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: Accepted)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::input;
use proconio::marker::Chars;

fn main() {
    input! {
        n_chars: Chars,
        k: usize,
    }

    let n = n_chars.iter().map(|&x| x.to_digit(10).unwrap() as usize).collect::<Vec<_>>();

    // dp[digit][less][0以外が出た回数]
    let mut dp = vec![vec![vec![0; 3+1]; 2]; n.len()+1];

    dp[0][0][0] = 1;

    for i in 1..n.len()+1 {
        let n_digit = n[i-1];
        let prev_i = i-1;

        for digit in 0..10 {
            if digit == 0 {
                if digit < n_digit {
                    for j in 0..3+1 {
                        dp[i][1][j] += dp[prev_i][0][j];
                        dp[i][1][j] += dp[prev_i][1][j];
                    }
                } else if digit == n_digit {
                    for j in 0..3+1 {
                        dp[i][0][j] += dp[prev_i][0][j];
                        dp[i][1][j] += dp[prev_i][1][j];
                    }
                }
            } else {
                if digit < n_digit {
                    for j in 0..3 {
                        dp[i][1][j+1] += dp[prev_i][0][j];
                        dp[i][1][j+1] += dp[prev_i][1][j];
                    }
                } else if digit == n_digit {
                    for j in 0..3 {
                        dp[i][0][j+1] += dp[prev_i][0][j];
                        dp[i][1][j+1] += dp[prev_i][1][j];
                    }
                } else {
                    for j in 0..3 {
                        dp[i][1][j+1] += dp[prev_i][1][j];
                    }
                }
            }
        }
    }

    println!("{}", dp[n.len()][0][k] + dp[n.len()][1][k]);
}
*/

fn main() {
    let exe = "/tmp/binDA8A4D93";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";