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>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef pair<ll, int> li;
typedef pair<int, ll> il;
typedef vector<ii> vii;
typedef vector<il> vil;
typedef vector<li> vli;
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define sz size()
#define EPS 1e-14
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1,-1, 1, -1, 0, 0, -1, 1};

int main()
{
    /*#ifndef ONLINE_JUDGE
    freopen ("in.txt", "r", stdin);
    freopen ("out.txt", "w", stdout);
    #endif*/
    int n, sum=0, m;
    cin>>n;
    m=n;
    while(n>0){
        sum+=(n%10);
        n/=10;
    }
    if(m%sum==0) cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
    return 0;
}