Write a loop that sets newscores to oldscores shifted once left, with element 0 copied to the end. ex: if oldscores = {10, 20, 30, 40}, then newscores = {20, 30, 40, 10}. note: these activities may test code with different test values. this activity will perform two tests, the first with a 4-element array (newscores = {10, 20, 30, 40}), the second with a 1-element array (newscores = {199}).

public class studentscores {
public static void main (string [] args) {
final int scores_size = 4;
int[] oldscores = new int[scores_size];
int[] newscores = new int[scores_size];
int i = 0;

oldscores[0] = 10;
oldscores[1] = 20;
oldscores[2] = 30;
oldscores[3] = 40;

/* your solution goes here */

for (i = 0; i < scores_size; ++i) {
system. out. print(newscores[i] + " ");
}
system. out. println();

return;
}

The program illustrates the use of loops.

In programming, loops are used for operations that are to be repeated  

Replace /* Your solution goes here */  with the following lines of code

for (i = 1; i < SCORES_SIZE; ++i) {

newScores[i-1] = oldScores[i];

}

newScores[3] = oldScores[0];

The above code segment with comments is as follows:

//This iterates through the oldScores array

for (i = 1; i < SCORES_SIZE; ++i) {

//This moves the second element to the fourth elements of oldScores to newScores

newScores[i-1] = oldScores[i];

}

//This moves the first element of oldScores to the last of newScores

newScores[3] = oldScores[0];

At the end of the iteration, the elements are shifted to newScore

Read more about loops at:

link

The code to this question can be given as:

Code:

int lastVector = newScores.size() -1; //define variable lastVector that holds updated size of newScores.

newScores = oldScores; //holds value.

for (i = 0; i < SCORES_SIZE - 1; i++) //define loop.

{  

newScores.at(i) = newScores.at(i+1); //holds value in newScores.

}

newScores.at(lastVector) = oldScores.at(0); //moving first element in last.

Explanation:

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explanation:

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