Problem:
Score : 200 points

Problem Statement

An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10.

Given an integer N, determine whether it is a Harshad number.

Constraints

1?N?10^8

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print Yes if N is a Harshad number; print No otherwise.

Sample Input 1

12

Sample Output 1

Yes

f(12)=1+2=3. Since 12 is divisible by 3, 12 is a Harshad number.

Sample Input 2

57

Sample Output 2

No

f(57)=5+7=12. Since 57 is not divisible by 12, 12 is not a Harshad number.

Sample Input 3

148

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repl(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repr(i, n) for(int i = n; i >= 0; i--)
#define reprl(i, n) for(ll i = n; i >= 0; i--)
#define fori(i, m, n) for(int i = m; i < (int)n; i++)
#define forl(i, m, n) for(ll i = m; i < (ll)n; i++)
#define all(v) v.begin(), v.end()
ll gcd(ll a,ll b) {
  return b? gcd(b,a%b): a;  
}
ll lcm(ll a,ll b) {
  return a*b/gcd(a,b);
}
bool isLowerCase(char c){
  return (c>='a'&&c<='z');
}
const string btos(bool ans){
  return (ans?"Yes":"No");
}
int main() {
  int n;cin>>n;
  bool ans;
  int base=n;
  int sum=0;
  while(n>0){
    sum+=n%10;
    n/=10;
  }
  cout << btos(base%sum==0)<<endl;
  return 0;
}