Print "Censored" if userInput contains the word "darn", else print userInput. End with newline. Ex: If userInput is "That darn cat.", then output is:CensoredEx: If userInput is "Dang, that was scary!", then output is:Dang, that was scary!Note: If the submitted code has an out-of-range access, the system will stop running the code after a few seconds, and report "Program end never reached." The system doesn't print the test case that caused the reported message.#include #include using namespace std;int main() {string userInput;getline(cin, userInput);int isPresent = userInput.find("darn");if (isPresent > 0){cout << "Censored" << endl; /* Your solution goes here */return 0;}

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MrRoyal MrRoyal · Answer 1 · 2020-03-05T05:35:53+01:00

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