Count Digit in C++ | Coding Ninga Striver A2z DSA

Problem statement You are given a number ’n’. Find the number of digits of ‘n’ that evenly divide ‘n’. Note: A digit evenly divides ‘n’ if it leaves no remainder when dividing ‘n’. Example: Input: ‘n’ = 336 Output: 3 Explanation: 336 is divisible by both ‘3’ and ‘6’. Since ‘3’ occurs twice it is counted two times. Note: You don’t need to print anything. Just implement the given function. Detailed explanation ( Input/output format, Notes, Images ) Sample Input 1: 35 Sample Output 1: 1 Explanation of sample output 1: 35 is only divisible by ‘5’, hence answer is 1. Sample Input 2: 373 Sample Output 2: 0 Explanation of sample output 2: There’s no digit in 373 that evenly divides it. Hence the output is 0. write code for this in cpp

 #include <iostream>

using namespace std;

int main() {

   int n = 35;

   int value = n;

  int count = 0;

    while (value != 0) {

        int ld = value % 10;

        if(n%ld == 0){

           count++; 

        }

       value = value / 10;

    }

    cout<<count;

    return 0;

}