Answer:
// program in C++.
#include <bits/stdc++.h>
using namespace std;
int main() {
// variable
int num;
cout<<"Enter the number between 20 and 98: ";
// read number
cin >> num;
while(num<20||num>98)
{
cout<<"Wrong input!!enter number between 20-98 only:";
cin>>num;
}
cout<<"The output is: ";
while(num % 10 != num /10)
{
// print numbers.
cout<<num<<" ";
// update num.
num--;
}
// display the number.
cout<<num<<endl;;
return 0;
}
Explanation:
Read a number from user and assign it to variable "num".Check if entered number is in between 20-98 or not.If input number is less than 20 or greater than 98 then ask again to enter a number between 20-98 until user enter a valid input.Then print the countdown from input number till both the digit of number are same.
Output:
Enter the number between 20 and 98: 99
Wrong input!!enter number between 20-98 only:12
Wrong input!!enter number between 20-98 only:93
The output is: 93 92 91 90 89 88
Enter the number between 20 and 98: 77
The output is: 77
Answer:
It has to be more specific and easier to understand because of its importance.
Explanation:
Answer:
Social networking platform and WindowsandroidIOS
Explanation:
Social media mostly uses Social networking platform and WindowsandroidIOS.
Hope this helps!
Feel free to ask if you have anymore questions!
<span>There are some network modeling tools that can scan the existing network.</span>
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
