loop problem?

[codeexpert123]

A program will calculate and display the sum of all even even numbers from 2 to n, where n is a positive number inputted by the user. What should I do and what is the expression of that???

[Homogenn]

Sounds a lot like homework, so a hint is looking at modulus (%).

[Shuja Ali]

You should write code for it.

Take a pen and paper, write down the steps that need to be performed, then write the same steps in C# code.

[torrud]

Interesting homework. You can solve it without using a loop. It is just a mathematical problem. The solution is quite simple so I believe you can solve it alone.

[Mutant_Fruit]

try this thing in your program
It works.

It doesn't work. Secondly, why are you trying to do this guys homework for him? Anyone with a week of programming experience in any language should be able to knock that out within 30 minutes in c#.

Let him at least put up his own attempt.

Also, don't fix your solution, at least make him do that.

[klamath23]

It doesn't work. Secondly, why are you trying to do this guys homework for him? Anyone with a week of programming experience in any language should be able to knock that out within 30 minutes in c#.

Let him at least put up his own attempt.

Also, don't fix your solution, at least make him do that.

How can you say it does'nt work. I have compiled it and seen the result. I should not do his homework, right i will delete for now. But if you say it does'nt work prove it to me , show me in private message. It very easy to just say it does'nt work.

[klamath23]

A program will calculate and display the sum of all even even numbers from 2 to n, where n is a positive number inputted by the user. What should I do and what is the expression of that???

This is one way you can do it .

int sum = 0;
for ( int i = 2; i <= n; i += 2 ) {
sum += i;
}

[Mutant_Fruit]

Well, the problem with your original code (from what i remember, i only glanced at it when i noticed the problem) was that you coded the loop like this:

for(int i=0; i <= n; n++)
    DoStuff();

So unless i misread it, it couldn't have worked. Maybe you transcribed it wrong, or maybe i misread. Either way, if that's what you did write, it was a mistake.

[klamath23]

Well, the problem with your original code (from what i remember, i only glanced at it when i noticed the problem) was that you coded the loop like this:

for(int i=0; i <= n; n++)
    DoStuff();

So unless i misread it, it couldn't have worked. Maybe you transcribed it wrong, or maybe i misread. Either way, if that's what you did write, it was a mistake.

i didnot write the full code because it was his homework problem , i just wanted to give him hint. I will send the complete code to by private message,
This problem is not new this is asked many times in c++, java etc.
klamath23

[torrud]

Because of post #7 from klamath23 the solution is out now. Therefore I can post my solution too:

public uint Calculate(uint n)
{
    //ensure the number is an even number
    if (n % 2 == 1)
       n -= 1;
    //calculate sum without a loop because it is a arithmetic progression
    return Convert.ToUInt32(n*(n/4 + 0.5));
}

The complexness is reduced to O(1) from O(n/2) when using a loop.

[Mutant_Fruit]

public uint Calculate(uint n)
{
    if (n % 2 == 1)
       n -= 1;

    return Convert.ToUInt32(n*(n/4 + 0.5));
}

Suppose n=6. The final answer should be 6 + 4 + 2 = 12.

In your code, if n=6, the final answer is 9. That's a bit of a problem

[Zaccheus]

6*(6/4+0.5) = 6 * 2 = 12

[klamath23]

Suppose n=6. The final answer should be 6 + 4 + 2 = 12.

In your code, if n=6, the final answer is 9. That's a bit of a problem

Try something like this ((n *n)/4 + (n* 0.5))
if n = 6
then ((6*6)/4 + (6 *0.5)) = 12

[Zaccheus]

Yes, but isn't n*(n/4 + 0.5) exactly the same as (n *n)/4 + (n* 0.5) ?

[klamath23]

6*(6/4+0.5) = 6 * 2 = 12

This works for human, this does'nt work in c# , so if you try something like

((n *n)/4 + ( n * 0.5)) = 12 in c#.