While: <span> the loop must repeat until a certain "condition" is met. If the "condition" is FALSE at the beginning of the loop, the loop is never executed. The "condition" may be determined by the user at the keyboard. The "condition" may be a numeric or an alphanumeric entry. <span>This is a good, solid looping process with applications to numerous situations.
</span></span><span>while:<span>int ctr = 1;
while (ctr < = 20)
{
cout<< ctr++ <<"\n";
}
</span><span>
</span><span>
</span><span>HOPE i HELPED! brainliest? :D </span></span>
Answer:
software as a service(SaaS) model
Explanation:
This is because the software applications are accessed over the internet meaning no need for infrastructure installation and maintenance.
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.
