Write a program that reads an integer, a list of words, and a character. The integer signifies how many words are in the list. The output of the program is every word in the list that contains the character at least once. Assume at least one word in the list will contain the given character. Ex: If the input is:
4 hello zoo sleep drizzle z
then the output is:
zoo
drizzle
To achieve the above, first read the list into a vector. Keep in mind that the character 'a' is not equal to the character 'A'.
5.23.1: LAB: Contains the character
#include
#include
using namespace std;
int main() {
/* Type your code here. */
return 0;
}

In C++:

#include<iostream>

#include<vector>

using namespace std;

int main() {

int len;

cout<<"Length: ";  cin>>len;

string inpt;

vector<string> vect;

for(int i =0;i<len;i++){

   cin>>inpt;

   vect.push_back(inpt); }

char ch;

cout<<"Input char: ";  cin>>ch;  

for(int i =0;i<len;i++){

   size_t found = vect.at(i).find(ch);  

       if (found != string::npos){

           cout<<vect.at(i)<<" ";

           i++;

       }

}  

return 0;

}

Explanation:

This declares the length of vector as integer

int len;

This prompts the user for length

cout<<"Length: ";  cin>>len;

This declares input as string

string inpt;

This declares string vector

vector<string> vect;

The following iteration gets input into the vector

for(int i =0;i<len;i++){

   cin>>inpt;

   vect.push_back(inpt); }

This declares ch as character

char ch;

This prompts the user for character

cout<<"Input char: ";  cin>>ch;  

The following iterates through the vector

for(int i =0;i<len;i++){

This checks if vector element contains the character

   size_t found = vect.at(i).find(ch);  

If found:

       if (found != string::npos){

Print out the vector element

           cout<<vect.at(i)<<" ";

And move to the next vector element

           i++;

       }

}  

its called a apple tv

explanation: