Get FREE 90 Days Trial of Kaspersky Internet Security 2014 License Keys

Kaspersky is known for its vast amount of World Class Security Products. Kaspersky Internet Security 2014 is one of the product that offers that computer users around the world to protect their PC from Internet Threats. Now You can get this KIS 2014 completely free for 90 days with Genuine Product key/ License Keys.

Kaspersky Internet Security 2014 Features

• Anti-Malware Protection -
• Internet Protection
• Identity Protection
• Anti-Phishing Protection
• Advanced Parental Control

KIS 2014 provides the full protection from Internet Threats and also It has the full Kaspersky Antivirus 2014 Package with itself. KIS 2014 provides complete protection for your computer and protect your data and money.
Download Free Kaspersky Internet Security 2014 License Keys for 3 Months

By Downloading the Genuine Trial version of Kaspersky Internet Security 2014, you can get the complete 3 months free Protection with other benefits like online shopping, banking and social networking Security.

Steps to Download Free License keys of Kaspersky Internet Security 2014

Just follow the below steps to get 90 days free Trial of Kaspersky Internet Security 2014.

1. First, Download the KIS 2014 from the following french website. Use the Google Translate feature translate website. If you are Firefox user, You can use this Auto-Translate Firefox Add-on to Translate this site.

2. Download the Latest version of Kaspersky Internet security 2014 that is KIS Version 2014 (14.0.0.4651ab).

3. Then launch the download file and install the Kis 2014 in your computer.

4. After you installed the KIS 2014, launch it and you will be asked to enter the Kaspersky Serial Keys.



Just Enter the following Serial/ License key – QCGUH-J8FF6-33WGA-UBY62


If the above KIS 2014 serial Key doesn’t work for you, use the following keys

• 4GGYH-S7HEJ-QEGXT-4C88H
• 
  
• 33WG QCGUH-J8FF6-A-UBY62
• 
  
• 4GGYH-S7HEJ-QEGXT-4C88H
• 
  
• QCGUH-J8FF6-33WGA-UBY62

(These KIS 2014 keys are Genuine License Keys provided by Kaspersky free for 90 days)

Now, You have the complete protection from Kaspersky Internet Security 2014 which will provide round the clock protection against all kinds of threats. In the mean time, you can also update the KIS 2014 with latest virus definitions that will provide the security against the latest threats.

If you have any issues in the installation of Kis 2014, leave your comment. We’ll try to get back you.

Coolest Windows & Keyboard Shorcuts

Working professionally with a computer has their hands on the keyboard most of the time. Reaching for the mouse can be an annoying disturbance and personally I often turn over my mouse in such situations. An easy solution is to simply keep the hands on the keyboard and complete as many tasks as possible with keyboard shortcuts only.

Apart from making you work more efficiently and faster, you can also impress your friends or colleagues by being able to work without a mouse. So I have some cool Windows 7 keyboard tricks to get you started. In the end you might never want to take your hands off the keyboard again.

Aero Shortcuts

• [Windows] + [Spacebar] (Aero Peek)

Make all open windows transparent to view gadgets and icons on desktop.

• [Windows] + [D] (Aero Peek)

Show or hide the desktop.

• [Windows] + [Home] (Aero Shake)

Minimize all but selected window. Reverse by clicking the key combination again.

• [Windows] + left arrow OR [Windows] + right arrow (Aero Snap)

Dock selected window to the left or right half of your screen.

• [Windows] + up arrow OR [Windows] + down arrow (Aero Snap)

Maximized and restores the selected window.

• [Windows] + [SHIFT] + up arrow OR [Windows] + [SHIFT] + down arrow (Aero Snap)

Maximizes and restores selected window in vertical dimension only.

• [Windows] + [Tab] (Aero Flip)

Launch 3D representation of open windows and click [Tab] key again to flip through them.

Windows & Taskbar

• [Alt] + [Ctrl] + [Tag] + left/right/up/down arrow

Flip window.

• [Alt] + [Tab]

Cycle through open windows.

• [Windows] + [T] OR [Windows] + [SHIFT] + [T]. Move focus to front or back of taskbar. Press [T] again while holding the [Windows] key to cycle through items in the taskbar from left to right or, with [SHIFT] button held too, from right to left.
• [Windows] + [B]

Puts focus on the ‘show hidden icons’ button on the system tray.

• [Windows] + [1] THROUGH [Windows] + [9]

Launch first through ninth icon on taskbar, including items pinned to taskbar.

• [Windows] + [SHIFT] + [1] THROUGH [Windows] + [SHIFT] + [9]

Starts new instance of respective taskbar icon.

• [Windows] + [Ctrl] + [1] THROUGH [Windows] + [Ctrl] + [9]

Cycles through multiple instances of the respective icon.

• [Windows] + [Alt] + [1] THROUGH [Windows] + [Alt] + [9]

Opens jump list for respective icon.

                    

Multiple Monitors

• [Windows] + [SHIFT] + right arrow OR [Windows] + [SHIFT] + left arrow

Move selected window from one monitor to another. They will remain in the same relative location.

• [Windows] + [P]

Select presentation display mode.

Magnifier

• [Windows] + [+] OR [Windows] + [-]

Activates Windows Magnifier to zoom in or out of screen.

• [Ctrl] + [Alt] + [D]

Switch to docked mode.

• [Ctrl] + [Alt] + [L]

Switch to lense mode.

• [Ctrl] + [Alt] + [F]

Switch from docked or lens mode back to full screen mode.

• [Ctrl] + [Alt] + [I]

Invert colors.

• [Windows] + [Esc]

Exist magnifier views.

Other

• [Windows] + [G]

Cycle through desktop gadgets.

• [Windows] + [X]

Launches Windows Mobility Center. Especially useful if you’re working on a laptop.

C program to calculate sum of Upper Triangular Elements in C

#include< stdio.h>
#include< conio.h>

void main()
{
int i,j,a[10][10],sum,m,n;

/* m - Number of rows 
   n - Number of Columns */

printf("nEnter the number of Rows : ");
scanf ("%d",&m);

printf("nEnter the number of Columns : ");
scanf ("%d",&n);

/* Accept the Elements in m x n Matrix */

for( i = 0 ; i < m ; i++ )
       for( j = 0 ; j < n ; j++ )
       {
       printf("Enter the Element a[%d][%d] : ", i , j);
       scanf("%d",&a[i][j]);
       }

/* Addition of all Diagonal Elements */

sum = 0;

for( i = 0 ; i < m ; i++ )
       for( j = 0 ; j < n ; j++ )
       {
        if ( i < j )             // Condition for Upper Triangle
        sum = sum + a[i][j];
       }

/* Print out the Result */

printf("nThe Addition of Upper Triangle Elements : %d",sum);

getch();

}

---

Output

Enter the number of Rows : 3

Enter the number of Columns : 3

Enter the Element a[0][0] : 1
Enter the Element a[0][1] : 2
Enter the Element a[0][2] : 3
Enter the Element a[1][0] : 2
Enter the Element a[1][1] : 1
Enter the Element a[1][2] : 1
Enter the Element a[2][0] : 1
Enter the Element a[2][1] : 2
Enter the Element a[2][2] : 1

The Addition of Upper Triangle Elements : 6

Explanation :

Considering above 3×3 matrix -

1. By Observing , it is clear that when i < j Condition is true then and then only we have to add the elements

C Program to find addition of Lower Triangular Elements in C Programming

#include<stdio.h>
#include<conio.h>

void main()
{
int i,j,a[10][10],sum,m,n;

/* m - Number of rows 
   n - Number of Columns */

printf("nEnter the number of Rows : ");
scanf ("%d",&m);

printf("nEnter the number of Columns : ");
scanf ("%d",&n);

/* Accept the Elements in m x n Matrix */

for( i = 0 ; i < m ; i++ )
       for( j = 0 ; j < n ; j++ )
       {
       printf("Enter the Element a[%d][%d] : ", i , j);
       scanf("%d",&a[i][j]);
       }

/* Addition of all Diagonal Elements */

sum = 0;

for( i = 0 ; i < m ; i++ )
       for( j = 0 ; j < n ; j++ )
       {
        if ( i > j )             // Condition for Lower Triangle
        sum = sum + a[i][j];
       }

/* Print out the Result */

printf("nThe Addition of Lower Triangle Elements : %d",sum);

getch();

}

---

Output :

Enter the number of Rows : 3

Enter the number of Columns : 3

Enter the Element a[0][0] : 1
Enter the Element a[0][1] : 2
Enter the Element a[0][2] : 3
Enter the Element a[1][0] : 2
Enter the Element a[1][1] : 1
Enter the Element a[1][2] : 1
Enter the Element a[2][0] : 1
Enter the Element a[2][1] : 2
Enter the Element a[2][2] : 1

The Addition of Lower Triangle Elements : 5

Explanation : 

Considering above 3×3 matrix  -

1. By Observing , it is clear that when i > j Condition is true then and then only we have to add the elements

Program to find Transpose of Given Square Matrix

#include<stdio.h>
#include<conio.h>
void main()
{
 int a[10][10],m,i,j,temp;
 /* actual size of matrix is m*n
 i,j  - for scanning of array
 temp - for interchanging of a[i][j] and a[j][i] */

 printf("n Enter the size of matrix :");
 scanf("%d",&m);

 /* Reading elements of matrix */

 printf("n Enter the values a:");

 for(i=0;i<m;i++)
  for(j=0;j<m;j++)
     scanf("%d",&a[i][j]);

 //to print original square matrix

 printf("nGiven square matrix is");

 for(i=0;i<m;i++)
  {
  printf("n");
     for(j=0 ; j 〈 m ; j++)
          printf("%dt",a[i][j]);
 }
 /* Find transpose */

 for(i=1;i<m;i++)
  for(j=0;j<i;j++)
   {
   temp=a[i][j];
   a[i][j]=a[j][i];
   a[j][i]=temp;
   }
 /* printing of all elements of final matrix */

 printf("nTranspose matrix is :");

 for(i=0;i 〈 m;i++)
 {
  printf("n");
  for(j=0;j<m;j++)
      printf("%dt",a[i][j]);
 }
 getch();
}

C Program to Check whether Matrix is Magic Square or Not ?

C Program to Check whether entered matrix is magic square or not ?

What is Magic Square :

1. A magic square is a simple mathematical game developed during the 1500.
2. Square is divided into equal number of rows and columns.
3. Start filling each square with the number from 1 to num ( where num = No of Rows X No of Columns )
4. You can only use a number once.
5. Fill each square so that the sum of each row is the same as the sum of each column.
6. In the example shown here, the sum of each row is 15, and the sum of each column is also 15.
7. In this Example : The numbers from 1 through 9 is used only once. This is called a magic square.

#include<stdio.h>
#include<conio.h>

void main()
{
int size,a[4][4];
int i,j=0;
int sum,sum1,sum2;
int flag=0;
clrscr();

printf("Enter matrix : ");
for(i=0;i<4;i++)
    {
    for(j=0;j<4;j++)
    scanf("%d",&a[i][j]);
    }

printf("Entered matrix is : nn");
for(i=0;i<4;i++)
{
    printf("n");
        for(j=0;j<4;j++)
        {
        printf("t%d",a[i][j]);
        }
}

//------for diagondal elements---------
sum=0;
for(i=0;i<4;i++)
    for(j=0;j<4;j++)
    {
    if(i==j)
    sum=sum+a[i][j];
    }
//-------------for rows--------------

for(i=0;i<4;i++)
{
    sum1=0;
    {
    for(j=0;j<4;j++)
    sum1=sum1+a[i][j];
    }
    if(sum==sum1)
        flag=1;
    else
        {
        flag=0;
        break;
        }
}
//-------------for colomns----------------
for(i=0;i<4;i++)
{
    sum2=0;
    for(j=0;j<4;j++)
    {
    sum2=sum2+a[j][i];
    }
    if(sum==sum2)
        flag=1;
    else
        {
        flag=0;
        break;
        }
}
//----------------------------------------
if(flag==1)
    printf("nntMagic square");
else
    printf("nntNo magic square");
//-----------------------------------------
getch();
}

Note :

Sum of Row1 = Sum of Row2 [Sum of All Rows must be Same]
Sum of Col1 = Sum of Col2 [Sum of All Cols must be Same]
Sum of Left Diagonal = Sum of Right Diagonal