If the vector oldData is the same as the vector newData, print "Data matches!" ended with a newline. Otherwise, assign oldData with newData. Ex: If oldData = {10, 12, 18} and newData = {25, 27, 29, 23}, then oldData becomes {25, 27, 29, 23}. Only enter the Solution Part
#include
#include
using namespace std;
int main() {
vector oldData(3);
vector newData(4);
unsigned int i = 0;
oldData. at(0) = 10;
oldData. at(1) = 12;
oldData. at(2) = 18;
newData. at(0) = 25;
newData. at(1) = 27;
newData. at(2) = 29;
newData. at(3) = 23;
// Enter Solution here
for (i = 0; i < oldData. size(); ++i) {
cout << oldData. at(i) << " ";
}
cout << endl;
return 0;
}

Following are the solution part to this question:

for(i=0;i< oldData.size()&&i< newData.size();i++) //defefing loop to assign newData value to oldData value              

    {

         if (oldData.at(i) != newData.at(i)) //defefing if block to check count value

         {

             oldData = newData; //assign newData value to oldData value

             break; //using break keyword

         }

         else //defefing else block  

         {  

           cout << "Data matches!" << endl;//print message  

           break; //using break keyword to exit from loop

         }

    }

Explanation:

The full Program to this question can be described as follows:

#include<vector>//defefing header file

#include<stdlib.h>//defefing header file

#include<iostream>//defefing header file

using namespace std;  

int main() //defefing main method

{  

vector<int> oldData(3); //defefing vector array oldData

vector<int> newData(4);//defefing vector array newData

unsigned int i = 0; //defefing integer variable

    oldData.at(0) = 10; //initilizing value oldData  

    oldData.at(1) = 12;//initilizing value oldData  

    oldData.at(2) = 18;//initilizing value oldData  

    newData.at(0) = 25;//initilizing value newData

    newData.at(1) = 27;//initilizing value newData

    newData.at(2) = 29;//initilizing value newData

    newData.at(3) = 23;//initilizing value newData

    for(i=0;i< oldData.size()&&i< newData.size();i++) //defefing loop to assign newData value to oldData value              

    {

         if (oldData.at(i) != newData.at(i)) //defefing if block to check count value

         {

             oldData = newData; //assign newData value to oldData value

             break; //using break keyword

         }

         else //defefing else block  

         {  

           cout << "Data matches!" << endl;//print message  

           break; //using break keyword to exit from loop

         }

    }

    for (i = 0; i < oldData.size(); ++i) //defining loop to print value

    {

         cout << oldData.at(i) << " ";//print oldData value

    }

    return 0;

}

Output:

25 27 29 23  

Description of the program  can be described as follows:

the correct answer is: a.)it is never appropriate to use acronyms and emoticons in business communication.