如何在1-100之间创建1000个随机整数的数组

Question

How do you program c# to make an array of 1000 random integers between 1-100.

And then how do you get when a person enters a number eg 68 how can you make the program say 68 appears so and so many times or that it doesn't work at all!

I am not asking for the complete answer I just need a hint where to get started.

Here is what I know:

I have to use the random function and an if but I dont know what to put where!

Answer 1

int[] iArray = new int[1000];
int counter = 0;
Random random = new Random();
for(int i = 0; i < 1000; i++){
   iArray[i] = random.Next(1, 101); //1 - 100, including 100
}
int number = Convert.ToInt32(Console.ReadLine());
foreach(int i in iArray){
  if(i == number)count++;
}
Console.WriteLine("The number "+ number+" appears "+count+" times!");

Answer 2

Start with a for loop , in each iterative you call the random function and put the result in a public list . After that you make an dialog for the user to type a number. You can serach with lambda expression in the list to see how many matches you get.

Answer 3

Make an array of 1000 random integers between 1-100 and when a person enters a number eg 68 how can you make the program say 68 appears so and so many times

I think you're looking for a method like this:

private static Random rnd = new Random();

public static IEnumerable<int> getRandomNumbers(int count, int lowerbound, int upperbound, int specialNumber = int.MinValue, int specialNumberCount = int.MinValue)
{
    List<int> list = new List<int>(count);
    HashSet<int> specialNumPositions = new HashSet<int>();

    if (specialNumberCount > 0)
    {
        // generate random positions for the number that must be create at least n-times
        for (int i = 0; i < specialNumberCount; i++)
        {
            while (!specialNumPositions.Add(rnd.Next(0, count)))
                ;
        }
    }

    while (list.Count < count)
    {
        if (specialNumPositions.Contains(list.Count))
            list.Add(specialNumber);
        else
            list.Add(rnd.Next(lowerbound, upperbound + 1));
    }
    return list;
}

which you can use in this way:

// ensure that 68 is generated at least 10 times
var list = getRandomNumbers(1000, 1, 100, 68, 10);

Demo

If you instead just want to know how often a number appears in the list, you can ue Linq:

int count = list.Count(i => i == 68);

Answer 4

 array = [...Array(1000).keys()].sort((x, y) => { let ren = Math.random() if (ren == 0.5) return 0; return ren > 0.5 ? 1 : -1 })