Question

Farmer John has recently acquired NN new cows (3≤N≤5×105)(3≤N≤5×105), each of whose breed is either Guernsey or Holstein.The cows are currently standing in a line, and Farmer John wants take a photo of every sequence of three or more consecutive cows. However, he doesn't want to take a photo in which there is exactly one cow whose breed is Guernsey or exactly one cow whose breed is Holstein --- he reckons this singular cow would feel isolated and self-conscious. After taking a photo of every sequence of three or more cows, he throws out all of these so-called "lonely" photos, in which there is exactly one Guernsey or exactly one Holstein.

Answer 1

//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;
}