Problem:
Score : 500 points

Problem Statement

Find the number of integers between 1 and N (inclusive) that contains exactly K non-zero digits when written in base ten.

Constraints

1 \leq N < 10^{100}

1 \leq K \leq 3

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the count.

Sample Input 1

100
1

Sample Output 1

19

The following 19 integers satisfy the condition:

1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100

Sample Input 2

25
2

Sample Output 2

14

The following 14 integers satisfy the condition:

11,12,13,14,15,16,17,18,19,21,22,23,24,25

Sample Input 3

314159
2

Sample Output 3

937

Sample Input 4

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
3

Sample Output 4

117879300

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

pub fn main() {
    let mut buf = String::new();
    let n: Vec<i32> = {
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim()
            .chars()
            .map(|ch| ch as i32 - '0' as i32)
            .collect()
    };
    let k: i32 = {
        buf.clear();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim().parse().unwrap()
    };
    let l = n.len() as i32;
    if l < k {
        println!("0");
        return;
    }
    let mut cnt = 0;
    if k == 1 {
        cnt += (l - 1) * 9;
        cnt += n[0];
    } else if k == 2 {
        cnt += (l - 1) * (l - 2) / 2 * 9 * 9;
        cnt += (n[0] - 1) * (l - 1) * 9;
        cnt += (l - 2) * 9;
        cnt += n[1];
    } else if k == 3 {
        cnt += (l - 1) * (l - 2) * (l - 3) / 6 * 9 * 9 * 9;
        cnt += (n[0] - 1) * (l - 1) * (l - 2) / 2 * 9 * 9;
        cnt += (l - 2) * (l - 3) / 2 * 9 * 9;
        if n[1] > 0 {
            cnt += (n[1] - 1) * (l - 2) * 9;
        }
        cnt += (l - 3) * 9;
        cnt += n[2];
    }
    println!("{}", cnt);
}