the answer is false bc digutal is easeir
Answer:
#include <string>
#include <iostream>
using namespace std;
int main() {
string userInput;
getline(cin, userInput);
// Here, an integer variable is declared to find that the user entered string consist of word darn or not
int isPresent = userInput.find("darn");
if (isPresent > 0){
cout << "Censored" << endl;
// Solution starts here
else
{
cout << userInput << endl;
}
// End of solution
return 0;
}
// End of Program
The proposed solution added an else statement to the code
This will enable the program to print the userInput if userInput doesn't contain the word darn
Answer:
The second one:
int sum = 0; for (int i = 0; i < values.length; i++) { if ((values[i] % 2) == 0) { sum += values[i]; } }
C. Healing brush
That’s the answer
