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Count down project

Count down project You need to make a countdown app. Given a number N as input, output numbers from N to 1 on separate lines. Also, when the current countdown number is a multiple of 5, the app should output "Beep". Sample Input: 12 Sample Output: 12 11 10 Beep 9 8 7 6 5 Beep 4 3 2 1 There's a code below this question please I need an explanation for it

30th Oct 2021, 5:36 PM
Emmanuel Osemudiamen
Emmanuel Osemudiamen - avatar
6 Réponses
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#include <iostream> using namespace std; int main() { int n; cin >> n; if (n<0){ cout << "Time cannot be negative"<< endl; cin >> n; } // minimum countdown value is 1. hence, 0 is not an acceptable output so condition to use is n > 0 or you can use n>=1. // for each countdown, you decrease the previous value by 1. written as n-- . for (n ; n> 0 ; n--) { // condition for beep (Number must be divisible by 5) // challenge requires that you display the value which is divisible by 5 before displaying the Beep. So we display both outputs if ( n%5 == 0 ){ cout << n << endl; cout << "Beep" << endl; } else { cout << n << endl; } } return 0; }
30th Oct 2021, 5:37 PM
Emmanuel Osemudiamen
Emmanuel Osemudiamen - avatar
0
There are some comments in the code that explain it. Which part precisely do you struggle with?
30th Oct 2021, 5:46 PM
Lisa
Lisa - avatar
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For each count down u reduce the number by one? Why?
30th Oct 2021, 6:05 PM
Emmanuel Osemudiamen
Emmanuel Osemudiamen - avatar
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And what numbers are divisible by 5 in this code?
30th Oct 2021, 6:08 PM
Emmanuel Osemudiamen
Emmanuel Osemudiamen - avatar
0
And what does multiple of 5 mean?
30th Oct 2021, 6:11 PM
Emmanuel Osemudiamen
Emmanuel Osemudiamen - avatar
0
n-- because it counts downwards, for example: 9, 8, 7, ... A number is divisible by 5 if (n % 5) == 0 % is the modulo operator, it gives us the remainder of a division. If the remainder equals 0 we know there is remainder, hence it is fully divisible by 5
30th Oct 2021, 6:32 PM
Lisa
Lisa - avatar