# I have a question

+5 votes
asked Nov 18, 2020
I have to make a program which: Enter an integer N. Enter N integers. Find N / 3 or more duplicate numbers between the entered numbers.

And I'm lost. I did something like this:

#include <stdio.h>

int main(int argc, char** argv)

{

int number;

printf("Enter an integer: ");

scanf("%d", &number);

}

struct eleCount {

int element;

int count;

};

struct eleCount temp[];

Could you please help?

## 1 Answer

+1 vote
answered Nov 20, 2020 by (22,620 points)

So far so good. You asked the user to enter the number N and you already read it.

The next step is to repeat reading N numbers. For that you should use a loop (for, while or do-while). In this loop, you ask for the numbers one by one (just like you did for N -- number -- above).

When done, you should look for duplicates. I leave this for you for think about how to do it. (There could be many ways).

I've created the skeleton of the program. Please, give it a try and try to fill the missing parts.

```#include <stdio.h>

int main()
{
// Instructions to the user + asking for N
printf("The program will read N number of integers then find duplicates in them.\n");
printf("Please, enter N (how many numbers you'll want to check for duplicates): ");
int count;
scanf("%d", &count);

int numbers[count]; // Now, that we know how many numbers we need, we can initialise the numbers array for the right count of elements

// Reading the N numbers
for (int i = 0; i < count; i++)
{
// TODO: Implement reading the numbers
//       1. Ask the user to enter the "i"th number.
//       2. Read the number and store it in the "numbers" array.
// Note: you can access the "i"th element of the array "numbers" as: numbers[i]
}

// Searching for duplicates
// TODO: Implement searching for duplicates
//       1. Find a duplicate.
//       2. Print it to the user.
//       3. Go back to 1. or finish, when there are no numbers left.

return 0;
}```

If you get stuck, please come back and I (or others) will be more than happy to help.

Good luck! :)