Even sum of digits

[FREEZX]

I need to write a program that will count how many numbers within an interval have an even sum of digits. you are given two numbers, a and b. i wrote a little program that is a little slow.. it checks for the sum of the digits for every number in the interval. Any suggestions about a better algorithm appreciated.

#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
    string c;
    stringstream ss;
    int a,b, rezx, count;
    cin>>a>>b;
    for(int i=a; i<b; i++)
    {
            ss<<i;
            c=ss.str();
            ss.str("");
            for(int x=0; x<c.length(); x++)
            {
                    rezx=rezx+c[x];
            }
            if(rezx%2==0)
            {
                         count++;
            }
            rezx=0;
    }
    cout<<count-1;
    system("pause");
    return 0;
}

[tux4life]

I think the following algorithm is a better one:

If you know the starting number of the interval:
-> Check whether the sum of it's digits are even (you could use the modulo operator for this)
-> If the sum was even, you automatically know the next number's sum of digits is odd

e.g: 450 => odd ; 451 => even ; 452 => odd ; 453 => even ; etc. ...

Edit:: So actually you don't have to check each number's sum of digits

Hope this helps ! :)

[FREEZX]

yeah but what when you get to a number like 159... 1+5+9=15 which is odd. than the next one 160 its odd again.. how do i get over these?

[tux4life]

The method described here is probably a more efficient one as yours as it isn't converting the number to a string each time ...

Edit:: I have to admit my previous post was a bad one ...

[ArkM]

It's not only the most slow algorithm (the original one) but it's a wrong algorithm. You have got not a sum of digits but a sum of char codes of digits. It's the other story that char code of '0' is even numbert in ASCII table; the C language does not specify this fact.

It's so easy to get the last digit of integer n: n%10 then get the last digit of n /= 10 and so on...

[FREEZX]

should i try to make an algorithm that checks if the first number's sum of digits is even and than to count the interval until 9 is reached and than to switch? example:
289-false
290-false
291-true
292-false
293-true
....
299-true
300-false
and than to count the true ones

[tux4life]

should i try to make an algorithm that checks if the first number's sum of digits is even and than to count the interval until 9 is reached and than to switch? example: 289-false 290-false 291-true 292-false 293-true .... 299-true 300-false and than to count the true ones

No, you only have to write a function which calculates the sum of the digits of a number (using the information mentioned in my previous post (look at the link) and ArkM's post) ...

[unbeatable0]

I don't understand - why do you have the numbers input as 'int'? Why don't you have the numbers input directly into a string and then use the operator [] to move through the digits and count values? I think it could be much more efficient - just run a loop like for(index=0;index<number.length();index++) and add every digit to a variable that holds the sum. If you assume ASCII, you won't even need to convert the character digits into digits, because their ASCII values are even for even digits.

[tux4life]

I don't understand - why do you have the numbers input as 'int'? Why don't you have the numbers input directly into a string and then use the operator [] to move through the digits and count values? I think it could be much more efficient - just run a loop like for(index=0;index<number.length();index++) and add every digit to a variable that holds the sum. If you assume ASCII, you won't even need to convert the character digits into digits, because their ASCII values are even for even digits.

I actually don't understand why you need a string for that purpose, you can do it using mathematical operations only.
Why making it difficult if it can be easy ?

Look at ArkM's post ! :)

[unbeatable0]

I actually don't understand why you need a string for that purpose, you can do it using mathematical operations only. Why making it difficult if it can be easy ?

Look at ArkM's post ! :)

When you don't need to do calculations with the number there's no reason for doing the mathematical operations on it - you have the possibility to separate the characters more easily usign strings.

[tux4life]

When you don't need to do calculations with the number there's no reason for doing the mathematical operations on it - you have the possibility to separate the characters more easily usign strings.

Not agreed, using strings is more processor intensive, dealing with numbers is actually a speed boost ...
BTW, You can get the last digit of a number using the modulo operator like this: lastdigit = number % 10;

[unbeatable0]

Running a loop for such a purpose will look like this using mathematical operations:

unsigned short int a,Sum=0;
a=12522;
   while(a)
   {
   Sum+=a%10;
   a/=10;
   }

Using string operations:

string a;
unsigned short int Sum=0,i;
a="12522";
for(i=0;i<a.length();i++) Sum+=a[i];

I personally prefer the second option when you assume ASCII and don't need a speedy algorithm - for example in this program. It looks clearer for me - more self-explanatory.
It doesn't change much, however. It doesn't worth an argument.

[tux4life]

Using string operations:

string a;
unsigned short int Sum=0,i;
a="12522";
for(i=0;i<a.length();i++) Sum+=a[i];

This code isn't returning the sum of the digits in the number !!
And that's the code you prefer ? Code which isn't working ??!!

Dear unbeatable, to make your code work change it to the following:

string a;
unsigned short int Sum=0,i;
a="12522";
for(i=0;i<a.length();i++) Sum+=a[i]-'0';

[unbeatable0]

It's working for determining if the sum is even or not. The Sum mod 2 remains the same.

[tux4life]

It's working for determining if the sum is even or not. The Sum mod 2 remains the same.

Can you explain we howit (your example using strings) was working ??

[unbeatable0]

ASCII '0' is 48, '1' is 49... '9' is 57 - '0' mod 2 remains 0, '1' mod 2 remains 1, ... , '9' mod 2 remains 1.

[tux4life]

Just look at this post and you'll see you've made a mistake in your code ...

[unbeatable0]

Nope. It doesn't change the answer of Sum%2 simply because '0' mod 2 = 0.
If x=y (mod n), and z=0 (mod n), x+z will still equal y (mod n).

[tux4life]

You can try if you don't believe me but this code:

#include <iostream>

using namespace std;

int main()
{
	unsigned short int a,Sum=0;
	a=12522;
	while(a)
	{
		Sum+=a%10;
		a/=10;
	}
	cout << Sum << endl;
}

Has another output than this code:

#include <iostream>

using namespace std;

int main()
{
	string a;
	unsigned short int Sum=0,i;
	a="12522";
	for(i=0;i<a.length();i++) Sum+=a[i];
	cout << Sum << endl;
}

Convinced ?

[unbeatable0]

We're talking about the result of the sum modulo 2. The result is the same.