integer userInput
integer i
integer mid
integer array(20) number
userInput = 1
for i = 0; userInput >= 0; i = i + 1
if number[i] > -1
userInput = Get next input
number[i] = userInput
i = i - 1
mid = i / 2
if i > 9
Put "Too many inputs" to output
elseif i % 2 == 0
Put number[mid - 1] to output
else
Put number[mid] to output
Answer:
The order of the efficiencies is as following:-
10,000 < 2n < nlog(n) < n5 < n!.
Explanation:
10,000 is constant time whatever will be the size of the problem the efficiency will remain the same.
2n this efficiency is linear it will grow proportionally as the size of the problem increases.
nlog(n) this efficiency is is a bit greater than 2n though it will grow faster than 2n but slower than n2 as the size of the problem increases.
n5 this efficiency is very poor.It is growing very rapidly as the size of the problem increases.
n! is the worst efficiency of them all.
n!=n*(n-1)*(n-2)*(n-3)*(n-4)*.......2*1.
It will grow beanstalk in jack and the beanstalk.
<span> The Bureaus and the Office of Engineering and Technology process applications for licenses and other filings, analyze complaints, conduct investigations, develop and implement regulatory programs and participate in hearings, among other things. </span>
Answer: the answer is A.
Explanation: hope this helps!
