Answer:
<em>C++</em>
////////////////////////////////////////////////////////////////////////////////////////////////////////
#include <iostream>
#include <vector>
using namespace std;
int main() {
vector<int> v;
int n = 1;
while (n != 0) {
cout<<"Enter an integer, the input ends if it is 0: ";
cin>>n;
v.push_back(n);
}
cout<<endl;
///////////////////////////////////////////////////////////
int sum = 0;
int num_positives = 0, num_negatives = 0;
for (int i=0; i<v.size()-1; i++) {
if (v[i] > 0)
++num_positives;
else
++num_negatives;
sum = sum + v[i];
}
//////////////////////////////////////////////////////////
cout<<"The number of positives is "<<num_positives<<endl;
cout<<"The number of negatives is "<<num_negatives<<endl;
cout<<"The total is "<<sum<<endl;
cout<<"The average is "<<(float)sum/(v.size()-1);
///////////////////////////////////////////////////////////
return 0;
}
Answer:
Table for Area codes are not missing;
See Attachment for area codes and major city I used
This program will be implemented using c++ programming language.
// Comments are used for explanatory purposes
// Program starts here
#include <iostream>
using namespace std;
int main( )
{
// Declare Variable area_code
int area_code;
// Prompt response from user
cout<<Enter your area code: ";
cin<<"area_code;
// Start switch statement
switch (area_code) {
// Major city Albany has 1 area code: 229...
case 229:
cout<<"Albany\n";
break;
// Major city Atlanta has 4 area codes: 404, 470 678 and 770
case 404:
case 470:
case 678:
case 770:
cout<<"Atlanta\n";
break;
//Major city Columbus has 2 area code:706 and 762...
case 706:
case 762:
cout<<"Columbus\n";
break;
//Major city Macon has 1 area code: 478...
case 478:
cout<<"Macon\n";
break;
//Major city Savannah has 1 area code: 912..
case 912:
cout<<"Savannah\n";
break;
default:
cout<<"Area code not recognized\n";
}
return 0;
}
// End of Program
The syntax used for the above program is; om
Answer:
they arent made to Express emotion? can you be a little more specific with the question?
Answer:
The function in C++ is as follows:
int isSorted(int ar[], int n){
if (
|| 
){
return 1;}
if (![ar[n - 1]](https://tex.z-dn.net/?f=ar%5Bn%20-%201%5D)
< ![ar[n - 2]](https://tex.z-dn.net/?f=ar%5Bn%20-%202%5D)
){
return 0;}
return isSorted(ar, n - 1);}
Explanation:
This defines the function
int isSorted(int ar[], int n){
This represents the base case; n = 1 or 0 will return 1 (i.e. the array is sorted)
if ( || ){
return 1;}
This checks if the current element is less than the previous array element; If yes, the array is not sorted
if ( < ){
return 0;}
This calls the function, recursively
return isSorted(ar, n - 1);
}
Answer:
Question was incomplete and continued the question
For each of the following scenarios, which of these choices would be best? Explain your answer.
BST
Sorted Array
Un-sorted Array
a) The records are guaranteed to arrive already sorted from lowest to highest (i.e., whenever a record is inserted, its key value will always be greater than that of the last record inserted). A total of 1000 inserts will be interspersed with 1000 searches.
b) The records arrive with values having a uniform random distribution (so the BST is likely to be well balanced). 1,000,000 insertions are performed, followed by 10 searches.
Explanation:
Answer for a: Un-sorted array or Un-sorted linked list : as mentioned in the question itself that the records are arriving in the sorted order and search will not be O(log n) and insert will be not be O(n).
Answer for b : Un-sorted array or Un-sorted linkedlist : Number of the items to be inserted is already known which is 1,000,000 but it is very high and at the same time search is low. Unsorted array or Unsorted linked list will be best option here.
