divide a number equal to the input

[virusisfound]

How to divide a number into multiple parts so that the resulting sum is equal to the input?
for ex :
input from textbox :

first input is(txt_val1) :5
second input is(txt_val2) :3

output :
3+2=5
2+3=5
4+1=5
1+4=5
or any other combinations like these.
both input are dynamic always.

[ddanbe]

Is your number integer only?
For instance if you allow complex numbers, (2 + 42i) + (3 - 42i) would be a solution.

[Ketsuekiame]

Why do you input two values?

[tinstaafl]

Something like this will work:

    Dim input As Integer = Val(TextBox1.Text)
    Dim output As New List(Of String)
    If input > 0 Then
        For i = 1 To input - 1
            For j = 1 To input - 1
                If i + j = input Then
                    output.Add(i.ToString.Trim + "," + j.ToString.Trim)
                End If
            Next
        Next
        For Each s As String In output
            Debug.WriteLine(s.Split(",")(0) + " + " + s.Split(",")(1) + " = " + input.ToString.Trim)
        Next
    End If

Or if you want to keep the parts separate something like this would work:

Public Class Form1    
    Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
        Dim input As Integer = Val(TextBox1.Text)
        Dim output As New List(Of EquationParts)
        If input > 0 Then
            For i = 1 To input - 1
                For j = 1 To input - 1
                    If i + j = input Then
                        Dim e As New EquationParts
                        e.Firstpart = i
                        e.SecondPart = j
                        output.Add(e)
                    End If
                Next
            Next
            For Each e As EquationParts In output
                Debug.WriteLine(e.Firstpart.ToString.Trim + " + " + e.SecondPart.ToString.Trim + " = " + input.ToString.Trim)
            Next
        End If
    End Sub
End Class
Public Class EquationParts
    Public Firstpart As Integer = 0
    Public SecondPart As Integer = 0
End Class

[ddanbe]

@tinstaafl: That is a fine piece of Visual Basic code on the C# forum. I think an "oops" would be appropriate here.

[tinstaafl]

Yep oops. let's try it again:

        Int32 input = 0;
        List<string> output = new List<string>();
        bool GoodInput = int.TryParse(textBox1.Text, out input);
        if (GoodInput && input >1)
        {
            for (int i = 1; i < input; i++)
            {
                for (int j = 1; j < input; j++)
                {
                    if (i + j == input)
                    {
                        output.Add(i.ToString().Trim() + "," + j.ToString().Trim());
                    }
                }
            }
            foreach (string s in output)
            {
                textBox2.AppendText(s.Split(',')[0] + " + " + s.Split(',')[1] + " = " + input.ToString().Trim());
            }
        }

And the code for saving the parts:

namespace WindowsFormsApplication1
{

    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {
            Int32 input = 0;
            List<EquationParts> output = new List<EquationParts>();
            textBox2.Clear();
            bool GoodInput = int.TryParse(textBox1.Text, out input);
            if (GoodInput && input >1)
            {
                for (int i = 1; i < input; i++)
                {
                    for (int j = 1; j < input; j++)
                    {
                        if (i + j == input)
                        {
                            EquationParts eq = new EquationParts();
                            eq.FirstPart = i;
                            eq.SecondPart = j;
                            output.Add(eq);
                        }
                    }
                }
                foreach (EquationParts eq in output)
                {
                        textBox2.AppendText(eq.FirstPart.ToString().Trim() + " + " + eq.SecondPart.ToString().Trim() + " = " + input.ToString().Trim() + "\r\n");
                }
            }

        }
    }
    public class EquationParts
    {
        public int FirstPart{get;set;}
        public int SecondPart{get;set;}
    }
}

[virusisfound]

hi..
thank you so much for your reply.

but my problem is :

i have two textboxs
txt_val1 and txt_val2
now i want to divide the numbers like : val1/val2 and the addition equals to val1
like:
5/3
then it should divide 5 three times. but the output addition must be 5
like:
input : 5/3
output :1+3+1 =5
input :7/5
output : 1+2+2+1+1=7 // it divide the values five times and addition of all number is 7
some think like this.

thank you once again.

[ddanbe]

Please explain what you mean by divide.
Example: 5/3 could be 1 or 1.666666666666666666666666666666666...

[tinstaafl]

This is starting to sound more and more like a homework assignment. Perhaps you should be showing what effort you're putting into this.

[virusisfound]

I got the solution.

here is code(console) :

using System;
using System.Collections.Generic;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int temp; 
             int number1 = 0;
            int sum = 0;
            Console.WriteLine("Please enter a number : ");
            number1 = Convert.ToInt32(Console.ReadLine());
            Console.WriteLine("Please enter a number to divide: ");
            int number2 = Convert.ToInt32(Console.ReadLine());
            if (number1 % number2 == 0)
            {
                temp = number1 / number2;
                int div = temp;
                Console.WriteLine(temp);
                for (int i = 1; i < number2; i++)
                {
                    temp = temp + div;
                    Console.WriteLine(temp);
                }
            }
            else
            {
                for (int i = 0; i < number2; i++)
                {
                    temp = number1 / number2;
                    sum = sum + temp;
                    if (sum >= number1)
                    {
                        break;
                    }
                    number1 = number1 + 1;
                    Console.WriteLine(temp); 
                    temp = 0;
                }

            }
            Console.ReadLine();

    }
        }
    }