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romanna [79]
3 years ago
11

You were just hired as an IT Specialist for Smalltown School District. Your first assignment is to review a problem area& in

student records processing. It seems that the program currently used to compute student grade point averages and class rankings runs terribly slow and as a result end of year reports are habitually late. Your are asked to come up with a list of items that should be checked in order to determine if a modification to code is in order. The IT department head is asking you to do this without the opportunity to actually see the current application/programs in use. What questions would you ask about the current code? What areas of code would you look at? What would you need to know about the student data?
Computers and Technology
1 answer:
umka21 [38]3 years ago
3 0

Answer:

Explanation:

There are various questions that you can ask in this scenario, such as

What grading policies are being implemented?

How many student grades are being calculated by the program?

What is the requirements for a student to pass?

All of these questions would allow you to get an idea of how extensive the code may be and its complexity. Once you know this you would look at the code revolving around actually looping through the data and doing the necessary calculations. You can then determine how to manipulate the code and make it much more efficient.

You would also need to know how the student data is being saved, which will help determine if it is the best data structure for saving this type of data or if it can be replaced in order to maintain the data secure while increasing the speed of the program. Mainly since this information needs to be continuously used from the data structure.

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hammer [34]

Answer:

In C++:

#include<iostream>

#include<vector>

using namespace std;

int main() {

int len;

cout<<"Length: ";  cin>>len;

string inpt;

vector<string> vect;

for(int i =0;i<len;i++){

   cin>>inpt;

   vect.push_back(inpt); }

char ch;

cout<<"Input char: ";  cin>>ch;  

for(int i =0;i<len;i++){

   size_t found = vect.at(i).find(ch);  

       if (found != string::npos){

           cout<<vect.at(i)<<" ";

           i++;

       }

}  

return 0;

}

Explanation:

This declares the length of vector as integer

int len;

This prompts the user for length

cout<<"Length: ";  cin>>len;

This declares input as string

string inpt;

This declares string vector

vector<string> vect;

The following iteration gets input into the vector

<em>for(int i =0;i<len;i++){ </em>

<em>    cin>>inpt; </em>

<em>    vect.push_back(inpt); } </em>

This declares ch as character

char ch;

This prompts the user for character

cout<<"Input char: ";  cin>>ch;  

The following iterates through the vector

for(int i =0;i<len;i++){

This checks if vector element contains the character

   size_t found = vect.at(i).find(ch);  

If found:

       if (found != string::npos){

Print out the vector element

           cout<<vect.at(i)<<" ";

And move to the next vector element

           i++;

       }

}  

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3 years ago
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Answer:

html

Explanation:

6 0
3 years ago
Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.

public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
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}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
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So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.

A quick explanation:

We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.

The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.


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Answer:

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Explanation:

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Answer:

I guessed D, taking it right now, sorry if it's wrong

Explanation:

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