need help in counting the elements

[Amin96]

how to make it count the even digits. and am I doing anything wrong ?

Scanner input = new Scanner (System.in);

int n; int even = 0;

System.out.print(" Enter a number with four digits: ");
n = input.nextInt();

if (n < 1000 ) {
System.out.print("number is less than 4 digits");}
else if (n > 9999){
System.out.print(" number is more than 4 digits");
}
else {
for(int i = 0; i < n ; i++) {
if((n%10)%2==0)
even++;

System.out.print("There’s" + even + "even number (s).");
}
}

}
}

[lastchance]

What language is that?

[thmm]

Looks like Java to me.

[Amin96]

yes it's java

[lastchance]

am I doing anything wrong ?

Might be more accurately written as "am I doing anything right?"

I'm pretty sure that if he/she puts 1000 in the result will be 1000.

But I've forgotten how to compile java.

[Amin96]

the problem is that I need to get how many even digits there is in a 4 digit integer

[jonnin]

and am I doing anything wrong ?
like posting java code in c++ forums? :P You may get some help anyway, but you would do better in ... a java forum...


also that for loop is checking N%10 every time.
I think you want to change that 10.
something like
for(stuff)
{
even+= 1 - n%2; //even, n%2 is 0, adds 1. odd, n%2 is 1, adds 0.
n = n%2; //if you need n later, this should be a copy of n...
}

or alternately try /1, 10,100 (start at 1, multiply by 10 each time).

I am not a chemist. Yet I constantly have to deal with elements and solutions. Does that seem right to you?

[lastchance]

#include <iostream>
using namespace std;

int countEven( int n ) { return ( n >= 10 ? countEven( n / 10 ) : 0 ) + 1 - n % 2; }

int main()
{
    int n = 0;
    while ( n < 1000 || n > 9999 ) { cout << "Input n: "; cin >> n; }
    cout << "Number of even digits = " << countEven( n );
}

[Amin96]

another topic can you give me head start to this one

Write a program that count until we got 5 numbers multiple of 3 (the program should ask the user to enter integer until we got 5 integers that are multiple of 3):

also in java

[lastchance]

also in java

You do know that this site is www.cplusplus.com, don't you?

[Amin96]

not really i thought it's just a name I"m sorry I'll disappear

[lastchance]

Oh, come back, Amin96!

C++ is a miles better language than Java! Don't go over to the dark side!

[Amin96]

i'll be back but it's for my university

[Ganado]

if((n%10)%2==0) is checking if something is divisible by 20.

[thmm]

Maybe like this:

import java.util.Scanner;

public class Main
{
	public static void main(String[] args) 
	{
    	Scanner input = new Scanner (System.in);
    
        int num, num_even = 0;
        System.out.print(" Enter a number with four digits: ");
        num = input.nextInt();
    
        if (num < 1000 ) 
        {
            System.out.print("number is less than 4 digits");
            return;
        }
        else if (num > 9999)
        {
            System.out.print(" number is more than 4 digits");
            return;
        }
        while (num > 0)
        {
            int last_digit = num % 10;
            if (last_digit % 2 == 0)
                num_even++;
            num = num / 10;
        }
        System.out.printf("There’s %d  even number(s).", num_even);
    }
}

[againtry]

Write a program that count until we got 5 numbers multiple of 3 (the program should ask the user to enter integer until we got 5 integers that are multiple of 3)

Forget about Java coding for a minute and concentrate on the process, using pseudocode

1. set a counter to zero
loop:
2. Enter a number
3. If the number is divisible by 3? (i.e. n%3 is zero), then add 1 to the counter
5. If the counter is < 5 loop back for another number, otherwise ...
...
6. ... THE END