Problem:
Score : 200 points

Problem Statement

Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

Constraints

1 \leq N \leq 10^4

1 \leq A \leq B \leq 36

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

Sample Input 1

20 2 5

Sample Output 1

84

Among the integers not greater than 20, the ones whose sums of digits are between 2 and 5, are: 2,3,4,5,11,12,13,14 and 20. We should print the sum of these, 84.

Sample Input 2

10 1 2

Sample Output 2

13

Sample Input 3

100 4 16

Sample Output 3

4554

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::stdin;

fn calc_ten_digit_num_sum(mut num: i32) -> i32 {
    let mut res = 0;
    while num > 0 {
        res += num % 10;
        num /= 10;
    }
    res
}

fn main() {
    let mut s = String::new();
    stdin().read_line(&mut s).ok();

    let nums: Vec<i32> = s
        .trim()
        .split_whitespace()
        .map(|x| x.parse().ok().unwrap())
        .collect();

    let n = nums[0];
    let a = nums[1];
    let b = nums[2];

    let mut total_sum = 0;

    // logic
    for i in 1..n+1 {
        let temp_sum = calc_ten_digit_num_sum(i);
        if a <= i && i <= b {
            total_sum += temp_sum;
        }
    }

    println!("{}", total_sum);
}