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SpyIntel [72]
3 years ago
10

What kind of resources can we share over the network? Plzzzzzzzz help!!!!!!

Computers and Technology
2 answers:
Misha Larkins [42]3 years ago
6 0
Security resources can be used
lubasha [3.4K]3 years ago
3 0

Sharing network resources requires abiding by certain constraints, as follows:

<span>Security: Organizations present ongoing opportunities for unauthorized shared resources. Security mechanisms should be implemented to provide efficient parameters.Compatibility: Various client-server operating systems may be installed, but the client must have a compatible OS or application to access shared resources. Otherwise, the client may encounter issues that create communication delays and requires troubleshooting.Mapping: Any shared OS hardware drive, file or resource may be accessed via mapping, which requires a shared destination address and naming conventions.<span>File Transfer Protocol (FTP) and File Sharing: FTP is not affected by shared resources because the Internet is FTP’s backbone. File sharing is an LAN concept.</span></span>
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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]);
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boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
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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.


4 0
3 years ago
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hjlf

Answer:

Java code is given below

Explanation:

import java.util.Random;

class Die{

private int sides;

public Die(){

sides = 6;

}

public Die(int s){

sides = s;

}

public int Roll(){

Random r = new Random();

return r.nextInt(6)+1;

}

}

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arr[r-1]++;

}

for(int i=0; i<6; i++)

System.out.println((i+1)+" was rolled "+arr[i]+" times.");

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}

8 0
3 years ago
When you open a program, the hard drive___
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Answer:

When you open a program, the hard drive <u>Registers the program and runs the program accordingly. </u>

I hope this helped!

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Answer:

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Explanation:

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Explanation:

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