Answer: virtual private network
Explanation:
A virtual private network, is normally refered to as a VPN is simply referred to as an encrypted connection which is done over the Internet. We should note that it usually take solace from a particular device to th network.
The function of the encrypted connection is to assist in the transmission of sensitive data. It works by using the Internet to relay communications and it maintains privacy through security procedures.
Answer:
Outdent.
Explanation:
The negative indent also called outdent is created by pulling the indent markers that are present on the ruler to the left of left margin and there is also one other way of doing it by setting a negative integer in the left box that is present in the paragraph group present on the page layout tab.
Programming Languages.
Explanation:
Programming languages create source code using words such as "IF", "IF NOT" which is similar to spoken language.
Examples of Programming languages are C++, COBOL and Java.
Answer:
Giving that: The following is a sequence of undo-log records written by 2 transactions T and U:
< START T >;
< T,A,10 >;
< START U >;
< U, B, 20 >;
< T, C, 30 >;
< U, D, 40 >;
< Commit U >;
< T, E, 50 >;
< Commit T >;
1. < START U >
Recovery action in this case will be undo(-1) and undo(0). All restored to its original Value
log records < T, A, 10 >, < T, abort >; as written out
2. < T, E, 50 >
Recovery action in this case will be undo(8) and redo(0). A and C is restored to its original value, B and D are set to 20 and 40
log records <T, C, 30 >, < T, A, 10 >, < T, abort > are written out
3. < Commit T >
Recovery action in this case will be redo(7) and redo(4). A and C are set to 10 and 30, B and D are set to 20 and 40
//Note: This program was compiled using turboc compiler. If you are using gcc or any other compiler, then you have to modify the code slightly
#include
#include
int main()
{
int field[110][110]; //FIELD SIZE
int N, i, j;
char dir[50]; //TO REPRESENT THE DIRECTIONS OF 50 COWS
int cow[50][1][2]; //TO STORE THE X,Y COORDINATES OF 50 COWS
int infinity[50]; //TO IDENTIFY, HOW MANY COWS VALUES WILL BE INFINITY AND VICE VERSA
int count[50]; //TO COUNT AMOUNT OF GRASS EATEN BY EVERY COW
int cont = 1; //FLAG VARIABLE, INITIALLY KEEP IT 1.
//CLEAR THE SCREEN
clrscr();
//MARK ALL FIELD POSITIONS AS 1, INDICATING THERE IS GRASS TO EAT
for (i=0; i<110; i++)
for (j=0; j<110; j++)
field[i][j]=1;
//FOR ALL COWS SET INFINITY AS 1 AND COUNT AS 0. I.E WE ASSUME THAT
//EVERY COW WILL EAT INFINITE GRASS AND INITIALLY IT HAS NOT EATEN ANYTHING
//SO COUNT IS SET AS 0
for (i=0 ; i<50 ; i++){
infinity[i] = 1;
count[i] = 0;
}
//READ N, I.E NO. OF COWS
scanf("%d", &N);
//FOR EACH COW, READ THE DIRECTION AND X, Y COORDINATES
for( i=0 ; i fflush(stdin);
scanf("%c%d%d", &dir[i], &cow[i][0][0], &cow[i][0][1]);
}
//REPEAT UNTIL CONT==1
while( cont==1 ){
//FOR EVERY COW CHECK ITS DIRECTION
for( i=0 ; i //IF THE COW IS FACING NORTH DIRECTION
if( dir[i] == 'N'){
//IF THE RESPECTIVE COW'S Y COORDINATE IS LESS THAN 109
if( cow[i][0][1] < 109 )
//IF THERE IS GRASS IN THE PARTICULAR LOCATION,
if( field[cow[i][0][0]][cow[i][0][1]] == 1 ){
//LET THE COW MOVES TO THE NEXT LOCATION, BY INCREMENTING THE VALUE ASSOCIATED WITHN Y-COORDINATE
cow[i][0][1]++;
//INCREMENT THE COUNT VALUE FOR THE RESPECTIVE COW
count[i]++;
}
else
//IF THERE IS NO GRASS IN THAT LOCATION, THEN MAKE THE INFINITY VALUE CORRESPONDING TO THIS COW AS 0
//COW STOPS
infinity[i] = 0;
}
//IF THE COW IS FACING EAST DIRECTION
if( dir[i] == 'E'){
//IF THE RESPECTIVE COW'S X COORDINATE POS IS LESS THAN 109
if( cow[i][0][0] < 109 )
//IF THERE IS GRASS IN THAT PARTICULAR LOCATION
if( field[cow[i][0][0]][cow[i][0][1]] == 1 ){
//LET THE COW MOVE ON TO THE NEXT LOCATION
cow[i][0][0]++;
//INCREMENT THE COUNT FOR THE RESPECTIVE COW
count[i]++;
}
else
//IF THERE IS NO GRASS THEN MAKE INFINTY AS 0 FOR THE PARTICULAR COW.
//THE COW STOPS
infinity[i] = 0;
}
}
/*IN THE PREVIOUS TWO LOOPS WE MADE THE COW TO MOVE TO THE NEXT LOCATION WITHOUT
EATING THE GRASS, BUT ACTUALLY THE COW SHOULD EAT THE GRASS AND MOVE. THE BELOW
LOOP ENSURES THAT. THIS TASK WAS SEPARTED, TO ALLLOW TO TWO COWS TO SHARE THE SAME
POSITION FOR EATING.*/
for( i=0 ; i //IF THE COW IS FACING NORTH
if( dir[i] == 'N'){
//IF CURRENT Y POSITION IS <= 109
if( cow[i][0][1] <= 109 )
//IF IN THE PREVIOUS Y POSITION, THERE IS GRASS
if( field[cow[i][0][0]][cow[i][0][1]-1] == 1 )
//REMOVE (EAT) THE GRASS
field[cow[i][0][0]][cow[i][0][1]-1]=0;
}
//IF THE COW IS FACING EAST
if( dir[i] == 'E'){
//IF CURRENT X POS IS <= 109
if( cow[i][0][0] <= 109 )
//IF IN THE PREVIOUS X POS, THERE IS GRASS
if( field[cow[i][0][0]][cow[i][0][1]] == 1 )
//REMOVE (EAT) THE GRASS
field[cow[i][0][0]-1][cow[i][0][1]]=0;
}
}
/*ASSUME THAT ALL THE COW STOPS EATING AS THERE IS NO GRASS IN THE CURRENT CELL
OR THE COW HAS REACHED THE LAST LOCATION*/
cont = 0;
//REPEAT FOR EVERY COW
for( i=0 ; i //IF THE COW IS FACING NORTH
if( dir[i] == 'N'){
//IF THE COW HAS NOT REACHED LAST Y LOC OR NOT STOPPED EATING
if( cow[i][0][1] < 109 && field[cow[i][0][0]][cow[i][0][1]] !=0 )
cont = 1; //MAKE COUNT AS 1, THE WHILE LOOP SHOULD REPEAT
}
//IF THE COW IS FACING EAST
if( dir[i] == 'E'){
//IF THE COW HAS NOT REACHED LAST X LOC OR NOT STOPPED EATING
if( cow[i][0][0] < 109 && field[cow[i][0][0]][cow[i][0][1]] !=0)
cont = 1; //MAKE COUNT AS 1
}
}
}
//DISPLAY THE OUTPUT
for( i=0 ; i {
//IF THE INFINITY VALUE OF THE RESPECTIVE COW IS 1, THEN DUSPLAY AS INFINITY
if( infinity[i] == 1 )
printf("\nInfinity");
else //ELSE DISPLAY THE RESPECTIVE COUNT VALUE
printf("\n%d", count[i]);
}
getch();
return 0;
}