Problem:
Score : 200 points

Problem Statement

Let S(n) denote the sum of the digits in the decimal notation of n.
For example, S(101) = 1 + 0 + 1 = 2.

Given an integer N, determine if S(N) divides N.

Constraints

1 \leq N \leq 10^9

Input

Input is given from Standard Input in the following format:

N

Output

If S(N) divides N, print Yes; if it does not, print No.

Sample Input 1

12

Sample Output 1

Yes

In this input, N=12.
As S(12) = 1 + 2 = 3, S(N) divides N.

Sample Input 2

101

Sample Output 2

No

As S(101) = 1 + 0 + 1 = 2, S(N) does not divide N.

Sample Input 3

999999999

Sample Output 3

Yes

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

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)  
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define llong long long
#define pb(a) push_back(a)
#define INF 999999999
using namespace std;
typedef pair<int, int> P;
typedef pair<llong, llong> LP;
typedef pair<int, P> PP;
typedef pair<llong, LP> LPP;

int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};

int main(){

  int n;
  cin>>n;
  int k = 0;
  int x = n;

  while(x>0){
    k += x%10;
    x /= 10;
  }

  if(n%k==0){
    cout<<"Yes"<<endl;
  }else{
    cout<<"No"<<endl;
  }

  return 0;
}