Alex1009
New member
- Joined
- Oct 19, 2023
- Messages
- 1
- Programming Experience
- Beginner
Hey everyone! I hope you're doing well. I'm still pretty new around here and not exactly a pro in the world of C#, but I love programming in my spare time. A few weeks back, I started with C# and now I'm working on this project for an app.
I'm tinkering with a program that evenly distributes different things (like 3 tomatoes, 4 bananas, and 6 apples) among a set number of bags (let's say 3 bags). The total count of each type of fruit and the overall count of items shouldn't differ by more than one, so it's as fair as possible for each bag.
Are the rounded numbers correctly stored in the 'results' array and the remainders, i.e., the decimal remainders, in the 'sumUp' array? Now, the only thing left is the distribution, but is this approach correct?"
C# Code link:
Or here:
I'm tinkering with a program that evenly distributes different things (like 3 tomatoes, 4 bananas, and 6 apples) among a set number of bags (let's say 3 bags). The total count of each type of fruit and the overall count of items shouldn't differ by more than one, so it's as fair as possible for each bag.
Are the rounded numbers correctly stored in the 'results' array and the remainders, i.e., the decimal remainders, in the 'sumUp' array? Now, the only thing left is the distribution, but is this approach correct?"
C# Code link:
Or here:
C#:
using System;
public class HelloWorld
{
public static void Main()
{
double[] objects = new double[]{4,4,2};
Pack(objects, 3);
}
public static void Pack(double[] things, int numBags)
{
double[] results = new double[]{0,0,0};
double[] sumUp = new double[]{0,0,0};
for(int i = 0; i < things.Length; i++)
{
double current = things[i] / numBags;
double nextnumber = (double)Math.Floor(current);
results[i] = current;
double part = current - nextnumber;
sumUp[i] = part;
Console.WriteLine(results[i] + "/" + Math.Floor(current) + "/" + part);
}
for(int b = 0; b < sumUp.Length; b++)
{
sumUp[b] = sumUp[b] * numBags;
Console.WriteLine(sumUp[b]);
}
}
}
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