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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.math.*;
 
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String n = sc.next();
		int k = sc.nextInt();
		int digit = n.length();
		int[][][] dp = new int[digit+1][k+1][2];
		dp[0][0][0] = 1;
		for(int i = 1; i <= digit; i++) {
			int x = Character.getNumericValue(n.charAt(i-1));
				for(int j = 0; j <= k; j++) {
					dp[i][j][1] = dp[i-1][j][1];
					if(j > 0)
						dp[i][j][1] += dp[i-1][j-1][1] * 9;
					if(x == 0)
						dp[i][j][0] = dp[i-1][j][0];
					else {
						dp[i][j][1] += dp[i-1][j][0];
						if(j > 0) {
							dp[i][j][0] = dp[i-1][j-1][0];
							dp[i][j][1] += dp[i-1][j-1][0] * (x-1);
						}
					}
				}
		}
		System.out.println(dp[digit][k][0] + dp[digit][k][1]);
 
	}
 
}