Question

I need some explanation for this code.
This is the example code given by others.

    for ( int i = 1; i <= 8; i++ )
    {
        if(check % 2 == 0)
            sum += i;
        else
            sum -= i;
        check /= 2;            <--- Need explanation for this line.
    }

But in the Pseudo code, there is no check /= 2; procedure.

Here is the full Pseudo code.

int binary = 0;
int sum;

while(binary<256)
sum = 0;
for(go through all 8 digits)
    if the i-th digit is 0
        sum += i
    if the i-th digit is 1
        sum -= i
end for

if sum == 0
    output

binary++
end while

So, what is the purpose for that line of code?

Since sum, binary, and check is initialize as 0.

I have written this code using the Pseudocode given above.
But seems like my code will duplicate the output and one more problem, the format.

I want the output be like this format:

Enter a number : 3

-1 -2 +3 = 0
1 +2 -3 = 0 

But my currently output is:

Enter a number : 3

-1 -2 3 = 0
1 2 -3 = 0

Here is my code:

CODE IS REMOVED!

Solved!

I’m too focus on the for-loop for the output part, hence miss the while-loop for the binary, because the pseudocode is for 256 possible solutions, hence, there will be same output for the front part, example:

1 - 2 - 3 + 4 = 0                             

1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 = 0

Hence, the pseudocode may give an same output. So, since the solution is in 2 ^ n where n = 1, 2, 3, ... form, so change the

 while( binary < 256 )   --->     while ( binary < Math.pow(2, input))

should solve it.

The format and the duplicate of the answer are solved.

Answer 1

This is the way to go through all digits. The most right digit is retrieved by check % 2, and after checking it, you shift check one digit (bit) to the right by check /= 2 (equals to check = check / 2;)