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How to take digits from two different numbers and form a new one

I have the following problem here:My input is several lines of 2 digit numbers and I need to make a new number using the second digit of the first number and the first of the next one.
Example:
int linesOfNumbers = Convert.ToInt32(Console.ReadLine());
for(int i = 0,i<linesOfNumbers,i++)
{
int numbers = Conver.ToInt32(Console.ReadLine());
//that's for reading the input
}
I know how to separate the numbers into digits.My question is how to merge them.
For example if your input is 12 and 21 the output should be 22.

I like oRole's answer, but I think they're missing a couple things with the example input that you provided in your comment. I'll also point out some of the errors in the code that you have.
First off, if you're only given the input 12,23,34,45, then you don't need to call Console.ReadLine within your for loop. You've already gotten the input, you don't need to get any more (from what you've described).
Secondly, unless you're doing mathematical operations, there is no need to store numerical data as ints, keep it as a string, especially in this case. (What I mean is that you don't store Zip Codes in a database as a number, you store it as a string.)
Now, onto the code. You had the right way to get your data:
var listOfNumbers = Console.ReadLine();
At that point, listOfNumbers is equal to "12,23,34,45". If you iterate on that variable as a string, you'll be taking each individual character, including the commas. To get each of the numbers to operate on, you'll need to use string.Split.
var numbers = listOfNumbers.Split(',');
This turns that list into four different two character numbers (in string form). Now, you can iterate over them, but you don't need to worry about converting them to numbers as you're operating on the characters in each string. Also, you'll need a results collection to put everything into.
var results = new List<string>();
// Instead of the regular "i < numbers.Length", we want to skip the last.
for (var i = 0; i < numbers.Length - 1; i++)
{
var first = numbers[i];
var second = numbers[i + 1]; // This is why we skip the last.
results.Add(first[1] + second[0]);
}
Now your results is a collection of the numbers "22", "33", and "44". To get those back into a single string, you can use the helper method string.Join.
Console.WriteLine(string.Join(",", results));

You could use the string-method .Substring(..) to achieve what you want.
If you want to keep int-conversion in combination with user input, you could do:
int numA = 23;
int numB = 34;
int resultAB = Convert.ToInt16(numA.ToString().Substring(1, 1) + numB.ToString().Substring(0, 1));
Another option would be to take the users input as string values and to convert them afterwards like that:
string numC = "12";
string numD = "21";
int resultCD = Convert.ToInt16(numC.Substring(1, 1) + numD.Substring(0, 1));

I hope this code snippet will help you combining your numbers. The modulo operator (%) means: 53 / 10 = 5 Rest 3
This example shows the computation of the numbers 34 and 12
int firstNumber = 34 - (34 % 10) // firstNumber = 30
int secondNumber = 12 % 10; // secondNumber = 2
int combined = firstNumber + secondNumber; // combined = 32
EDIT (added reading and ouput code):
boolean reading = true;
List<int> numbers = new ArrayList();
while(reading)
{
try
{
int number = Convert.ToInt32(Console.ReadLine());
if (number > 9 && number < 100) numbers.Add(number);
else reading = false; // leave reading process if no 2-digit-number
}
catch (Exception ex)
{
// leave reading process by typing a character instead of a number;
reading = false;
}
}
if (numbers.Count() > 1)
{
List<int> combined = new ArrayList();
for (int i = 1; i <= numbers.Count(); i++)
{
combined.Add((numbers[i-1] % 10) + (numbers[i] - (numbers[i] % 10)));
}
//Logging output:
foreach (int combination in combined) Console.WriteLine(combination);
}

As you mention, if you already have both numbers, and they are always valid two digit integers, following code should work for you.
var num1 = 12;
var num2 = 22;
var result = (num2 / 10)*10 + (num1 % 10);
num2/10 returns the first digit of second number, and num1 % 10 returns the second digit of the first number.

The % and / signs are your savior.
If you want the 'ones' digit of a number (lets call it X), simply do X%10 - the remainder will be whatever number is in the 'ones' digit. (23%10=3)
If, instead, the number is two digits and you want the 'tens' digit, divide it by ten. (19/10=1).
To merge them, multiply the number you want to be in the 'tens' digit by ten, and add the other number to it (2*10+2=22)

There are other solutions like substring, etc and many one have already given it above. I am giving the solution VIA LINQ, note that this isn't efficient and it's recommended only for learning purpose here
int numA = 12;
int numB = 21 ;
string secondPartofNumA = numA.ToString().Select(q => new string(q,1)).ToArray()[1]; // first digit
string firstPartofNumB = numB.ToString().Select(q => new string(q,1)).ToArray()[0]; // second digit
string resultAsString = secondPartofNumA + firstPartofNumB;
int resultAsInt = Convert.ToInt32(resultAsString);
Console.WriteLine(resultAsString);
Console.WriteLine(resultAsInt);

How to get count of numbers in int and how to split a number without making a string

I have a number like 601511616
If all number's length is multiple of 3, how can a split the number into an array without making a string
Also, how can I count numbers in the int without making a string?
Edit: Is there a way to simply split the number, knowing it's always in a multiple of 3... good output should look like this: {616,511,601}

You can use i % 10 in order to get the last digit of integer.
Then, you can use division by 10 for removing the last digit.
1234567 % 10 = 7
1234567 / 10 = 123456
Here is the code sample:
int value = 601511616;
List<int> digits = new List<int>();
while (value > 0)
{
digits.Add(value % 10);
value /= 10;
}
// digits is [6,1,6,1,1,5,1,0,6] now
digits.Reverse(); // Values has been inserted from least significant to the most
// digits is [6,0,1,5,1,1,6,1,6] now
Console.WriteLine("Count of digits: {0}", digits.Count); // Outputs "9"
for (int i = 0; i < digits.Count; i++) // Outputs "601,511,616"
{
Console.Write("{0}", digits[i]);
if (i > 0 && i % 3 == 0) Console.Write(","); // Insert comma after every 3 digits
}
IDEOne working demonstration of List and division approach.
Actually, if you don't need to split it up but only need to output in 3-digit groups, then there is a very convenient and proper way to do this with formatting.
It will work as well :)
int value = 601511616;
Console.WriteLine("{0:N0}", value); // 601,511,616
Console.WriteLine("{0:N2}", value); // 601,511,616.00
IDEOne working demonstration of formatting approach.

I can't understand your question regarding how to split a number into an array without making a string - sorry. But I can understand the question about getting the count of numbers in an int.
Here's your answer to that question.
Math.Floor(Math.Log10(601511616) + 1) = 9
Edit:
Here's the answer to your first question..
var n = 601511616;
var nArray = new int[3];
for (int i = 0, numMod = n; i < 3; numMod /= 1000, i++)
nArray[i] = numMod%1000;
Please keep in mind there's no safety in this operation.
Edit#3
Still not perfect, but a better example.
var n = 601511616;
var nLength = (int)Math.Floor(Math.Log10(n) + 1)/ 3;
var nArray = new int[nLength];
for (int i = 0, numMod = n; i < nLength; numMod /= 1000, i++)
nArray[i] = numMod%1000;
Edit#3:
IDEOne example http://ideone.com/SSz3Ni
the output is exactly as the edit approved by the poster suggested.
{ 616, 511, 601 }

Using Log10 to calculate the number of digits is easy, but it involves floating-point operations which is very slow and sometimes incorrect due to rounding errors. You can use this way without calculating the value size first. It doesn't care if the number of digits is a multiple of 3 or not.
int value = 601511616;
List<int> list = new List<int>();
while (value > 0) // main part to split the number
{
int t = value % 1000;
value /= 1000;
list.Add(t);
}
// Convert back to an array only if it's necessary, otherwise use List<T> directly
int[] splitted = list.ToArray();
This will store the splitted numbers in reverse order, i.e. 601511616 will become {616, 511, 601}. If you want the numbers in original order, simply iterate the array backwards. Alternatively use Array.Reverse or a Stack

Since you already know they are in multiples of 3, you can just use the extracting each digit method but use 1000 instead of 10. Here is the example
a = 601511616
b = []
while(a):
b.append(a%1000)
a = a//1000
print(b)
#[616, 511, 601]

How to pick one random string from given strings?

How can I make a program that will pick one random string from given strings like this:
int x;
x = Random.Range(0,2);
string[] Quest0 = {"You","Are","How","Hello"};
string[] Quest1 = {"Hey","Hi","Why","Yes"};
string[] Quest2 = {"Here","Answer","One","Pick"};
I would like to print out like this:
if x = 2 it would print out Quest2 and so on.
Thank you!

List<String[]> quests = new ArrayList<String[]>();
quests.add(0, new string[]{"You","Are","How","Hello"});
quests.add(1, new string[]{"Hey","Hi","Why","Yes"});
quests.add(2, new string[]{"Here","Answer","One","Pick"});
int x = new Random().nextInt((2 - 0) + 1);
System.out.println(quests.get(x).toString());

Fistly you need to declare a random variable.
Random random = new Random();
this will create a variable in which you can now get random numbers from. to get random numbers you will use random.next(x,y) or in your case random.next(0,3) because the final argument is exclusive, so if you want 0, 1 or 2, you must use (0,3).
you then need to make some conditional statments, i would use If statments, to accomplish your goal use something like this:
if (x == 2)
{
foreach (string s in Quest2)
{
Console.WriteLine(s);
}
}
Do this for each possible outcome and it will print out all of the values in your array of strings. Hope I have been helpful, thanks.
Also if you new become familiar with these links:
http://msdn.microsoft.com/en-us/library/system.random%28v=vs.110%29.aspx
http://msdn.microsoft.com/en-gb/library/aa288453%28v=vs.71%29.aspx

dont want a random number to show twice

I am trying to save random numbers in a list, the same number can not come twice.
It´s a BINGO game where this method is used to display ex B12, then the user hits enter and a new number will show. This will keep on until the user writes q.
This works, BUT the number can show up twice...
static void bingo()
{
Random rnd =new Random();
List<int> check = new List<int>();
string choice = "";
while (choice != "Q")
{
int number = rnd.Next(1, 76);
while (!check.Contains(number))
{
kontroll.Add(number);
}
if (number <=15)
{
choice = Interaction.InputBox("B" + number);
choice = choice.ToUpper();
}
else if(number <= 30)
etc.

Something like this should work (if I'm reading your question correctly)
Enumerable.Range(1,76).OrderBy(n => rnd.NextDouble())

There are a couple of ways to do this:
Keep track of what numbers have been "called" - if a number is in the list, pick a different one
Remove numbers that have been called from the original list, then pick a new one at random each time.
Sort the list of possible values by a random number and just work though the list

The easiest way to accomplish this is to use a HashSet.
var usedNumbers = new HashSet<int>();
...
int number;
do {
number = rnd.Next(1, 76);
} while (usedNumbers.Contains(number));
usedNumbers.Add(number);

Performance issue with generation of random unique numbers

I have a situation where by I need to create tens of thousands of unique numbers. However these numbers must be 9 digits and cannot contain any 0's. My current approach is to generate 9 digits (1-9) and concatenate them together, and if the number is not already in the list adding it into it. E.g.
public void generateIdentifiers(int quantity)
{
uniqueIdentifiers = new List<string>(quantity);
while (this.uniqueIdentifiers.Count < quantity)
{
string id = string.Empty;
id += random.Next(1,10);
id += random.Next(1,10);
id += random.Next(1,10);
id += " ";
id += random.Next(1,10);
id += random.Next(1,10);
id += random.Next(1,10);
id += " ";
id += random.Next(1,10);
id += random.Next(1,10);
id += random.Next(1,10);
if (!this.uniqueIdentifiers.Contains(id))
{
this.uniqueIdentifiers.Add(id);
}
}
}
However at about 400,000 the process really slows down as more and more of the generated numbers are duplicates. I am looking for a more efficient way to perform this process, any help would be really appreciated.
Edit: - I'm generating these - http://www.nhs.uk/NHSEngland/thenhs/records/Pages/thenhsnumber.aspx

As others have mentioned, use a HashSet<T> instead of a List<T>.
Furthermore, using StringBuilder instead of simple string operations will gain you another 25%. If you can use numbers instead of strings, you win, because it only takes a third or fourth of the time.
var quantity = 400000;
var uniqueIdentifiers = new HashSet<int>();
while (uniqueIdentifiers.Count < quantity)
{
int i=0;
i = i*10 + random.Next(1,10);
i = i*10 + random.Next(1,10);
i = i*10 + random.Next(1,10);
i = i*10 + random.Next(1,10);
i = i*10 + random.Next(1,10);
i = i*10 + random.Next(1,10);
i = i*10 + random.Next(1,10);
i = i*10 + random.Next(1,10);
i = i*10 + random.Next(1,10);
uniqueIdentifiers.Add(i);
}
It takes about 270 ms on my machine for 400,000 numbers and about 700 for 1,000,000. And this even without any parallelism.
Because of the use of a HashSet<T> instead of a List<T>, this algorithm runs in O(n), i.e. the duration will grow linear. 10,000,000 values therefore take about 7 seconds.

This suggestion may or may not be popular.... it depends on people's perspective. Because you haven't been too specific about what you need them for, how often, or the exact number, I will suggest a brute force approach.
I would generate a hundred thousand numbers - shouldn't take very long at all, maybe a few seconds? Then use Parallel LINQ to do a Distinct() on them to eliminate duplicates. Then use another PLINQ query to run a regex against the remainder to eliminate any with zeroes in them. Then take the top x thousand. (PLINQ is brilliant for ripping through large tasks like this). If needed, rinse and repeat until you have enough for your needs.
On a decent machine it will just about take you longer to write this simple function than it will take to run it. I would also query why you have 400K entries to test when you state you actually need "tens of thousands"?

The trick here is that you only need ten thousand unique numbers. Theoretically you could have almost 9,0E+08 possibilities, but why care if you need so many less?
Once you realize that you can cut down on the combinations that much then creating enough unique numbers is easy:
long[] numbers = { 1, 3, 5, 7 }; //note that we just take a few numbers, enough to create the number of combinations we might need
var list = (from i0 in numbers
from i1 in numbers
from i2 in numbers
from i3 in numbers
from i4 in numbers
from i5 in numbers
from i6 in numbers
from i7 in numbers
from i8 in numbers
from i9 in numbers
select i0 + i1 * 10 + i2 * 100 + i3 * 1000 + i4 * 10000 + i5 * 100000 + i6 * 1000000 + i7 * 10000000 + i8 * 100000000 + i9 * 1000000000).ToList();
This snippet creates a list of more than a 1,000,000 valid unique numbers pretty much instantly.

Try avoiding checks making sure that you always pick up a unique number:
static char[] base9 = "123456789".ToCharArray();
static string ConvertToBase9(int value) {
int num = 9;
char[] result = new char[9];
for (int i = 8; i >= 0; --i) {
result[i] = base9[value % num];
value = value / num;
}
return new string(result);
}
public static void generateIdentifiers(int quantity) {
var uniqueIdentifiers = new List<string>(quantity);
// we have 387420489 (9^9) possible numbers of 9 digits in base 9.
// if we choose a number that is prime to that we can easily get always
// unique numbers
Random random = new Random();
int inc = 386000000;
int seed = random.Next(0, 387420489);
while (uniqueIdentifiers.Count < quantity) {
uniqueIdentifiers.Add(ConvertToBase9(seed));
seed += inc;
seed %= 387420489;
}
}
I'll try to explain the idea behind with small numbers...
Suppose you have at most 7 possible combinations. We choose a number that is prime to 7, e.g. 3, and a random starting number, e.g. 4.
At each round, we add 3 to our current number, and then we take the result modulo 7, so we get this sequence:
4 -> 4 + 3 % 7 = 0
0 -> 0 + 3 % 7 = 3
3 -> 3 + 3 % 7 = 6
6 -> 6 + 6 % 7 = 5
In this way, we generate all the values from 0 to 6 in a non-consecutive way. In my example, we are doing the same, but we have 9^9 possible combinations, and as a number prime to that I choose 386000000 (you just have to avoid multiples of 3).
Then, I pick up the number in the sequence and I convert it to base 9.
I hope this is clear :)
I tested it on my machine, and generating 400k unique values took ~ 1 second.

Meybe this will bee faster:
//we can generate first number wich in 9 base system will be between 88888888 - 888888888
//we can't start from zero becouse it will couse the great amount of 1 digit at begining
int randNumber = random.Next((int)Math.Pow(9, 8) - 1, (int)Math.Pow(9, 9));
//no we change our number to 9 base, but we add 1 to each digit in our number
StringBuilder builder = new StringBuilder();
for (int i=(int)Math.Pow(9,8); i>0;i= i/9)
{
builder.Append(randNumber / i +1);
randNumber = randNumber % i;
}
id = builder.ToString();

Looking at the solutions already posted, mine seems fairly basic. But, it works, and generates 1million values in approximate 1s (10 million in 11s).
public static void generateIdentifiers(int quantity)
{
HashSet<int> uniqueIdentifiers = new HashSet<int>();
while (uniqueIdentifiers.Count < quantity)
{
int value = random.Next(111111111, 999999999);
if (!value.ToString().Contains('0') && !uniqueIdentifiers.Contains(value))
uniqueIdentifiers.Add(value);
}
}

use string array or stringbuilder, wjile working with string additions.
more over, your code is not efficient because after generating many id's your list may hold new generated id, so that the while loop will run more than you need.
use for loops and generate your id's from this loop without randomizing. if random id's are required, use again for loops and generate more than you need and give an generation interval, and selected from this list randomly how much you need.
use the code below to have a static list and fill it at starting your program. i will add later a second code to generate random id list. [i'm a little busy]
public static Random RANDOM = new Random();
public static List<int> randomNumbers = new List<int>();
public static List<string> randomStrings = new List<string>();
private void fillRandomNumbers()
{
int i = 100;
while (i < 1000)
{
if (i.ToString().Contains('0') == false)
{
randomNumbers.Add(i);
}
}
}

I think first thing would be to use StringBuilder, instead of concatenation - you'll be pleasantly surprised.
Antoher thing - use a more efficient data structure, for example HashSet<> or HashTable.
If you could drop the quite odd requirement not to have zero's - then you could of course use just one random operation, and then format your resulting number the way you want.

I think #slugster is broadly right - although you could run two parallel processes, one to generate numbers, the other to verify them and add them to the list of accepted numbers when verified. Once you have enough, signal the original process to stop.
Combine this with other suggestions - using more efficient and appropriate data structures - and you should have something that works acceptably.
However the question of why you need such numbers is also significant - this requirement seems like one that should be analysed.

Something like this?
public List<string> generateIdentifiers2(int quantity)
{
var uniqueIdentifiers = new List<string>(quantity);
while (uniqueIdentifiers.Count < quantity)
{
var sb = new StringBuilder();
sb.Append(random.Next(11, 100));
sb.Append(" ");
sb.Append(random.Next(11, 100));
sb.Append(" ");
sb.Append(random.Next(11, 100));
var id = sb.ToString();
id = new string(id.ToList().ConvertAll(x => x == '0' ? char.Parse(random.Next(1, 10).ToString()) : x).ToArray());
if (!uniqueIdentifiers.Contains(id))
{
uniqueIdentifiers.Add(id);
}
}
return uniqueIdentifiers;
}