Write a program to implement function overloading as follows:
a.display(int n)-Takes the number from main() and displays whether the number is a Buzz number or not. (A number is a Buzz number if it’s last digit is 7 or if the number is divisible by 7. Ex. 21 and 37 are Buzz numbers.)
b.display()-Displays the following series
1 + 2 + 3 + 4 +……..+ 20
1! 2! 3! 4! 20!
Answers
Answer:
Input : 63
Output : Buzz Number
Explanation: 63 is divisible by 7, one
of the condition is satisfied.
Input : 72
Output : Not a Buzz Number
Explanation: 72 % 7 != 0, 72 is neither
divisible by 7 nor it ends with 7 so
it is not a Buzz Number.
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
// C++ program to check whether the
// given number is Buzz Number or not.
#include <cmath>
#include <iostream>
using namespace std;
// function to check BUzz number.
bool isBuzz(int num)
{
// checking if the number
// ends with 7 and is divisible by 7
return (num % 10 == 7 || num % 7 == 0);
}
// Driver method
int main(void)
{
int i = 67, j = 19;
if (isBuzz(i))
cout << "Buzz Number\n";
else
cout << "Not a Buzz Number\n";
if (isBuzz(j))
cout << "Buzz Number\n";
else
cout << "Not a Buzz Number\n";
}