loop problem?

Quote:

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Originally Posted by Zaccheus

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

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if you know algebra then you should know this
c * (a + b) = c * a + c * b;
2 * (3 + 5) = 2 * 3 + 2 *5;
16 = 16 ;

Indeed.

Quote:

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Originally Posted by klamath23

This works for human, this does'nt work in c#.

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You are correct, because it is integer division!

It would wouk if he wrote n / 4.0.
:D

Quote:

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Originally Posted by Zaccheus

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

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Ah, but you're performing integer division, not floating point division.

6/4 == 1

EDIT: It won't actually be completly accurate if you just change it to 4.0. You may end up with an answer like 11.99999999 instead of 12.

To round a floating point number to the *nearest* integer what you need to do is add 0.5 then just cast the float to int.

i.e.

double result = n * (n / 4.0 + 0.5);
int finalAnswer = (int)(result + 0.5);

Alternatively one of the other forms of the equation would be a better way of doing it.

Well, as it happens 1.5, can be accurately reprepresented in binary - but I do take your point. Rounding errors are always an issue in floating point calculations.
:thumb:

Quote:

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Originally Posted by Zaccheus

Well, as it happens 1.5, can be accurately reprepresented in binary - but I do take your point. Rounding errors are always an issue in floating point calculations.
:thumb:

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There are better ways to do this problem
like this

Code:

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int sumEven = 0;
for(int i =2; i<=n;n++)
{
if((i %  2)== 0)
sumEven+=i;
}

Console.WriteLine(" the sum of even numbers till n is:"+sumEven);

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Quote:

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Originally Posted by Zaccheus

Indeed.



You are correct, because it is integer division!

It would wouk if he wrote n / 4.0.
:D

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You are right. It was just a typo mistake of me because I did not test the code. It means n / 4.0 on that part.

But I am understand the point of view for a rounding error. There is a simple solution. You have only to eliminate the fraction inside.

Code:

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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 n*(n + 2) >> 2;
}

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That's it. There should no rounding error any longer. :wave:

thanks for the help but I myself solved it:D