VirtualBox and fillscreen

Full screen not full? Make sure that the minimal amount of Video memory is sufficient.

Netstat hack

Long long time ago I was not quite happy with the default behaviour of netstat for Linux. Those were the days when I was very fond of Gentoo Linux. I am still fond of it, but re-compiling everything from scratch became quite tiring. The program displays only first 20 characters of FQDN. This may be inconvinient, when we try to monitor data flow between our host and the net. I downloaded the latest source and grepped for the number 20. Finally I found the interesting line and changed it to 180.

Podzbiory

We are going to solve another simple task.

Here's the source: http://ki.staszic.waw.pl/task.php?name=podzbiory

And this is a rough translation:

Print all subsets of set A consisting of n elements. Note that each subset can be represented as a sequence of 0 and 1. If B is an array of booleans, B[i] shows whether element i is present in subset A or not.

For example: A = {2, 3, 5} can be represented as:
..................
B.=.{0, 1, 1, 0, 1}
........^..^.....^
........2..3.....5
The program reads one integer n = cardinality of A

For the following input:

3

The program is supposed to print:

000
001
010
011
100
101
110
111

How to solve it? We can easily see that the example explains more than the task itself. We are asked to simply print all numbers from 0 to 1 << n in binary form, where "<<" means Shift Left.

Let's solve the task using masks.

Let's consider the following integer:

011100110

How do we see if n-th bit of an integer is on or off? We simply multiply the integer with appropriate mask. We want to see if second bit (from the right) is on.

  11100110
* 00000010
__________
  00000010

If the answer != 0, the bit is on.

We now want to check whether the first bit is on.

  11100110
* 00000001
__________
  00000000

No, it isn't.

We can simply test it by calculating (i & (1<<j)) where i is the number we want to check and j is the bit we are interested of.

Solving the task with this knowledge is a piece of cake. We simply iterate through all numbers from 0 to 1 << n, and print the output in binary form.

Znowu liczby fibonaciego

We are going to solve a simple task.

The task is here: http://ki.staszic.waw.pl/task.php?name=znowu

And here goes the transcript: Write a program that calculates n-th Fibbonaci number mod 123456789. It is as clear as daylight that we are not going to calculate every single Fibbonaci's number as this function is periodical. How to find the period? We calculate Fibbonaci's numbers like this: Each n-tn number = n-2-th number + n-1-th number.
You cad do it recursively but it will crash for large n. Instead, let's do it by iteration. for (int i = 2; i < n; i++) { a = b; b = c; c = (a + b);