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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
#define all(x) x.begin(),x.end()
#define rep(i,j,n) for (long long i = j; i < (long long)(n); i++)
#define _GLIBCXX_DEBUG
#define MOD 1000000007
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
//name[i][j]-> vvl name(i,vl(j))
//map<int,int>mp
//余りを切り上げる =>  (a + b - 1) / b 

signed main(){
	ll n; cin>>n;
	ll cnt=0;
	string s=to_string(n);
	int m=s.size();
	rep(i,0,m){
		cnt+=ll(s[i]-'0');
	}
	//cout<<cnt<<endl;
	if(n%cnt==0)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
	return 0;
}