Solution :
int f(int x){
;
}*/
int f(
){
int stack[2] = {
};
;
;
;
R1 = R1 * C; /*R1 =
*/
;
* R2; /* R0 = bx */
; /*R0 =
*/
;
; /*R0 = a+bx+cx^2 */
}
/*
int s=0;
{
;)
*/
AREA SumSquares, code, readWrite
ENTRY
;loop
MOV r1, #0 ; s = 0
Loop
;calculate i*i
;s = s+ i*i
, #1 ; i = i+1
,#10 ; test for end
Loop ;
END
Designers use both isometric and perspective sketches to
maintain an object’s visual proportions. Isometric sketches are 3D objects
drawn with lines that have no vanishing points while perspective sketches are 3D
objects drawn with one or more vanishing points.
Designers choose to use an
isometric sketch because it is more accurate than perspective drawings and
makes the design look more professional. Isometric sketches are much easier to
handle because the length of each measure is same as the plan. This is especially
for designers drawing pictures by hand. It is time saving and you do not need
to calculate the angle of every edge.
Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
Answer:
Refer to the troubleshooting checklist.
Explanation:
I got it right in edge:)
have you tried www.khanacademy.org