number of times each number in array occurs?
 

Hi everyone! I need to write program that will find the average of 20 element array and also display the number of each number in the array occurs.
I have sorted average number but straggle with "number of occurrences". Please help!!!!

#include<stdio.h>

#include<math.h>

 

main()

{

float avg,sum=0;

int i,j;

int numbers[20]; /*array declaration*/

for(i=0;i<20;i++)

{

printf("\nEnter number: ");

scanf("%d",&numbers[i]); /*store data in array*/

}

for(i=0;i<20;i++)

sum=sum+numbers[i]; /*read data from array*/

avg = sum/20;

printf("\n\nAverage number of this 20 numbers is %5.2f\n\n\n",avg);

Re: number of times each number in array occurs?
 

First of all use int main( void ) and not just main( ) since its not standard.

Don't hard code values in your program, if you want to process 20 values make a preprocessor defination at the start of your program which can easily be modified if you are asked to change the requirements. Using magic numbers as such causes a lot of confusion.

Can you give us a sample run or a dummy example of what kind of output you are expecting ? Are you required to store the occurances or just display them?

Re: number of times each number in array occurs?
 

Thank you for your reply!

This program should be displayed on the screen in two columns, i.e col.1: the number and col.2: the number of occurences.

Re: number of times each number in array occurs?
 

Hmm..if thats the case then all you are required to do is to use nested loops. Here is a short algorithm:

• Create a variable count which will keep track of the occurances of the numbers.
• Start an outer loop with i as the counter or index, initialize it with 0 and continue looping till i is less than the number of elements or till all the elements have been visited.
• Create an inner loop with j as the counter, initialize it in the same manner as the previous loop.
• Inside the inner loop check if [I]my_array equals my_array[j] and if it does, increment the counter by one.
• Close the inner for loop and after the loop completion, print out the value of counter which should be the number of times the element [I]my_arrayhas occured along with the element under consideration.
• Reset the counter value back to zero for the remaining elements.
• Keep looping the outer loop till all the elements have been visited.
• End.

Hope it helped, bye.

Re: number of times each number in array occurs?
 

thanks a lot!

Just one more question. What do you mean by "and if it does, increment the counter by one"?

Re: number of times each number in array occurs?
 

counter = counter + 1

Re: number of times each number in array occurs?
 

What I mean is :

// if the element of array matches with another element in the same 

// array, then increment the counter.



if( my_array[i] == my_array[j] ) 

{

     ++counter ; // counter = counter + 1

}



// my_array = {1, 2, 1, 3, 4, 5 }

// my_array[i] = my_array[0] = 1 

// counter = 2 ( since 1 has a match at 0th and 2nd position )

Hope it helped, bye.

Storing numbers in an array, please help!
 

#include <conio.h>

#include <fstream.h>

#include <iomanip.h>

#include <iostream.h>

  

ofstream fout;                                    // output textfile declaration

using namespace std;



// =============================================================================



int main ()



  {

    int i, count;

    char reply;

    double x[100];



    fout.open ("array.txt");                    // prepares output textfile



    // introduction



    for (i = 1; i <= 9; i++)

      {

        cout << endl;

      }

    cout << setw (52) << "STORING DATA IN AN ARRAY" << endl;

    cout << endl << endl;

    cout << setw (66) << "This program will prompt the user for a set of numbe";

    cout << "rs:" << endl;

    cout << endl << endl;

    cout << setw (56) << " Press ENTER to continue ... ";

    cin.get ();

    

    // prompting the user for a set of numbers

    

    count = 0;



    fout << setw (52) << "Data stored in array";

    fout << endl << endl;



    do // while there is more input data

       {

         count++;

     

    do // while data is incorrect

       {

         clrscr ();

          

         for (i = 1; i <= 11; i++)

              {

                cout << endl;

              }



         cout << setw (56) << "Enter data " << count << "or enter -1 to end";

         cout << " data entry:";

         cin >> x [count];



    if (count % 25 == 0)



        {

         

         fout << "\f";

         fout << setw (45) << "Values Stored in an Array";

         fout << endl << endl;

         

        } 

            do // while response is invalic

              {

                clrscr ();



                for (i = 1; i <= 11; i++)

                  {

                    cout << endl;

                  }



            while (y != -1);

              }

   do // record verified data in output textfile

      {

        fout << setw (32) << "#" << count << ")" << setw (5) << count;

        fout << endl << endl;



      do // while response is invalid

          {

            clrscr ();



            for (i = 1; i <= 9; i++)

              {

                cout << endl;

              }

          }

   while (count != -1)

       }



    cin.get ();

    fout.close ();                                     // closes output textfile

    return 0;



  }

Re: number of times each number in array occurs?
 

Use code tags when you post code -- otherwise it will be an unreadable mess.

This is also the C programming forum, not the C++ one, so C++ code is out of place.

It's also old-style C++ code. You're using <iostream.h> and <fstream.h> etc, which are pre-standard. Use <iostream> and <fstream> instead. Also, you're declaring an ofstream before the using namespace std, which shouldn't work if you use the right header files. Declare it afterwards.