Counting Occurrence of numbers

[nickd1085]

Hey guys,

Just got this new assignment and really don't know where to start. Any help would be great. I'm not a Computer Science major. I have a Bachelors in Graphic Design and am taking this as a pre-requisite for an HTML course over the summer I'm planning to take. Heres the assignemnt:

Write a Program that reads the integers between 1 and 100 and counts the occurrence of each number. Assume the input ends with 0.

**Note** if a number occurs more than one time, the plural word "times" is used in the output.


Thanks guys!

[nickd1085]

PS -- it has to written in C++

[ciphermagi]

What is the source of the input, and how is it formatted?

[nickd1085]

Hey man,

all the codes start off like this...

#include <iostream>
using namespace std;

[nickd1085]

In the book, here is the sample run of the program:

Enter the integers between 1 and 100: 2 5 6 5 4 3 23 43 2 0

2 occurs 2 times
3 occurs 1 time
4 occurs 1 time
5 occurs 2 times
6 occurs 1 time
23 occurs 1 time
43 occurs 1 time

[nickd1085]

Ok, heres some code I came up with. I know it needs some hacking though cause its not working:

#include <iostream>
#include <ctime>
#include "InsertionSort.h"
using namespace std;

int main()
{

	double list[100];
	double integers[100];
	int numberOfIntegers = 1;

	cout<<"Enter some integers between 1 and 100: "; //Prompts the user to enter integers between 1 and 100

	for (int i = 0; i < numberOfIntegers; i++)
	{
		cin>>list[i]; //gets integers from user
		numberOfIntegers++;

	for (int j = 0; j<100; j++)
	{
		if (list[i] = j)
		integers[j]++;
		else
			integers[j] = 0;
	
	for (int j = 0; j<100; j++)
	{
		// the following code displays the results
		if (integers[j] != 1 && integers[j] != 0){
			cout<< integers[i] << " occurs " << integers[j] << " times"<<endl;
			break;}
		else if (integers[j] == 1 && integers[j] != 0){
			cout<< integers[i] << " occurs 1 time"<<endl;
			break;}
		}
	}
	}
	int pause;
	cin>>pause;
	return 0;
}

[ciphermagi]

It's a homework assignment, so I'm not going to show you code, but that said, I can point you in the right direction.

Try using a loop to parse all of the integers entered into a list, use an iterator to page through the list, and count the integers that are in there, then display any values that occur.

[ciphermagi]

You're starting off on the right track. Here's some comments that might help you out.

#include <iostream>
#include <ctime>
#include "InsertionSort.h"
using namespace std;

int main()
{

	double list[100]; // Don't use doubles to store int
	double integers[100]; // Don't use doubles to store int
	int numberOfIntegers = 1;

	cout<<"Enter some integers between 1 and 100: "; //Prompts the user to enter integers between 1 and 100

	for (int i = 0; i < numberOfIntegers; i++) // This loop will never, ever end.
	{
		cin>>list[i]; //gets integers from user
		numberOfIntegers++;

	for (int j = 0; j<100; j++)
	{
		if (list[i] = j)
		integers[j]++;
		else
			integers[j] = 0; // You just reset your count. The program will only return the very last number counted.
	
	for (int j = 0; j<100; j++)
	{
		// the following code displays the results
		if (integers[j] != 1 && integers[j] != 0){
			cout<< integers[i] << " occurs " << integers[j] << " times"<<endl;
			break;}
		else if (integers[j] == 1 && integers[j] != 0){ // By definition, if something == 1, then the same thing != 0. This is a redundancy.
			cout<< integers[i] << " occurs 1 time"<<endl;
			break;}
		}
	}
	}
	int pause;
	cin>>pause;// Use cin.get();
	return 0;
}

[nickd1085]

#include <iostream>
#include <cstdlib>
#include <string>

using namespace std;

int main (void)
{
    string num = "";
    char search = '0';
    int i = 0;
    int count = 0;
    
    cout << endl << "Enter Number: ";
    cin >> num;

    cout << endl << "Enter Search Value: ";
    cin >> search;
    
    for (i = 0; i < num.length(); i++)
    {
        //cout << num[i] << endl;
        if ( num[i] == search )
        {
             count++;
        }
    }
    
    cout << endl << "Number of occurences: " << count << endl;
    
    system("PAUSE");
    return 0;
}

I know Im closer but should this work?

[ciphermagi]

You are a bit closer, but you'll never get a for() loop to work right. Look up while() loops.

[zb0S216C]

but you'll never get a for() loop to work right

Care to explain why?

[ciphermagi]

He's trying to parse a single line, which means the end of the for loop has to be defined at runtime, and unless he's willing to take argc and argv[] under consideration, there's no way for the compiler (that I know of) to predefine a variable for the max iterations of a for loop.

[Computergeek01]

@ ciphermagi: Do you mean at compile time? You are contridicting yourself right now. That's irrelevant though since a for loop will indeed allow you to use a variable that isn't defined until the program is executed (run time) for the end condition. I do this constantly to iterate through vectors like this for(int i = 0; i < MyVector.size(); i++).

EDIT: Also, as long as you keep in mind that you are comparing chars then his for loop could work.

@ OP: Do you know how to use structures yet? I just did something like this as a personal project and they help keep things straight.

[ciphermagi]

@Computergeek01: Sure, he can use a vector to do it, but without a vector, there's not a way of terminating the for.

[nickd1085]

I know this is wrong but could someone show me where I went wrong:

#include <iostream>
#include <cstdlib>
#include <string>

using namespace std;

#define max_amount_of_numbers 100

#define biggest_allowed_number 100
#define lowest_allowed_number 1

#define end_of_input "0"

int main ()
{
    string num[max_amount_of_numbers];
    string search;

    cout << endl << "Enter Numbers: " << endl;

    int i = 0;
    while(cin>>num[i]){
      if(num[i]==end_of_input) break;

      if((atoi(num[i].c_str())>=lowest_allowed_number)&&(atoi(num[i].c_str())<=biggest_allowed_number))
        i++;
      else
        cout << "Imput out of range ("
             << lowest_allowed_number << " - " << biggest_allowed_number
             << "). Please try again"
             << endl;
    }

    cout << "Enter the search value: ";
    cin  >> search;
    cout << endl;

    int count = 0;
    while(i>-1){
      if( num[i].compare(search)==0 )count++;
      i--;
    }

    if(count==0)cout << "Given value did not occur.";
      else if(count==1)cout << "Given value occurred 1 time.";
        else cout << "Given value occurred: " << count << " times.";


    cin.get();
    return 0;
}

[ciphermagi]

That looks identical to the code madredcake posted in http://www.cplusplus.com/forum/beginner/40000/#msg215958
...you should study up how to do it yourself. You might learn something.

[Computergeek01]

@ nickd1085: Do you know how to use structures or not? This will help me show you how to do this in a way that won't drive you insane.

[nickd1085]

No, actually. Should I? My teacher is not the greatest for this class.

Thanks Computergeek01!