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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use std::cmp;

// read a line(without line feed)
fn read_line() -> String {
    let mut in_str = String::new();
    io::stdin().read_line(&mut in_str).unwrap();
    return in_str.trim().to_string();
}

// read wods on a line
pub fn read_words_in_line() -> Vec<String> {
    let s = read_line();
    return s.split_whitespace().map(|x| x.to_string()).collect::<Vec<String>>()
}

// read type t on a line
pub fn read_t_in_line<T : std::str::FromStr>() -> Vec<T> {
    let mut v : Vec<T> = Vec::new();
    for e in read_words_in_line().iter().map(|x| x.parse::<T>()) {
        if let Ok(r) = e {
            v.push(r);
        }
        else{
            return Vec::new();
        }
    }
    return v;
}

// sort slice
pub fn sort<T : std::cmp::Ord>(slice : &mut [T]) {
    slice.sort();
}
// sort slice by reverse order
pub fn sort_rev<T : std::cmp::Ord>(slice : &mut [T]) {
    slice.sort_by(|x, y| y.cmp(x));
}

pub fn min<T : std::cmp::Ord>(a : T, b : T) -> T{
    return cmp::min(a, b);
}

pub fn max<T : std::cmp::Ord>(a : T, b : T) -> T{
    return cmp::max(a, b);
}


fn main() {
    let n = read_line();
    let n_int = n.parse::<i32>().unwrap();
    let mut dig_sum = 0;
    for c in n.chars() {
        let diff = (c as i32) - ('0' as i32);
        dig_sum += diff;
    }
    if n_int % dig_sum == 0 {
        println!("Yes");
    }
    else{
        println!("No");
    }

}