Answer:
Check the explanation
Explanation:
An integer (int) is of two different bytes and each page has 200 bytes in length. What this means is that each row of array A (100 int) will fits perfectly in a page.
(a) For the initial or first array-initialization loop, one column is processed at a time, so a page fault will be generated at every inner loop iteration, with a total of 100*100=10,000 page faults.
(b) And when it comes to the second array-initialization loop, one row is processed at a time, and a page fault is generated at every outer loop iteration, with a total of 100 page faults.
Hence second array-initialization loop, has better spatial locality.
Select Query
I hope this helps! :)
Answer:
- common = []
- num1 = 8
- num2 = 24
- for i in range(1, num1 + 1):
- if(num1 % i == 0 and num2 % i == 0):
- common.append(i)
- print(common)
Explanation:
The solution is written in Python 3.
Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).
Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).
Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i will be zero and the if block will run to append the current i value to common list (Line 6-8).
After the loop, print the common list and we shall get [1, 2, 4, 8]
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