Answer:
RAX = 333000h (16 bits with preceding zeros removed)
RDX = 20h (also 16 bits with preceding zeros removed)
Explanation:
The "div" opcode in the assembly language source code is used to divide operands. It accepts a divisor ( the denominator) and divides the content of the AX register. The result is saved in the AX register while the remainder (if any) is saved in the DX register. If the DX register holds any data, the data is replaced with the divisor remnant.
The code above divides the content of the RAX register with the divisor variable and saves the result and remainder in the RAX and RDX respectively.
Answer:
O(n^2)
Explanation:
The number of elements in the array X is proportional to the algorithm E runs time:
For one element (i=1) -> O(1)
For two elements (i=2) -> O(2)
.
.
.
For n elements (i=n) -> O(n)
If the array has n elements the algorithm D will call the algorithm E n times, so we have a maximum time of n times n, therefore the worst-case running time of D is O(n^2)
Answer:
Given that:
A= 40n^2
B = 2n^3
By given scenario:
40n^2=2n^3
dividing both sides by 2
20n^2=n^3
dividing both sides by n^2 we get
20 = n
Now putting n=20 in algorithms A and B:
A=40n^2
= 40 (20)^2
= 40 * (400)
A= 16000
B= 2n^3
= 2 (20)^3
= 2(8000)
B= 16000
Now as A and B got same on n = 20, then as given:
n0 <20 for n =20
Let us take n0 = 19, it will prove A is better than B.
We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.
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