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1- Beautiful Strings
Download Source Code :
2-Balanced Smileys
John knows the following: for each index i, where k <= i < n, m[i] is the minimum non-negative integer which is *not* contained in the previous *k* values of m.
For example, if k = 3, n = 4 and the known values of m are [2, 3, 0], he can figure out that m[3] = 1.
John is very busy making the world more open and connected, as such, he doesn't have time to figure out the rest of the array. It is your task to help him.
Given the first k values of m, calculate the nth value of this array. (i.e. m[n - 1]).
Because the values of n and k can be very large, we use a pseudo-random number generator to calculate the first k values of m. Given positive integers a, b, c and r, the known values of m can be calculated as follows:
m[0] = a
m[i] = (b * m[i - 1] + c) % r, 0 < i < k
Input
The first line contains an integer T (T <= 20), the number of test cases.
This is followed by T test cases, consisting of 2 lines each.
The first line of each test case contains 2 space separated integers, n, k (1 <= k <= 105, k < n <= 109).
The second line of each test case contains 4 space separated integers a, b, c, r (0 <= a, b, c <= 109, 1 <= r <= 109).
Output
For each test case, output a single line containing the case number and the nth element of m.
Example input:
Example output
Solutions by Vikash verma (@vikashvverma) &Sobre Krolzaum
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[PHP] Download config File Killer and Mannu Shell (updated) by Team Indishell
[PHP] Download config File Killer and Mannu Shell (updated Version) Coded By Team Indishell
Configration File Killer
Latest Fuctions :
- Genrate php.ini
- Mannual Symlinking
- Automated Mass Symlink
- perl Based Symlink
Mannu Shell
New Features :
- command execution
- CGI Telnet
- CMS based symlink,
- VBulletin,wordpress and Joomla admin panel password changer
- PERL Back connect
- Python Back connect
- Can't read /etc/named.conf" bypasser+auto symlink public_html directory
- username (ls /etc/valiases)
Innobuzz Cloud Campus : New Age Learning with a difference [Review]
On the internet Education and learning has been around for many years. But, providing in new ways of studying in this area shall make surf. This year Innobuzz places out to quickly improve the internet studying design through its lately released reasoning university. Although this area is populated by start-ups, Innobuzz reasoning university or the exclusive lab seems to progress of others in near upcoming with its entertaining and interesting studying actions. As of now, it has studying segments on Moral Coughing and PHP.
I tried its 2 day test system for free to see how the lab performs. I discovered it awesome and believed to make a evaluation on it.
I tried its 2 day test system for free to see how the lab performs. I discovered it awesome and believed to make a evaluation on it.
The reasoning university has an awesome UI with several difficulties, projects, useful video clips, assessments, gamification and much more. It creates studying so much fun and interesting and that too at a little cost. It actually assisted me develop real-time abilities that I can apply in my job or individual venture. All you need is a good internet access and you can choose up abilities at any time anywhere.
Innobuzz Reasoning university or exclusive lab has an simple, sharp and simple to use interface with 4 different choices to choose lab atmosphere as per your relaxation. There are Hundreds of difficulties and training guides to make you essentially work on course segments and practice you to be the best. There is whole lot of content available in the electronic collection. I had to be able to look at a stay web seminar by an Market Professional. The cloud university guarantees that the studying atmosphere is fun and aggressive by allowing you to generate badges after finishing difficulties. I could discuss them electronically on community systems with buddies, potential companies, colleagues, etc. and let them know about the new abilities that I obtained. A great way indeed to advertise yourself at the profession front! After finishing difficulties, you can give online exam, get instant report card that can be again shared digitally and apply for jobs in various companies at a click.
In every sense, Innobuzz clearly depicts a live campus that is online. It amazes me at every step and I go back to it everyday for learning something new. If you are getting bored of those run-of-the-mill study patterns and guides, you ought to try this out. It's quite interactive and engaging. The only area of improvement that I shall suggest Innobuzz is to develop more online courses that are sought after by Industry. However, I wouldn't be surprised to see this promising and unique cloud based training application at the top in future.
In every sense, Innobuzz clearly depicts a live campus that is online. It amazes me at every step and I go back to it everyday for learning something new. If you are getting bored of those run-of-the-mill study patterns and guides, you ought to try this out. It's quite interactive and engaging. The only area of improvement that I shall suggest Innobuzz is to develop more online courses that are sought after by Industry. However, I wouldn't be surprised to see this promising and unique cloud based training application at the top in future.
Reverse Engineering Camera Firmware (IP CAM)
In this
tutorial we are going to understand the ip camera firmware software like what
actually running inside ip camera by doing some reverse engineering on the
firmware of ip camera.
But looking
at the hexdump we didn’t figure out any new interesting things only we can see
html and other commands cgi scripts callings.
For reverse
engineering we need to have camera firmware image you can get from you camera
vendor website for flush or upgrade the camera firmware.
Camera
firmware is nothing but the complied or compressed .bin file for flashing the
camera or upgrades the firmware.
Here the
interface of the ip camera so that we can browse the .bin firmware file and
upgrade the camera firmware.
But here are
not going to upgrade the firmware but reverse engineer it for fun and profit.
Now we need
to get the camera firmware from the vendor.
Here the
required the bin camera file or firmware.
First we
need to analyze that what this file is all about (using Linux file utility)
So using
file utility we did not find any interesting thing. Let move further
Now we are
going to use hexdump and string utility against the firmware for more
information.
So first we
need to look at the strings output and analyze it.
Here we can
see that firmware is based on LINUX OPERATING SYSTEM for embedded devices. So
let also see the hexdump may be it can reveal more interesting things for us.
But looking
at the hexdump we didn’t figure out any new interesting things only we can see
html and other commands cgi scripts callings.
Now we are
going to use binwalk against the firmware image it may give some false result.
It give very
useful information like the compressesion method used is gzip and the file
system is squashfs little endian version 3 and more interesting information.
Now we know
that firmware is based on Linux and the file system is squashfs and the compressed
data format is gzip now we are going to use another utility so that we can
extract and decompress the data.
Here the
firmware has been extracted using the firmware mod kit utility.
Now we can
also decompress the data and we find very interesting things inside it
(decompression can be done using decompression utility for squashfs system
included in firmware mod kit)
Here we can
see that the camera firmware is from sercommn (vendor)
Now we can
see all the file system inside the firmware image and everything
Here we can
see each n every cgi and other files and folder inside the camera's webserver.
Here is the
index html file which is running inside the camera, model is 8061.
So finally
you reverse engineered the camera image or firmware hack it according to your
own wish (make required manipulation according to need).
About The Autor : This Post was Written By Zero Cool
Paypal in Process to Remove some sites from Bug Bounty Program
Today I was Checking my Mails and I found out that Paypal have Mailed me
In that they had Included the List of Site they are Removing/Excluding from the Bug Bounty Programme
Email I got was =
So Below is the List of Site are going to get Excluded in coming months form the Bounty Submision
In that they had Included the List of Site they are Removing/Excluding from the Bug Bounty Programme
Email I got was =
Hello-This message is to notify you of a current update to sites that are eligible for Bug Bounty Submissions.Our second party hosted sites (www.paypal-*.com) are mainly marketing based sites that are not part of the core paypal domains (*paypal.com) and are managed by hosting vendor companies. They do not retain as long a lifecycle as the core domains and can have a more volatile timeline as many are tied to projects and regional initiatives. Due to their many times frequent updates, the list will not be maintained on the terms and conditions page. For your own reference, I have provided you a list of sites currently in process of being decommissioned and therefore not eligible for Bug Bounty processing.
So Below is the List of Site are going to get Excluded in coming months form the Bounty Submision
Sites to be decommissioned/Excluded in coming months:
paypal-deutschland.de
|
paypal-danmark.dk
|
paypal-promo.es
|
paypal-europe.com
|
paypal-france.fr
|
paypal-nederland.nl
|
paypal-norge.no
|
paypal-marketing.pl
|
paypal-sverige.se
|
paypal-turkiye.com
|
paypal-business.co.uk
|
paypal-marketing.co.uk
|
paypal-shopping.co.uk
|
paypal-australia.com.au
|
paypal-biz.com
|
paypal-business.com.hk
|
paypal-marketing.com.hk
|
paypal-offers.com.hk
|
paypal-shopasia.com
|
paypal-japan.com
|
paypal-apac.com
|
paypal-plaza.com
|
thepaypalblog.com
|
paypal-marketing.ca
|
The sites in scope of the program are:
Other PayPal Companies
o Zong
o BillMeLater
o Where
o Microplace
o Card.io
o Billsafe
· Partner Sites
=============================================================
Screen Shot of the Mail wich I had Received :-
It Was Bit Annoying to me seeing that they Excluded So Many sites from the Progarmme ,Hope this Article Helps you guys Thanks
Download Hulk PHP DDOSer Script
Learn How to Hack Facebook Accounts via ARP Poisoning
Hello, friends ! Compromising Facebook Account Via ARP Poisoning is e-Book written By Deep, this book will explain “ARP Poisoning Attack” or “Man in the Middle Attack”… In this book we use a packet sniffer called “Wireshark” to capture the packets ie coockie. Here we will see how Wireshark sniffs the packets and finally captured facebook’s authentication coockie and replaced the victims authentication coockie with our own authentication coockie allow us to compromise a facebook account easily. In this book/white paper we will see how we can hack a facebook account over a LAN with ARP Poisoning or MitMA
Retail Makes a Smart Career Choice
In view of the recent global financial meltdown that has impacted almost all the verticals of the industry, most sectors that appeared to be a smart career choice five years ago are losing their shine. However, one industry that has emerged the least affected from the fiscal tornado is the retail sector. Today, an increasing number of graduates are eager to jumpstart their career in this field. If you are also thinking about the same, you may be on the right track, as this sector offers several career benefits that are absent in other fields. For instance, you can hone your skills in various departments in a retail company, be it procurement or accounting, and later on specialize in a particular area as per your penchant.
Scope of the retail industry in India
The rising neo-middle class in India has high disposable income that it does not mind spending on branded merchandise such as food and beverages, watches, apparels, shoes, cosmetics, and skin care products. Products that were considered a luxury over a decade back have now become lifestyle products that most households can afford. This has open up new avenues of growth for the retail industry players who can cash in on the new market by simply concentrating on their branding, marketing, and packaging. With more and more players entering this vertical, the scope of the retail industry is likely to increase further.
Retail industry offers multiple areas of specialization
As already mentioned above, you can choose to specialize in any career stream in this industry. Thus, whether you find procurement interesting or marketing more challenging, you can take it up in a retail company. Some other areas of specializations that you can choose while working in the retail sector are e-commerce, financing, branding, and human resource. You can enter any of these fields based on your academic qualifications and then build a career by learning on the job.
Job benefits in the retail sector
With a flurry of brands in the market, it has become impossible to count them all on your finger. Since this industry thrives on branding and marketing, taking up a job in this sector will be nothing like a regular nine to five desk job that you can find in other industries. Retail jobs can be quite exciting and come with a fair share of challenges that are bound to help you learn better and faster. Furthermore, the massive employee discounts that you get from the brand you work for will be a major bonus for you. After all, everybody needs to shop.
Final word
To sum up, a career in the retail industry is appearing to be one of the most promising choices in the current economic climate and with international retail brands coming to India; the sky is the limit when it comes to career growth in this industry. Apply for one of the Bharti Walmart jobs or Tesco jobs listed in the classifieds section today and you may get your dream break soon.
Facebook Hackercup 2013 Qualification Round Solutions
1- Beautiful Strings
The problem statement is given below:
When John was
a little kid he didn't have much to do. There was no internet, no Facebook, and
no programs to hack on. So he did the only thing he could... he evaluated the
beauty of strings in a quest to discover the most beautiful string in the
world.
Given a string s, little Johnny defined the
beauty of the string as the sum of the beauty of the letters in it.
The beauty of
each letter is an integer between 1 and 26, inclusive, and no two letters have
the same beauty. Johnny doesn't care about whether letters are uppercase or
lowercase, so that doesn't affect the beauty of a letter. (Uppercase 'F' is
exactly as beautiful as lowercase 'f', for example.)
You're a
student writing a report on the youth of this famous hacker. You found the
string that Johnny considered most beautiful. What is the maximum possible
beauty of this string?
Input
The input file consists of a single integer m followed by m lines.
Output
Your output should consist of, for each test case, a line containing the
string "Case #x: y" where x is
the case number (with 1 being the first case in the input file, 2 being the
second, etc.) and y is the maximum beauty for that test case.
Constraints
5 ≤ m ≤ 50
2 ≤ length of s ≤ 500
2 ≤ length of s ≤ 500
Sample Input
5
ABbCcc
Good luck in the Facebook Hacker Cup this year!
Ignore punctuation, please :)
Sometimes test cases are hard to make up.
So I just go consult Professor Dalves
Sample Output
Case #1: 152
Case #2: 754
Case #3: 491
Case #4: 729
Case #5: 646
Here goes my solution for the “Beautiful Strings” problem:
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.LineNumberReader;
import java.util.ArrayList;
import java.util.Collections;
/**
*
* @author VIK
*/
public class Problem1 {
private static ArrayList < Integer > integers;
private static File f=new File("Solution1.txt");
public static void main(String[] args) throws Exception {
FileReader fr = new FileReader("beautiful_stringstxt.txt");
LineNumberReader lnr = new LineNumberReader(fr);
int t = Integer.parseInt(lnr.readLine().trim());
FileWriter fw=new FileWriter(f);
for (int i = 0; i < t; i++) {
int sum= getMax(lnr.readLine().trim());
if ( i < t-1 )
fw.write ( "Case #" + ( i+1 ) + ": " + sum+ "\n" );
else
fw.write( "Case #" + (i+1) + ": " + sum );
fw.flush();
}
}
private static int getMax ( String string ) throws Exception {
integers = new ArrayList < Integer > ();
for ( int i = 65, j = 97; i <= 90 && j <= 123 ; i++, j++ ) {
String s = string.replaceAll ( "[^" + (char) i + (char) j + "]", "" );
s.trim();
integers.add(s.length());
}
int sum = 0;
Collections.sort(integers);
for ( int i = integers.size() - 1, j = 26; i >= 0 && j >= 1; --i, --j) {
sum += integers.get(i) * j;
}
return sum;
}
}
Download Source Code :
2-Balanced Smileys
The problem statement is given
below:
Your friend John
uses a lot of emoticons when you talk to him on Messenger. In addition to being
a person who likes to express himself through emoticons, he hates unbalanced
parenthesis so much that it makes him go :(
Sometimes he puts
emoticons within parentheses, and you find it hard to tell if a parenthesis
really is a parenthesis or part of an emoticon.
A message has
balanced parentheses if it consists of one of the following:
1. An empty
string ""
2. One or more of the following characters: 'a' to
'z', ' ' (a space) or ':' (a colon)
3. An open
parenthesis '(', followed by a message with balanced parentheses, followed by a
close parenthesis ')'.
4. A message with balanced parentheses followed by
another message with balanced parentheses.
5. A smiley face ":)" or a frowny face
":("
Write a program
that determines if there is a way to interpret his message while leaving the
parentheses balanced.
Input
The
first line of the input contains a number T (1 ≤ T ≤ 50), the number of test cases.
The following T lines each contain a message of length s that you got from John.
The following T lines each contain a message of length s that you got from John.
Output
For
each of the test cases numbered in order from 1 to T, output "Case
#i: " followed by a string stating whether or not it is possible that the
message had balanced parentheses. If it is, the string should be
"YES", else it should be "NO" (all quotes for clarity only)
Constraints
1 ≤ length
of s ≤ 100
Sample Input
5
:((
i am sick today
(:()
(:)
hacker cup: started
:):)
)(
Sample Output
Case #1: NO
Case #2: YES
Case #3: YES
Case #4: YES
Case #5: NO
Here goes my solution for the “Beautiful Strings” problem:
//Problem2.java
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.LineNumberReader;
import java.util.ArrayList;
/**
*
* @author VIK
*
*/
public class Problem2 {
private static File f = new File("Solution2.txt");
public static void main(String[] args) throws Exception {
FileReader fr = new FileReader("balanced_smileystxt.txt");
LineNumberReader lnr = new LineNumberReader(fr);
int t = Integer.parseInt(lnr.readLine().trim());
ArrayList < String > list = new ArrayList < String >();
FileWriter fw = new FileWriter(f);
for (int i = 0; i < t; i++) {
list.add(lnr.readLine().trim());
String res = printResult(list.get(i), i + 1);
if ( i < t - 1 ) {
fw.write("Case #" + (i + 1) + ": " + res + "\n");
} else {
fw.write("Case #" + (i + 1) + ": " + res);
}
fw.flush();
}
}
private static String printResult ( String string, int index) {
String s = string.replaceAll ( "[^:\\(\\)]", "" );
for (int i = 0; i < s.length() / 2; i++) {
String lead = "";
String trail = "";
if (s.charAt(i) == '(' && s.charAt(s.length() - i - 1) == ')') {
if (i > 0) {
lead = s.substring(0, i);
trail = s.substring (s.length() - i, s.length());
}
s = lead + s.substring ( i + 1, s.length() - i - 1 ) + trail;
--i;
}
}
s = s.replaceAll( ":\\(", "" );
s = s.replaceAll( ":\\)", "" );
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '(' || s.charAt(i) == ')') {
return "NO";
}
}
return "YES";
}
}
Download Source Code :
3- Find the Min
After sending smileys, John decided to play with arrays. Did you know
that hackers enjoy playing with arrays? John has a zero-based index
array, m, which contains n non-negative integers. However, only the first k values of the array are known to him, and he wants to figure out the rest.John knows the following: for each index i, where k <= i < n, m[i] is the minimum non-negative integer which is *not* contained in the previous *k* values of m.
For example, if k = 3, n = 4 and the known values of m are [2, 3, 0], he can figure out that m[3] = 1.
John is very busy making the world more open and connected, as such, he doesn't have time to figure out the rest of the array. It is your task to help him.
Given the first k values of m, calculate the nth value of this array. (i.e. m[n - 1]).
Because the values of n and k can be very large, we use a pseudo-random number generator to calculate the first k values of m. Given positive integers a, b, c and r, the known values of m can be calculated as follows:
m[0] = a
m[i] = (b * m[i - 1] + c) % r, 0 < i < k
Input
The first line contains an integer T (T <= 20), the number of test cases.
This is followed by T test cases, consisting of 2 lines each.
The first line of each test case contains 2 space separated integers, n, k (1 <= k <= 105, k < n <= 109).
The second line of each test case contains 4 space separated integers a, b, c, r (0 <= a, b, c <= 109, 1 <= r <= 109).
Output
For each test case, output a single line containing the case number and the nth element of m.
Example input:
5
97 39
34 37 656 97
186 75
68 16 539 186
137 49
48 17 461 137
98 59
6 30 524 98
46 18
7 11 9 46
Case #1: 8
Case #2: 38
Case #3: 41
Case #4: 40
Case #5: 12
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