Answer:
The answer is (C)
Let’s run the algorithm on a small input to see the working process.
Let say we have an array {3, 4, 1, 5, 2, 7, 6}. MAX = 7
- Now for i=0, i < 7/2, here we exchange the value at ith index with value at (MAX-i-1)th index.
- So the array becomes {6, 4, 1, 5, 2, 7, 3}. //value at 0th index =3 and value at (7-0-1)th index is 6.
- Then for i=1, i < 7/2, the value at index 1 and (7-1-1)=5 are swapped.
- So the array becomes {6, 7, 1, 5, 2, 4, 3}.
- Then for i=2, i < 7/2, the value at index 2 and (7-2-1)=4 are swapped.
- So the array becomes {6, 7, 2, 5, 1, 4, 3}.
- Then for i=3, i not < 7/2, so stop here.
- Now the current array is {6, 7, 2, 5, 1, 4, 3} and the previous array was{3, 4, 1, 5, 2, 7, 6}.
Explanation:
So from the above execution, we got that the program reverses the numbers stored in the array.
The answer is Smart technology :)
Answer:
Well when you think about it they both do the same thing but they are diffirent, they have the same perpose but diffirent parts in them so, when you compair them the phone is smaller and slower, the bigger the computer the better it is depending on the amount of money spent on parts.
Answer:
Following are the response to the given question:
Explanation:
The glamorous objective is to examine the items (as being the most valuable and "cheapest" items are chosen) while no item is selectable - in other words, the loading can be reached.
Assume that such a strategy also isn't optimum, this is that there is the set of items not including one of the selfish strategy items (say, i-th item), but instead a heavy, less valuable item j, with j > i and is optimal.
As 
, the i-th item may be substituted by the j-th item, as well as the overall load is still sustainable. Moreover, because 
and this strategy is better, our total profit has dropped. Contradiction.
Answer:
dax and dear god, also my heart hurts
Explanation:
they are great so is he u should listen to him
