Develop Reference

I am trying to find a value of multiple factorials . both values will be divided by like 100!/98! =?

static void Main(string[] args)
{
Console.WriteLine("Enter your number: ");
int number= Convert.ToInt32(Console.ReadLine());
int number2 = Convert.ToInt32(Console.ReadLine());
double factorial = Factorial(number,number2);
Console.WriteLine("Factorial of " + number +" / "+ number2 + " = " + factorial );
Console.ReadKey();
}
//Factorial function added
public static double Factorial(int number, int number2)
{
if (number == 1 && number2 ==1 )
{
return 1;
}
double factorial = 1;
double factorial1 = 1;
double factorial2 = 1;
for (int i = number, j = number2; i >= 1 && j >= 1; i--, j--)
{
factorial1 = (factorial * i);
factorial2 = (factorial * j);
factorial = factorial1 / factorial2;
}
return factorial;
}

Your attempted solution is simply so overcomplicated, I wouldn't know where to begin. This usually happens when you don't stop to think about how you'd resolve this problem by hand:
So, the question is, whats 5!/3!? Ok, lets write it out:
(5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)
Wow, that looks like it can be simplified simply to 5 * 4.
The key insight here is that m! % n! = 0 if n is less or equal to m. In other words, m! is always divisible by n! because there is always an integer r such that r * n! = m!, and you don't need to evaluate m! or n! to figure out what r is, you simple do:
r = m * (m - 1) * (m - 2) * ... * (n + 1); // m >= n
If n > m, r is zero unless you are looking for a real number solution in which case you would simply evaluate r as n! / m! and then return 1.0 / r because m! / n! = 1 / (n! / m!).
How to evaluate r?
public static long DivideFactorials(int m, int n)
{
if (n > m)
return 0;
var r = 1L;
for (var k = m; k > n; k--)
r *= k;
return r;
}
Or the real number solution:
public static double DivideFactorials(int m, int n)
{
if (n > m)
return 1 / DivideFactorials(n, m);
var r = 1.0;
for (var k = m; k > n; k--)
r *= k;
return r;
}

If I had to save your try:
public static double Factorial(int number, int number2)
{
if (number == 1 && number2 == 1)
{
return 1;
}
double facNum = 1;
double facNum2 = 1;
// counting up is easier, we start at 2 as we initialized to 1
// we count up to the max of both numbers
for (int i = 2; i <= Math.Max(number, number2); i++)
{
if (i <= number)
facNum *= i; // we mult this until we reached number
if (i <= number2)
facNum2 *= i; // we mult this until we reach number2
}
// return the devision of both - this wont handle number < number2 well!
return facNum / facNum2; // do this outside the loop
}
If I had to create a solution:
Factorial division of integers has 3 outcomes (I can think of):
N! / O! with N == O:
let N=3, O=3
N! = 1*2*3
O! = 1*2*3
N! / O! = 1*2*3/(1*2*3) == 1
N! / O! with N > O:
let N=5, O=3
N! = 1*2*3*4*5
O! = 1*2*3
N! / O! == 1*2*3*4*5/(1*2*3) == 4*5 == 20
N! / O! with N < O:
let N=3, O=5
N! = 1*2*3
O! = 1*2*3*4*5
N! / O! == 1*2*3/(1*2*3*4*5) == 1/(4*5) == 1/20
Based on this I would model the problem like that:
using System;
using System.Collections.Generic;
using System.Linq;
internal class Program
{
public static decimal CalcFactDivision(int n1, int n2)
{
// calclulate the division of a factorial by another, num1 must be >= num2
IEnumerable<int> getRemaining(int num1, int num2)
{
// special cases: div by 0 and 0 div something
if (num2 == 0)
num2 = 1; // 0! == 1
else if (num1 == 0)
return new[] { 0 };
// get all numbers that make up the factorial in one step
// I can guarantee that num1 will always be bigger then num2
// by how I call this
return Enumerable.Range(num2 + 1, num1 - num2);
}
// calculate the product of an ienumerable of ints
int product(IEnumerable<int> nums) => nums.Aggregate((a, b) => a * b);
if (n1 == n2)
return 1;
else if (n1 > n2) // use product(...) to calc
return product(getRemaining(n1, n2));
else // flip them and use 1/product(...) to calc
return (decimal)1 / product(getRemaining(n2, n1));
}
static void Main(string[] args)
{
foreach (var a in Enumerable.Range(1, 10))
Console.WriteLine($"{a}! / {10 - a}! = {CalcFactDivision(a, 10 - a)} ");
Console.ReadLine();
}
}
Output:
1! / 9! = 0,0000027557319223985890652557
2! / 8! = 0,0000496031746031746031746032
3! / 7! = 0,0011904761904761904761904762
4! / 6! = 0,0333333333333333333333333333
5! / 5! = 1
6! / 4! = 30
7! / 3! = 840
8! / 2! = 20160
9! / 1! = 362880
10! / 0! = 3628800

Why aren't 10 and 100 considered Kaprekar numbers?

I'm trying to do the Modified Kaprekar Numbers problem (https://www.hackerrank.com/challenges/kaprekar-numbers) which describes a Kaprekar number by
Here's an explanation from Wikipedia about the ORIGINAL Kaprekar
Number (spot the difference!): In mathematics, a Kaprekar number for a
given base is a non-negative integer, the representation of whose
square in that base can be split into two parts that add up to the
original number again. For instance, 45 is a Kaprekar number, because
45² = 2025 and 20+25 = 45.
and what I don't understand is why 10 and 100 aren't Kaprekar numbers.
10^2 = 1000 and 10 + 00 = 10
Right?
So my solution
// Returns the number represented by the digits
// in the range arr[i], arr[i + 1], ..., arr[j - 1].
// If there are no elements in range, return 0.
static int NumberInRange(int[] arr, int i, int j)
{
int result = 0;
for(; i < j; ++i)
{
result *= 10;
result += arr[i];
}
return result;
}
// Returns true or false depending on whether k
// is a Kaprekar number.
// Example: IsKaprekar(45) = true because 45^2=2025 and 20+25=45
// Example: IsKaprekar(9) = false because the set of the split
// digits of 7^2=49 are {49,0},{4,9} and
// neither of 49+0 or 4+9 equal 7.
static bool IsKaprekar(int k)
{
int square = k * k;
int[] digits = square.ToString().Select(c => (int)Char.GetNumericValue(c)).ToArray();
for(int i = 0; i < digits.Length; ++i)
{
int right = NumberInRange(digits, 0, i);
int left = NumberInRange(digits, i, digits.Length);
if((right + left) == k)
return true;
}
return false;
}
is saying all the Kaprekar numbers between 1 and 100 are
1 9 10 45 55 99 100
whereas the "right" answer is
1 9 45 55 99

In 100+00 the right is 00, which is wrong because in a kaprekar number the right may start with zero (ex: 025) but cannot be entirely 0.
Therefore you can put a condition in the loop that
if(right==0)
return false;

The reason is because 10 x 10 = 100. Then you substring the right part with a length equals d = 2, that is digit count of original value (10), then the left part would be 1.
So l = 1 and r = 00, l + r = 1, that is not equals to 10.
The same for 100. 100 x 100 = 10000. l = 10, r = 000, so l + r = 10 not equal 100.
Here is my solution in JAVA.
static void kaprekarNumbers(int p, int q) {
long[] result = IntStream.rangeClosed(p, q).mapToLong(Long::valueOf)
.filter(v -> {
int d = String.valueOf(v).length();
Long sq = v * v;
String sqSt = sq.toString();
if (sqSt.length() > 1) {
long r = Long.parseLong(sqSt.substring(sqSt.length() - d));
long l = Long.parseLong(sqSt.substring(0, sqSt.length() - d));
return r + l == v;
} else return v == 1;
}).toArray();
if (result.length > 0) {
for (long l : result) {
System.out.print(l + " ");
}
} else {
System.out.println("INVALID RANGE");
}
}

How about something like this.
static bool IsKaprekar(int k)
{
int t;
for (int digits = new String(k).length(); digits > 0; digits--, t *= 10);
long sq = k * k;
long first = sq / t;
long second = sq % t;
return k == first + second;
}
find a number to divide and mod the square with in order to split it. This number should be a factor of 10 based on the number of digits in the original number.
calculate the square.
split the square.
compare the original to the sum of the splits.

How can I determine if a certain number can be made up from a set of numbers?

So I have an integer, e.g. 1234567890, and a given set of numbers, e.g. {4, 7, 18, 32, 57, 68}
The question is whether 1234567890 can be made up from the numbers given (you can use a number more than once, and you don't have to use all of them). In the case above, one solution is:38580246 * 32 + 1 * 18
(Doesn't need to give specific solution, only if it can be done)
My idea would be to try all solutions. For example I would try1 * 4 * + 0 * 7 + 0 * 18 + 0 * 32 + 0 * 57 + 0 * 68 = 42 * 4 * + 0 * 7 + 0 * 18 + 0 * 32 + 0 * 57 + 0 * 68 = 83 * 4 * + 0 * 7 + 0 * 18 + 0 * 32 + 0 * 57 + 0 * 68 = 12.....308 641 972 * 4 * + 0 * 7 + 0 * 18 + 0 * 32 + 0 * 57 + 0 * 68 = 1234567888308 641 973 * 4 * + 0 * 7 + 0 * 18 + 0 * 32 + 0 * 57 + 0 * 68 = 1234567892 ==> exceeds0 * 4 * + 1 * 7 + 0 * 18 + 0 * 32 + 0 * 57 + 0 * 68 = 71 * 4 * + 1 * 7 + 0 * 18 + 0 * 32 + 0 * 57 + 0 * 68 = 112 * 4 * + 1 * 7 + 0 * 18 + 0 * 32 + 0 * 57 + 0 * 68 = 15and so on...
Here is my code in c#:
static int toCreate = 1234567890;
static int[] numbers = new int[6] { 4, 7, 18, 32, 57, 68};
static int[] multiplier;
static bool createable = false;
static void Main(string[] args)
{
multiplier = new int[numbers.Length];
for (int i = 0; i < multiplier.Length; i++)
multiplier[i] = 0;
if (Solve())
{
Console.WriteLine(1);
}
else
{
Console.WriteLine(0);
}
}
static bool Solve()
{
int lastIndex = 0;
while (true)
{
int comp = compare(multiplier);
if (comp == 0)
{
return true;
}
else if (comp < 0)
{
lastIndex = 0;
multiplier[multiplier.Length - 1]++;
}
else
{
lastIndex++;
for (int i = 0; i < lastIndex; i++)
{
multiplier[multiplier.Length - 1 - i] = 0;
}
if (lastIndex >= multiplier.Length)
{
return false;
}
multiplier[multiplier.Length - 1 - lastIndex]++;
}
}
}
static int compare(int[] multi)
{
int osszeg = 0;
for (int i = 0; i < multi.Length; i++)
{
osszeg += multi[i] * numbers[i];
}
if (osszeg == toCreate)
{
return 0;
}
else if (osszeg < toCreate)
{
return -1;
}
else
{
return 1;
}
}
The code works fine (as far as I know) but is way too slow. It takes about 3 secs to solve the example, and there may be 10000 numbers to make from 100 numbers.

I have a recursive solution. It solves the OP's original problem in about .005 seconds (on my machine) and tells you the calculations.
private static readonly int Target = 1234567890;
private static readonly List<int> Parts = new List<int> { 4, 7, 18, 32, 57, 68 };
static void Main(string[] args)
{
Console.WriteLine(Solve(Target, Parts));
Console.ReadLine();
}
private static bool Solve(int target, List<int> parts)
{
parts.RemoveAll(x => x > target || x <= 0);
if (parts.Count == 0) return false;
var divisor = parts.First();
var quotient = target / divisor;
var modulus = target % divisor;
if (modulus == 0)
{
Console.WriteLine("{0} X {1}", quotient, divisor);
return true;
}
if (quotient == 0 || parts.Count == 1) return false;
while (!Solve(target - divisor * quotient, parts.Skip(1).ToList()))
{
if (--quotient != 0) continue;
return Solve(target, parts.Skip(1).ToList());
}
Console.WriteLine("{0} X {1}", quotient, divisor);
return true;
}
Basically, it goes through each number to see if there is a possible solution "below" it given the current quotient and number. If there isn't, it subtracts 1 from the quotient and tries again. It does this until it exhausts all options for that number and then moves on to the next number if available. If all numbers are exhausted, there is no solution.

Don't have the means test the solution, but the following should do.
Given a target number target and a set numbers of valid numbers:
bool FindDecomposition(int target, IEnumerable<int> numbers, Queue<int> decomposition)
{
foreach (var i in numbers)
{
var remainder = target % i;
if (remainder == 0)
{
decomposition.Enqueue(i);
return true;
}
if (FindDecomposition(remainder, numbers.Where(n => n < i), decomposition))
{
return true;
}
}
return false
}
Building up n from decomposition is pretty straightforward.

You could always try using the modulo function in conjunction with LINQ expressions to solve the problem.
You would have a list and a running modulo variable to keep track of where you are at in your iteration. Then simply use recursion to determine whether or not you have meet the conditions.
One example would be the following:
static int toCreate = 1234567890;
static List<int> numbers = new List<int> { 4, 7 };
static void Main(string[] args)
{
numbers.Sort();
numbers.Reverse();
Console.WriteLine(Solve(numbers,toCreate).ToString());
}
static bool Solve(List<int> lst1, int runningModulo)
{
if (lst1.Count == 0 && runningModulo != 0)
return false;
if (lst1.Count == 0 || runningModulo == 0)
return true;
return numbers.Any(o => o < (toCreate % lst1.First())) ? //Are there any in the remaining list that are smaller in value than the runningModulo mod the first element in the list.
Solve(lst1.Where(o => o != lst1.First()).ToList(), runningModulo % lst1.First()) //If yes, then remove the first element and set the running modulo = to your new modulo
: Solve(lst1.Where(o => o != lst1.First()).ToList(), toCreate); //Otherwise, set the running modulo back to the original toCreate value.
}

An algorithm for a number divisible to n

At first user gives a number (n) to program, for example 5.
the program must find the smallest number that can be divided to n (5).
and this number can only consist of digits 0 and 9 not any other digits.
for example if user gives 5 to program.
numbers that can be divided to 5 are:
5, 10, 15, 20, 25, 30, ..., 85, 90, 95, ...
but 90 here is the smallest number that can be divided to 5 and also consist of digits (0 , 9). so answer for 5 must be 90.
and answer for 9 is 9, because it can be divided to 9 and consist of digit (9).
my code
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
Int64 i ,j=1,y=x;
bool t = false;
for (i = x + 1; t == false; i++)
{
if (i % 9 == 0 && i % 10 == 0 && i % x == 0)
{
j = i;
for (; (i /= 10) != 0; )
{
i /= 10;
if (i == 0)
t = true;
continue;
}
}
}
lblAnswer.Text = Convert.ToString(j);

If you're happy to go purely functional then this works:
Func<IEnumerable<long>> generate = () =>
{
Func<long, IEnumerable<long>> extend =
x => new [] { x * 10, x * 10 + 9 };
Func<IEnumerable<long>, IEnumerable<long>> generate2 = null;
generate2 = ns =>
{
var clean = ns.Where(n => n > 0).ToArray();
return clean.Any()
? clean.Concat(generate2(clean.SelectMany(extend)))
: Enumerable.Empty<long>();
};
return generate2(new[] { 9L, });
};
Func<long, long?> f = n =>
generate()
.Where(x => x % n == 0L)
.Cast<long?>()
.FirstOrDefault();
So rather than iterate through all possible values and test for 0 & 9 and divisibility, this just generates only numbers with 0 & 9 and then only tests for visibility. It is much faster this way.
I can call it like this:
var result = f(5L); // 90L
result = f(23L); //990909L
result = f(123L); //99999L
result = f(12321L); //90900999009L
result = f(123212L); //99909990090000900L
result = f(117238L); //990990990099990990L
result = f(1172438L); //null == No answer
These results are super fast. f(117238L) returns a result on my computer in 138ms.

You can try this way :
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
int counter;
for (counter = 1; !isValid(x * counter); counter++)
{
}
lblAnswer.Text = Convert.ToString(counter*x);
code above works by searching multiple of x incrementally until result that satisfy criteria : "consist of only 0 and or 9 digits" found. By searching only multiple of x, it is guaranteed to be divisible by x. So the rest is checking validity of result candidate, in this case using following isValid() function :
private static bool isValid(int number)
{
var lastDigit = number%10;
//last digit is invalid, return false
if (lastDigit != 0 & lastDigit != 9) return false;
//last digit is valid, but there is other digit(s)
if(number/10 >= 1)
{
//check validity of digit(s) before the last
return isValid(number/10);
}
//last digit is valid, and there is no other digit. return true
return true;
}
About strange empty for loop in snippet above, it is just syntactic sugar, to make the code a bit shorter. It is equal to following while loop :
counter = 1;
while(!isValid(input*counter))
{
counter++;
}

Use this simple code
int inputNumber = 5/*Or every other number, you can get this number from input.*/;
int result=1;
for (int i = 1; !IsOk(result,inputNumber); i++)
{
result = i*inputNumber;
}
Print(result);
IsOk method is here:
bool IsOk(int result, int inputNumber)
{
if(result%inputNumber!=0)
return false;
if(result.ToString().Replace("9",string.Empty).Replace("0",string.Empty).Length!=0)
return false;
return true;
}

My first solution has very bad performance, because of converting a number to string and looking for characters '9' and '0'.
New solution:
My new solution has very good performance and is a technical approach since of using Breadth-first search(BFS).
Algorithm of this solution:
For every input number, test 9, if it is answer print it, else add 2 child numbers (90 & 99) to queue, and continue till finding answer.
int inputNumber = 5;/*Or every other number, you can get this number from input.*/
long result;
var q = new Queue<long>();
q.Enqueue(9);
while (true)
{
result = q.Dequeue();
if (result%inputNumber == 0)
{
Print(result);
break;
}
q.Enqueue(result*10);
q.Enqueue(result*10 + 9);
}
Trace of number creation:
9
90,99
900,909,990,999
9000,9009,9090,9099,9900,9909,9990,9999
.
.
.

I wrote this code for console, and i used goto command however it is not prefered but i could not write it with only for.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace main
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Enter your number");
Int64 x = Convert.ToInt64(Console.ReadLine());
Int64 y, j, i, k, z = x;
x = x + 5;
loop:
x++;
for (i = 0, y = x; y != 0; i++)
y /= 10;
for (j = x, k = i; k != 0; j /= 10, k--)
{
if (j % 10 != 9)
if (j % 10 != 0)
goto loop;
}
if (x % z != 0)
goto loop;
Console.WriteLine("answer:{0}",x);
Console.ReadKey();
}
}
}

C# ModInverse Function

Is there a built in function that would allow me to calculate the modular inverse of a(mod n)?
e.g. 19^-1 = 11 (mod 30), in this case the 19^-1 == -11==19;

Since .Net 4.0+ implements BigInteger with a special modular arithmetics function ModPow (which produces “X power Y modulo Z”), you don't need a third-party library to emulate ModInverse. If n is a prime, all you need to do is to compute:
a_inverse = BigInteger.ModPow(a, n - 2, n)
For more details, look in Wikipedia: Modular multiplicative inverse, section Using Euler's theorem, the special case “when m is a prime”. By the way, there is a more recent SO topic on this: 1/BigInteger in c#, with the same approach suggested by CodesInChaos.

int modInverse(int a, int n)
{
int i = n, v = 0, d = 1;
while (a>0) {
int t = i/a, x = a;
a = i % x;
i = x;
x = d;
d = v - t*x;
v = x;
}
v %= n;
if (v<0) v = (v+n)%n;
return v;
}

The BouncyCastle Crypto library has a BigInteger implementation that has most of the modular arithmetic functions. It's in the Org.BouncyCastle.Math namespace.

Here is a slightly more polished version of Samuel Allan's algorithm. The TryModInverse method returns a bool value, that indicates whether a modular multiplicative inverse exists for this number and modulo.
public static bool TryModInverse(int number, int modulo, out int result)
{
if (number < 1) throw new ArgumentOutOfRangeException(nameof(number));
if (modulo < 2) throw new ArgumentOutOfRangeException(nameof(modulo));
int n = number;
int m = modulo, v = 0, d = 1;
while (n > 0)
{
int t = m / n, x = n;
n = m % x;
m = x;
x = d;
d = checked(v - t * x); // Just in case
v = x;
}
result = v % modulo;
if (result < 0) result += modulo;
if ((long)number * result % modulo == 1L) return true;
result = default;
return false;
}

There is no library for getting inverse mod, but the following code can be used to get it.
// Given a and b->ax+by=d
long[] u = { a, 1, 0 };
long[] v = { b, 0, 1 };
long[] w = { 0, 0, 0 };
long temp = 0;
while (v[0] > 0)
{
double t = (u[0] / v[0]);
for (int i = 0; i < 3; i++)
{
w[i] = u[i] - ((int)(Math.Floor(t)) * v[i]);
u[i] = v[i];
v[i] = w[i];
}
}
// u[0] is gcd while u[1] gives x and u[2] gives y.
// if u[1] gives the inverse mod value and if it is negative then the following gives the first positive value
if (u[1] < 0)
{
while (u[1] < 0)
{
temp = u[1] + b;
u[1] = temp;
}
}