Author: Rupanjan Hari
Tester: Rupanjan Hari
Given an integer N, traverse its digits (d1,d2,…,dn) and determine how many digits evenly divide N(i.e.: count the number of times N divided by each digit di has a remainder of 0). Print the number of evenly divisible digits.
Note: Each digit is considered to be unique, so each occurrence of the same evenly divisible digit should be counted (i.e.: for N=111, the answer is 3).
To make the problem look easy, I have divided the program into functions of appropriate name.
Arrays:a - The primary array that receives all the numbers
Functions:int count(long long int n) - This function counts the digits that evenly divide the number.
In the main function just call the count function for each number.
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