Given a sorted list of integers, output the middle integer. Assume the number of integers is always odd. Ex: If the input is:
2 3 4 8 11 -1
(a negative indicates the end), the output is:
4
The maximum number of inputs for any test case should not exceed 9. If exceeded, output "Too many inputs".
Hint: First read the data into a vector. Then, based on the number of items, find the middle item.
#include
#include // Must include vector library to use vectors
using namespace std;
int main() {
/* Type your code here. */
return 0;
}

Complete the program as thus:

 vector<int> myvector;

 int myinput;  

 cout << "Input Numbers:\n";  cin >> myinput;

 do {

   myvector.push_back (myinput);

   if(myvector.size()==9){

       cout<<"Too many inputs"<<endl;

       myinput = -1;

   }else{

   cin >> myinput;}

 } while (myinput>0);

 int mid = (myvector.size()+1)/2;

 cout<<"Mid Item: "<< myvector.at(mid-1);

 return 0;

Explanation:

Declare a vector of integer

 vector<int> myvector;

Declare an integer variable for input

 int myinput;  

Prompt the user for integer input

 cout << "Input Numbers:\n";  cin >> myinput;

The following iteration is repeated until a negative input is entered

 do {

This inserts the integer input into the vector

   myvector.push_back (myinput);

This ensures that the vector size do not exceed 9

   if(myvector.size()==9){

       cout<<"Too many inputs"<<endl;

       myinput = -1;

  }else{

   cin >> myinput;}

 } while (myinput>0);

This calculates the mid index

int mid = (myvector.size()+1)/2;

This prints the mid item

 cout<<"Mid Item: "<< myvector.at(mid-1);  

the answer is d. splog

1. a

2. a

3. d

explanation: