArray);
private static unsafe void UIntStringReverse(ref char[] arr)
{
uint Temp;
uint Temp2;
fixed (char* arrPtr = &arr[0])
{
uint* p, q;
p = (uint*)(arrPtr);
q = (uint*)(arrPtr + arr.LongLength - 2);
if (arr.LongLength == 2)
{
Temp = *p;
*p = ((Temp & 0xFFFF0000) >> 16) | ((Temp & 0x0000FFFF) << 16);
return;
}
while (p < q)
{
Temp = *p;
Temp2 = *q;
*p = ((Temp2 & 0xFFFF0000) >> 16) | ((Temp2 & 0x0000FFFF) << 16);
*q = ((Temp & 0xFFFF0000) >> 16) | ((Temp & 0x0000FFFF) << 16);
p++;
q--;
}
}
}

try this too:
http://weblogs.sqlteam.com/mladenp/archive/2006/03/19/9350.aspx

string Reverse(string s)
{
return new string(s.ToCharArray().Reverse().ToArray());
}

Using ggf31416's FastReverse function, here is the solution to Project Euler's Problem #4 which completes on my computer in 47ms.
using System;
using System.Diagnostics;
namespace Euler_Problem_4
{
class Program
{
static void Main(string[] args)
{
Stopwatch s = new Stopwatch();
s.Start();
int t = 0;
for (int i = 999; i > 99; i--)
{
for (int j = i; j > 99; j--)
{
if (i*j == FastReverse(i*j))
{
if (i * j > t)
{
t = i * j;
}
}
}
}
Console.WriteLine(t);
s.Stop();
Console.WriteLine("{0}mins {1}secs {2}ms", s.Elapsed.Minutes, s.Elapsed.Seconds, s.Elapsed.Milliseconds);
Console.ReadKey(true);
}
private static int FastReverse(int num)
{
int res = 0;
int q = (int)((214748365L * num) >> 31);
int rm = num - 10 * q;
num = q;
if (rm == 0) return -1;
res = res * 10 + rm;
while (num > 0)
{
q = (int)((214748365L * num) >> 31);
rm = num - 10 * q;
num = q;
res = res * 10 + rm;
}
return res;
}
}
}

The Stopwatch class needs reset after each run. the code below has been corrected
var d = s.ToCharArray();
Array.Reverse(d);
return s == new string(d);
using System;
using System.Diagnostics;
namespace longeststring_codegolf
{
class Program
{
static void Main(string[] args)
{
int t = 0, v = 0;
var sw = new Stopwatch();
sw.Start();
for (int i = 999; i > 99; i--)
for (int j = 999; j > 99; j--)
if ((v = i * j) > t && IsPalindromicMine(v.ToString()))
t = v;
sw.Stop();
var elapsed = sw.Elapsed;
var elapsedMilliseconds = sw.ElapsedMilliseconds;
var elapsedTicks = sw.ElapsedTicks;
Console.WriteLine("Ticks: " + elapsedTicks.ToString());//~189000
Console.WriteLine("Milliseconds: " + elapsedMilliseconds.ToString()); //~9
sw = Stopwatch.StartNew();
for (int i = 999; i > 99; i--)
for (int j = 999; j > 99; j--)
if ((v = i * j) > t && IsPalindromic(v.ToString()))
t = v;
sw.Stop();
var elapsed2 = sw.Elapsed;
var elapsedMilliseconds2 = sw.ElapsedMilliseconds;
var elapsedTicks2 = sw.ElapsedTicks;
Console.WriteLine("Ticks: " + elapsedTicks2.ToString());//~388000
Console.WriteLine("Milliseconds: " + elapsedMilliseconds2.ToString());//~20
}
static bool IsPalindromicMine(string s)
{
var d = s.ToCharArray();
Array.Reverse(d);
return s == new string(d);
}
static bool IsPalindromic(string s)
{
int len = s.Length;
int half = len-- >> 1;
for (int i = 0; i < half; i++)
if (s[i] != s[len - i])
return false;
return true;
}
}
}