Answer:
In the website host’s cPanel, there is a set of tools for organizing the website.
A design mode in IDE lets the website creator type directly into the web pages or specific fields without knowing HTML code.
All website files need to reside in the HTDOCS or the Document root folder .
The content folder is where all of the content will be stored.
Explanation:
The highlighted are answers.
The content folder contains all the content of the website. And the website files are being stored at the htdocs or the Document root folder. And the cPanel has a set of tools for organizing the website. And the IDE does the job as being explained above in answer section.
I found this loop. But it ends like this: After the loop terminates, <span>it prints out the sum of all the even integers read. Declare any variables that are needed.</span>
<span>int sum=0;
int num=1;
while(num > 0){
cin >> num;
if ((num % 2)==0 & (num>0)){
sum+=num;
}
}
cout << sum;
I'm not familiar with coding but I think you can work on this loop and edit it according to your requirement.</span>
What type of program would have a class named Student with objects called fullTime and partTime?
A. machine language program
B. object-oriented program
C. markup language program
D. procedural language program
Answer:
B. object-oriented program
Explanation:
An object-oriented program or OOP is a type of program that uses the concepts of objects and methods.
Although they are quite broad, they also make use of classes and types.
Java, for instance makes use of OOP as they use classes and objects under those classes and name them anyhow they want.
Therefore, the correct answer is B
Cmd.exe
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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.
