PersistentBugger.cpp

#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include <string>
#include <sstream>
#include <climits>
using namespace std;
int persistence(long long n) {
	int count = 0;
	while (n >= 10) {
		int sum = 1;
		count++;
		while (n > 0) {
			sum *= n % 10;
			n /= 10;
		}
		n = sum;
	}
	return count;

}
int main()
{
	cout << persistence(999);
	return 0;
}
/*Description:

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

For example (Input --> Output):

39 --> 3 (because 3*9 = 27, 2*7 = 14, 1*4 = 4 and 4 has only one digit, there are 3 multiplications)
999 --> 4 (because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2, there are 4 multiplications)
4 --> 0 (because 4 is already a one-digit number, there is no multiplication)

Fundamentals
Mathematics*/