Answer:
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:
- In the given C++ program there are two vector array is defined that are "oldScores and newScores". The oldScores array holds elements that are "10, 20, 30, 40".
- In the above code, we remove the array element at first position and add it to the last position. To this process, an integer variable "lastVector" is defined.
- This variable holds the size of the newScores variable and uses and assigns all vector array elements from oldScores to newScores. In the loop, we use the at function the removes element form first position and add in the last position.
- Then we use another for loop for print newScores array elements.
Answer:
The campaign could be improved by 78% if the listed recommendations are followed.
Explanation:
While conducting the following Search advertising program for a few months, Meredith has announced that revenues of its branded goods are beginning to slow. She reviews her Google Advertising Suggestions webpage which states that her campaign's performance ranking is 22 points.
Thus, the campaign will be increased by 78% if the above recommendations are implemented to inform Meredith regarding its Google Search Advertising plan.
What did you do? Write that as the first part
void minMax(int a, int b, int c, int*big, int*small)
{
if(a>b && a >c){
*big = a;
if(b>c)
*small = c;
else
*small = b;
}
else if (b>a && b>c){
*big = b;
if(a>c)
*small = c;
else
*small = a;
}
else{
*big = c;
if(a>b)
*small = b;
else
*small = a;
}
}
