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san4es73 [151]
3 years ago
6

// Exercise 4.16: Mystery.java

Computers and Technology
1 answer:
nignag [31]3 years ago
8 0

Answer:

385

Explanation:

There was no question so I simply run the program for you and included the output.

The program seems to calculate: \sum\limits_{x=1}^{10} x^2

You might be interested in
Is iPhone better than android
andriy [413]

Answer:

It's up to personal preference.

Explanation:

iPhone/Apple gives more compliments to security than Android.

Android has more friendly user GUI and dev tools.

4 0
3 years ago
Read 2 more answers
Which of the following recently developed technologies has NOT greatly increased the speed of information dissemination?
Luba_88 [7]
<span>The option which hasn't greatly increased the speed of information dissemination is C. typewriter. Information can be easily and quickly shared through smartphones that have Internet access, through emails which you can send at any time, and through teleconferencing, which is basically using your camera to talk to people. The only technology which isn't really helpful nowadays is the typewriter - there are better things, such as computers, that do a better job today.</span>
3 0
3 years ago
Read 2 more answers
Write a Java method to delete a record from a B-tree of order n.
max2010maxim [7]

Answer:

Deleting a record on a B-tree consists of three main events:

- Searching the node where the key to be deleted exists

- Deleting the key

- Balancing the tree if required

Explanation:

q = NULL;

   p = tree;

   while (p) {

       i = nodesearch(p, key);

       q = p;

       if (i < used(p) -1 && key == k(p,i)) {

           found = TRUE;

           position = i;

           break;

       }

       p = son(p,i);

   }

   if (!found)

   else if (subtree(p)) {

         if (used(p) > ((n-1)/2)+1)

         delkey (p, position, key);

       else {

           replace (p, position, fsucc(p));

           p0 r1 p1 r2 p2 r3 ……. pn-1 rn-1 pn

           q = &fsucc(p);

           qpos = index of fsucc;

           if (used(rbrother(p)) > ((n-1)/2)+1)

               replace (q, qpos, sonsucc(q));

           else

               while (q && used(q) < (n-1)/2) {

                   concatenate(q, brother(q));

                   q = father(q);

               }

       }

   }

   else

   delkey(p, position, key);

}

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 {
j++;
}
}
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;
}


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
4 years ago
1. Print out the string length of s1 2. Loop through characters in s2 with charAt() and display reversed string 3. Compare s2, s
konstantin123 [22]

Answer:

See explaination

Explanation:

public class StringLab9 {

public static void main(String args[]) {

char charArray[] = { 'C', 'O', 'S', 'C', ' ', '3', '3', '1', '7', ' ', 'O', 'O', ' ', 'C', 'l', 'a', 's', 's' };

String s1 = new String("Objected oriented programming language!");

String s2 = "COSC 3317 OO class class";

String s3 = new String(charArray);

// To do 1: print out the string length of s1

System.out.println(s1.length());

// To do 2: loop through characters in s2 with charAt and display reversed

// string

for (int i = s2.length() - 1; i >= 0; --i)

System.out.print(s2.charAt(i));

System.out.println();

// To do 3: compare s2, s3 with compareTo(), print out which string (s2 or s3)

// is

// greater than which string (s2 or s3), or equal, print the result out

if (s2.compareTo(s3) == 0)

System.out.println("They are equal");

else if (s2.compareTo(s3) > 0)

System.out.println("s2 is greater");

else

System.out.println("s3 is greater");

// To do 4: Use the regionMatches to compare s2 and s3 with case sensitivity of

// the first 8 characters.

// and print out the result (match or not) .

if (s2.substring(0, 8).compareTo(s3.substring(0, 8)) == 0)

System.out.println("They matched");

else

System.out.println("They DONT match");

// To do 5: Find the location of the first character 'g' in s1, print it out

int i;

for (i = 0; i < s2.length(); ++i)

if(s2.charAt(i)=='g')

break;

System.out.println("'g' is present at index " + i);

// To do 6: Find the last location of the substring "class" from s2, print it

// out

int index = 0, ans = 0;

String test = s2;

while (index != -1) {

ans = ans + index;

index = test.indexOf("class");

test = test.substring(index + 1, test.length());

}

System.out.println("Last location of class in s2 is: " + (ans + 1));

// To do 7: Extract a substring from index 4 up to, but not including 8 from

// s3, print it out

System.out.println(s3.substring(4, 8));

} // end main

} // end class StringLab9

7 0
3 years ago
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