C++ program to calculate sum of two-digit even integers

Question

0.00/5 (No votes)

See more:

I have to calculate the sum of two-digit even numbers and i have done it but i want to find a shorter way possible to do it. If anyone can help.

What I have tried:

#include <iostream>
using namespace std;

int main(){
    int counter = 0, sum = 0;
    

   do{
   	
        if(counter != 0 && counter != 2 && counter != 4 &&counter != 6 && counter != 8){
            cout<<"\n" <<counter;
            sum += counter;
        }
    } while ((counter += 2) <= 50);
    cout<<"\n sum of even numbers:"<< sum;
    
    return 0;
}

Posted 7-Nov-17 12:25pm

Nabeel Munir

Updated 7-Nov-17 15:04pm

Add a Solution

Comments

Peter_in_2780 7-Nov-17 18:35pm

start with
int counter = 10;
then throw out the "if"

Rick York 7-Nov-17 18:38pm

Peter_in_2780 is on the right track. Also consider what the range of two-digit numbers is. Aren't there some beyond 50?

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Kenneth Haugland · Answer 1 · 2017-11-07T12:38:00

There is a formula for that, fo there is no need for a for loop but you can also do that (C# code though):

    int num = 50;
    int sum = 0;
    for (int i = 0; i < num+1; i += 2)
        sum += i;

    int sum2 = GetEvenSum(num);
}

static int GetEvenSum(int UpperNumber)
{
    if (UpperNumber % 2 != 0)
         UpperNumber -= 1;

    int HalfLimit = UpperNumber / 2;

    return HalfLimit * (1 + HalfLimit);
}

Patrice T · Answer 2 · 2017-11-07T15:04:00

Quote:
i have done it but i want to find a shorter way possible to do it.

are you sure 2 digits numbers stops at 50 ?

You start counter at 0 and then do a test to remove 0, 2, 4, 6, 8.
You are lucky you were not requested to counter 4 digits even numbers!
Think about it, what is the reason why your counter starts ar 0 ?

You do ... while loop is a little complicated, why don't you use the syntax especially made for loops with counters ?

Have read on this, it can apply to your problem with a little adaptation.
1 + 2 + 3 + 4 + ⋯ - Wikipedia[^]

C++ program to calculate sum of two-digit even integers

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0